Jekyll2021-08-18T08:24:39+00:00https://jongyeong.github.io/feed.xmlJongyeong’s blogArchive of what I have done. Jongyeong Leelee@ms.k.u-tokyo.ac.jpRandom math: 392021-08-17T00:00:00+00:002021-08-17T00:00:00+00:00https://jongyeong.github.io/rm/rm39<p>Source of most content: <a href="https://www.youtube.com/channel/UCB_AbuIVIG8I3K3_T_j0FSw/">Gunhee Cho</a> and Topology by Munkres.</p> <h1 id="separation-axiom">Separation Axiom</h1> <p class="info"><code class="language-plaintext info highlighter-rouge">Def.</code> Given a T.S. $(X,T)$, $X$ is <strong>T1</strong> if any finite set is a closed set.</p> <p class="info"><code class="language-plaintext info highlighter-rouge">Def.</code> Given a T.S. $(X,T)$, $X$ is <strong>T2 (Hausdorff)</strong> if $\forall x,y \in X$ with $x\ne y$, $\exists$ nbhds $U_x, U_y$ of $x,y$ respectively such that $V_x \cap V_y= \emptyset$.</p> <p class="info"><code class="language-plaintext info highlighter-rouge">Def.</code> Given a T.S. $(X,T)$, $X$ is <strong>T3 (regular)</strong> if $X$ is T1 and $\forall x \in X$ and $\forall$ closed set $A\subseteq X$ with $x\not\in A$, $\exists$ nbhd $V_x, V_A$ s.t. $V_x \cap V_A = \emptyset$.</p> <p class="info"><code class="language-plaintext info highlighter-rouge">Def.</code> Given a T.S. $(X,T)$, $X$ is <strong>T4 (normal)</strong> if $X$ is T1 and $\forall$ closed sets $A,B\subseteq X$ with $A \cap B = \emptyset$, $\exists$ nbhd $V_A, V_B$ s.t. $V_A \cap V_B = \emptyset$.</p> <p>NB1; If $X$ is a metric space, then T1 $\Leftrightarrow$ T2 $\Leftrightarrow$ T3 $\Leftrightarrow$ T4.</p> <h2 id="propositions">Propositions</h2> $\text{Given a T1 T.S. } (X,T), \nonumber$ <ol> <li>$X$ is regular iff $\forall x \in X$, $\forall$ nbhd $U_x$ of $x$, $\exists$ nbhd $V$ of $x$ s.t. $x\in \bar{V} \subseteq U_x$.</li> <li>$X$ is normal iff $\forall$ closed set $A$ in $X$, $\forall$ nbhds $U_A$ of $A$, $\exists$ nbhd $V$ of $A$ s.t. $A \subseteq \bar{V} \subseteq U_A$.</li> </ol> <h3 id="proof">Proof</h3> <ul> <li>(1 $\Rightarrow$): Let $x\in U_x \in T$, i.e., $U_x^c$ is closed in $X$. Since $X$ is regular, $\exists$ nbhds $V_x, V_{U_x^c}$ s.t. $V_x\cap V_{U_x^c}=\emptyset$. Then, $\overline{V_x}\cap V_{U_x^c}=\emptyset$ since if $y \in U_x^c$, $V_{U_X^c}$ is a nbhd of $y$ disjoint from $V_x$. Therefore, $\overline{V_x}\cap U_x^c = \emptyset$, i.e., $\overline{V_x} \subseteq U_x$.</li> <li>(1 $\Leftarrow$): Take any $x \in X$ and closed set $A$ with $x \not\in A$, i.e., $x \in A^c$ for open $A^c$. By assumption, $\exists$ nbhd $V_x$ s.t. $x \in \overline{V_x} \subseteq A^c$, i.e., $A \subseteq (\overline{V_x})^c$. This implies that $(\overline{V_x})^c$ is a nbhd of $A$. Therefore, $\overline{V_x} \cap (\overline{V_x})^c = \overline{V_x} \cap V_A = \emptyset$.</li> <li>(2): One can apply the proof of (1) by replacing $x$ by the closed set $A$.</li> </ul> <hr /> $\text{If } X \text{ is regular (or T2), then } \nonumber$ <ol> <li>Subspace $Y$ of $X$ is regular (or T2).</li> <li>$\prod_i X_i$ is regular (or T2) in product space.</li> </ol> <h3 id="proof-1">Proof</h3> <ul> <li>(1-T2): Take $x, y \in Y$. If $U_x$ and $U_y$ are disjoint nbhds in $X$, then $U_x\cap Y$ and $U_y \cap Y$ are disjoint nbhds of $x$ and $y$ in $Y$.</li> <li>(1-T3): Take $x \in Y$ and let $A$ be a closed subset of $Y$ with $x \not\in A$. Since $\bar{A} \cap Y = A$, $x \not\in \bar{A}$ holds. From the regularity, $\exists$ nbhds $U_x, U_{\bar{A}}$ of $x$ and $\bar{A}$ in $X$. Then, $U_x \cap Y$ and $U_{\bar{A}}$ are disjoint open sets in $Y$.</li> <li>(2-T2): <a href="/rm/rm25#proposition">RM25</a></li> <li>(2-T3): Let $X=\prod_\alpha X_\alpha$. Then, $X$ is T2 by (2-T2) since regular space is T2, so one-point sets are closed in $X$. Let $\mathbf{x} = (x_\alpha)$ be a point of $X$ and let $U=\prod_\alpha U_\alpha$ be a basis element about $\mathbf{x}$ contained in $X$. For each $\alpha$, choose a nbhd $V_\alpha$ of $x_\alpha \in X_\alpha$ s.t. $\overline{V_\alpha} \subset U_\alpha$. If $U_\alpha = X_\alpha$, then choose $V_\alpha = X_\alpha$. Since $\bar{V} \stackrel{(*)}{=} \prod_\alpha \overline{V_\alpha}$, it holds that $\bar{V} \subset \prod U_\alpha \subseteq U$, i.e., $X$ is regular.</li> </ul> <hr /> $(*): \text{ Let } (X_\alpha) \text{ be an indexed family of spaces and } \forall \alpha, \, A_\alpha \subset X_\alpha. \nonumber$ $\text{If } \prod X_\alpha \text{ is given either the product or the box topology, then } \prod \overline{A_\alpha} = \overline{\prod A_\alpha}. \nonumber$ <h3 id="proof-2">Proof</h3> <ul> <li>$(\subseteq)$: Take $\mathbf{x} = (x_\alpha) \in \prod \overline{A_\alpha}$ and let $U=\prod U_\alpha$ be a basis element for either the box or product topology that contains $\mathbf{x}$. Since $x_\alpha \in \overline{A_\alpha}$, i.e., $U_\alpha \cap A_\alpha \ne \emptyset$ one can choose a point $y_\alpha \in U_\alpha \cap A_\alpha$. Then, $\mathbf{y} = (y_\alpha)$ belongs to both $U$ and $\prod A_\alpha$. Since $U$ is arbitrary, $\mathbf{x}$ belongs to $\overline{\prod A_\alpha}$.</li> <li>$(\supseteq)$: Take $\mathbf{x} = (x_\alpha) \in \overline{\prod A_\alpha}$. Let $U _\alpha$ be a nbhd of $x _\alpha$. Since $\pi _\alpha ^{-1}(U _\alpha)$ is open in $\prod X _\alpha$, it contains a point $\mathbf{y}$ of $\prod A _\alpha$. Then, $y _\alpha \in U _\alpha \cap A _\alpha$, i.e., $x _\alpha \in \bar{A} _\alpha$, which concludes the proof.</li> </ul> <h2 id="some-examples">Some examples</h2> <ol> <li>$(X, T_{cofinite})$ is T1 but not T2.</li> <li>$\mathbb{R} _k := (\mathbb{R}, T _{k})$ is T2 but not regular since nbhd of $0$ must contain some points of $K:= \{\frac{1}{n}: n \in \mathbb{N} \}$.</li> </ol> <p>NB2; If $X$ is a metric space, 2nd countalbe $\Leftrightarrow$ Lindelof $\Leftrightarrow$ separble.</p> <hr /> $\mathbb{R}_L := (\mathbb{R}, T_{\text{lower limit top}}), \text{ where } \mathcal{B} = \{ [a,b) : a,b \in \mathbb{R}, a&lt; b \}. \nonumber$ <p>Note that $\bar{\mathbb{Q}}=\mathbb{R}$.</p> <ul> <li>1st countable: <br /> Let $x\in \mathbb{R}$ and $B_x := \{ [x, x+\frac{1}{n}) : n \in \mathbb{N} \}$. Then, $\forall$ open $U_x$ of $x$, $\exists B_{x,n} \in B_x$ s.t. $x \in B_{x,n} \subseteq U_x$, i.e., $\mathbb{R}_L$ is 1st countable.</li> <li>2nd countable: <br /> Let $B_x = [x, b)$ for any $b &gt; x$. Then, $B_x$ is open and $\inf B_x = x$ by definition. If $x\ne y$ for $x,y \in \mathbb{R}$, then $\inf B_x \ne \inf B_y$. Then, we can define an injective function $\Phi: \mathbb{R} \to \mathcal{B}$ ($x \mapsto [x, \cdot)$). Since $\Phi$ is injective, $\mathcal{B}$ must be uncountable, i.e., $\mathbb{R}_L$ is not 2nd countable.</li> <li><a href="/rm/rm38#propositions">Lindelof</a> (existence of countable subcover): <br /> Take an open cover $O = \{ [a_\alpha, b_\alpha) : a_\alpha, b_\alpha \in \mathbb{R}, a_\alpha &lt; b_\alpha \} = \{ I_\alpha : \alpha \in I \}$. Consider $J_\alpha := (a_\alpha ,b_\alpha)$. Then, $\mathbb{R} = \cup _\alpha I _{\alpha}$. Take any $x\in \mathbb{R} \setminus \cup _\alpha J _{\alpha}$. Then $x= a _\alpha$ for some $\alpha \in I$. By density of rationals, there exists $q _x \in [x, b) \subset \mathbb{Q}$. Define $\Phi: \mathbb{R} \setminus \cup _\alpha J _{\alpha} \to \mathbb{Q}$ $(x \mapsto q _x)$. Take $x,y \in \mathbb{R} \setminus \cup _\alpha J _{\alpha}$ with $x &lt; y$. If $q _x \geq q _y$, i.e., $y \in (x, b _\alpha)$, then $y \in \cup _\alpha J _{\alpha}$, which contradicts to assumption. Thus, $\mathbb{R} \setminus \cup _\alpha J _{\alpha}$ is at most countable. Hence, $q _x &lt; q _y$, i.e., $q _x \ne q _y$. Topologize $\cup _\alpha J _{\alpha}$ as a subspace of $(\mathbb{R}, T _{std})$. Then, $\exists (\alpha _k)$ s.t. $\cup _\alpha J _{\alpha} = \cup _{k=1}^\infty J _{ \alpha _k}$. Therefore, $\mathbb{R} = \left(\mathbb{R} \setminus \cup _{k=1}^\infty J _{ \alpha _k} \right) \bigcup \cup _{k=1}^\infty J _{ \alpha _k}$. Thus, $O’ := \{ [a _ {\alpha _k}, b _ {\alpha _k}), [a _ {r _k}, b _ {r _k}) : a _i, r _i \in \mathbb{N} \}$ is a <strong>countable</strong> subcover, i.e., $\mathbb{R} _L$ is Lindelof.</li> <li>T4: <br /> Take any disjoint closed set $A,B$ in $\mathbb{R} _L$. $\forall x \in A$, $\exists$ basis element s.t. $[x, r _x) \cap B = \emptyset$. Therefore, $\{ [x, r _x) : x \in A \}$ and $\{ [y, r _y) : y \in B \}$ are open covers of $A, B$ respectively. Take $U _A = \cup _{x \in A} [ x, r_x)$ and $U _B = \cup _{y \in B} [ y, r _y)$, which are open. Since $U _A \cap U _B = \emptyset$, $\mathbb{R} _L$ is T4.</li> </ul> <hr /> $\mathbb{R}_L \times \mathbb{R}_L \text{ is} \nonumber$ <ol> <li>Not Lindelof: <br /> Note that base element of $\mathbb{R}_L \times \mathbb{R}_L$ is $[a, b) \times [c, d)$ and $\{ [x, b) \times [-x, d) : x \in \mathbb{R} \}$ is a open cover of $\mathbb{R}_L \times \mathbb{R}_L$. Consider a set $S:= \{ (x, -x) : x \in \mathbb{R} \} \subset \mathbb{R}_L \times \mathbb{R}_L$, which is a closed set. Then, $(\mathbb{R}_L \times \mathbb{R}_L) \setminus S$ is open and thus $\{ [x, b) \times [-x, d) , (\mathbb{R}_L \times \mathbb{R}_L) \setminus S : x \in \mathbb{R} \}$ is also a open cover. Since we cannot cover the uncoutable choices from $(\mathbb{R}_L \times \mathbb{R}_L) \setminus S$ by countable number, $\mathbb{R}_L \times \mathbb{R}_L)$ is not Lindelof.</li> <li>T3 but not T4: <br /> Since $\mathbb{R}_L$ is T3 from its definition, its product $\mathbb{R}_L \times \mathbb{R}_L$ is also T3. Suppose $\mathbb{R}_L$ is T4. Since $S$ is closed and $S$ is discrete, every subsets of $S$ is open and closed. For each closed $A,B$ in $S$, those are closed in $\mathbb{R}_L \times \mathbb{R}_L$. If $A\cap B = \emptyset$, then $\exists$ nbhds $U_A, U_B$ with $U_A \cap U_B = \emptyset$. Note that $\mathbb{Q}$ is dense in $\mathbb{R}_L$, $\mathbb{Q} \times \mathbb{Q}$ is also dense in $\mathbb{R}_L \times \mathbb{R}_L$. Let define $\Phi : 2^S \to 2^{\mathbb{Q}\times \mathbb{Q}}$ s.t. $A (\ne S) \mapsto U_A \cap (\mathbb{Q}\times \mathbb{Q})$. Note that $A$ and $S\setminus A$ are closed, so $\exists U_A, V_A$ nbhds and $\Phi(S) = \mathbb{Q}\times \mathbb{Q}$ and $\Phi(\emptyset) = \emptyset$. Since $\Phi$ is injective, $2^S \cong \Phi( 2^{\mathbb{Q}\times \mathbb{Q}}) \cong 2^N \cong \mathbb{R}$, i.e., $2^S \cong S$, which is a contradiction. Therefore, $\mathbb{R}_L \times \mathbb{R}_L$ is not T4.</li> </ol>Jongyeong Leelee@ms.k.u-tokyo.ac.jpSource of most content: Gunhee Cho and Topology by Munkres. Separation Axiom Def. Given a T.S. $(X,T)$, $X$ is T1 if any finite set is a closed set. Def. Given a T.S. $(X,T)$, $X$ is T2 (Hausdorff) if $\forall x,y \in X$ with $x\ne y$, $\exists$ nbhds $U_x, U_y$ of $x,y$ respectively such that $V_x \cap V_y= \emptyset$. Def. Given a T.S. $(X,T)$, $X$ is T3 (regular) if $X$ is T1 and $\forall x \in X$ and $\forall$ closed set $A\subseteq X$ with $x\not\in A$, $\exists$ nbhd $V_x, V_A$ s.t. $V_x \cap V_A = \emptyset$. Def. Given a T.S. $(X,T)$, $X$ is T4 (normal) if $X$ is T1 and $\forall$ closed sets $A,B\subseteq X$ with $A \cap B = \emptyset$, $\exists$ nbhd $V_A, V_B$ s.t. $V_A \cap V_B = \emptyset$. NB1; If $X$ is a metric space, then T1 $\Leftrightarrow$ T2 $\Leftrightarrow$ T3 $\Leftrightarrow$ T4. Propositions $\text{Given a T1 T.S. } (X,T), \nonumber$ $X$ is regular iff $\forall x \in X$, $\forall$ nbhd $U_x$ of $x$, $\exists$ nbhd $V$ of $x$ s.t. $x\in \bar{V} \subseteq U_x$. $X$ is normal iff $\forall$ closed set $A$ in $X$, $\forall$ nbhds $U_A$ of $A$, $\exists$ nbhd $V$ of $A$ s.t. $A \subseteq \bar{V} \subseteq U_A$. Proof (1 $\Rightarrow$): Let $x\in U_x \in T$, i.e., $U_x^c$ is closed in $X$. Since $X$ is regular, $\exists$ nbhds $V_x, V_{U_x^c}$ s.t. $V_x\cap V_{U_x^c}=\emptyset$. Then, $\overline{V_x}\cap V_{U_x^c}=\emptyset$ since if $y \in U_x^c$, $V_{U_X^c}$ is a nbhd of $y$ disjoint from $V_x$. Therefore, $\overline{V_x}\cap U_x^c = \emptyset$, i.e., $\overline{V_x} \subseteq U_x$. (1 $\Leftarrow$): Take any $x \in X$ and closed set $A$ with $x \not\in A$, i.e., $x \in A^c$ for open $A^c$. By assumption, $\exists$ nbhd $V_x$ s.t. $x \in \overline{V_x} \subseteq A^c$, i.e., $A \subseteq (\overline{V_x})^c$. This implies that $(\overline{V_x})^c$ is a nbhd of $A$. Therefore, $\overline{V_x} \cap (\overline{V_x})^c = \overline{V_x} \cap V_A = \emptyset$. (2): One can apply the proof of (1) by replacing $x$ by the closed set $A$. $\text{If } X \text{ is regular (or T2), then } \nonumber$ Subspace $Y$ of $X$ is regular (or T2). $\prod_i X_i$ is regular (or T2) in product space. Proof (1-T2): Take $x, y \in Y$. If $U_x$ and $U_y$ are disjoint nbhds in $X$, then $U_x\cap Y$ and $U_y \cap Y$ are disjoint nbhds of $x$ and $y$ in $Y$. (1-T3): Take $x \in Y$ and let $A$ be a closed subset of $Y$ with $x \not\in A$. Since $\bar{A} \cap Y = A$, $x \not\in \bar{A}$ holds. From the regularity, $\exists$ nbhds $U_x, U_{\bar{A}}$ of $x$ and $\bar{A}$ in $X$. Then, $U_x \cap Y$ and $U_{\bar{A}}$ are disjoint open sets in $Y$. (2-T2): RM25 (2-T3): Let $X=\prod_\alpha X_\alpha$. Then, $X$ is T2 by (2-T2) since regular space is T2, so one-point sets are closed in $X$. Let $\mathbf{x} = (x_\alpha)$ be a point of $X$ and let $U=\prod_\alpha U_\alpha$ be a basis element about $\mathbf{x}$ contained in $X$. For each $\alpha$, choose a nbhd $V_\alpha$ of $x_\alpha \in X_\alpha$ s.t. $\overline{V_\alpha} \subset U_\alpha$. If $U_\alpha = X_\alpha$, then choose $V_\alpha = X_\alpha$. Since $\bar{V} \stackrel{(*)}{=} \prod_\alpha \overline{V_\alpha}$, it holds that $\bar{V} \subset \prod U_\alpha \subseteq U$, i.e., $X$ is regular. $(*): \text{ Let } (X_\alpha) \text{ be an indexed family of spaces and } \forall \alpha, \, A_\alpha \subset X_\alpha. \nonumber$ $\text{If } \prod X_\alpha \text{ is given either the product or the box topology, then } \prod \overline{A_\alpha} = \overline{\prod A_\alpha}. \nonumber$ Proof $(\subseteq)$: Take $\mathbf{x} = (x_\alpha) \in \prod \overline{A_\alpha}$ and let $U=\prod U_\alpha$ be a basis element for either the box or product topology that contains $\mathbf{x}$. Since $x_\alpha \in \overline{A_\alpha}$, i.e., $U_\alpha \cap A_\alpha \ne \emptyset$ one can choose a point $y_\alpha \in U_\alpha \cap A_\alpha$. Then, $\mathbf{y} = (y_\alpha)$ belongs to both $U$ and $\prod A_\alpha$. Since $U$ is arbitrary, $\mathbf{x}$ belongs to $\overline{\prod A_\alpha}$. $(\supseteq)$: Take $\mathbf{x} = (x_\alpha) \in \overline{\prod A_\alpha}$. Let $U _\alpha$ be a nbhd of $x _\alpha$. Since $\pi _\alpha ^{-1}(U _\alpha)$ is open in $\prod X _\alpha$, it contains a point $\mathbf{y}$ of $\prod A _\alpha$. Then, $y _\alpha \in U _\alpha \cap A _\alpha$, i.e., $x _\alpha \in \bar{A} _\alpha$, which concludes the proof. Some examples $(X, T_{cofinite})$ is T1 but not T2. $\mathbb{R} _k := (\mathbb{R}, T _{k})$ is T2 but not regular since nbhd of $0$ must contain some points of $K:= \{\frac{1}{n}: n \in \mathbb{N} \}$. NB2; If $X$ is a metric space, 2nd countalbe $\Leftrightarrow$ Lindelof $\Leftrightarrow$ separble. $\mathbb{R}_L := (\mathbb{R}, T_{\text{lower limit top}}), \text{ where } \mathcal{B} = \{ [a,b) : a,b \in \mathbb{R}, a&lt; b \}. \nonumber$ Note that $\bar{\mathbb{Q}}=\mathbb{R}$. 1st countable: Let $x\in \mathbb{R}$ and $B_x := \{ [x, x+\frac{1}{n}) : n \in \mathbb{N} \}$. Then, $\forall$ open $U_x$ of $x$, $\exists B_{x,n} \in B_x$ s.t. $x \in B_{x,n} \subseteq U_x$, i.e., $\mathbb{R}_L$ is 1st countable. 2nd countable: Let $B_x = [x, b)$ for any $b &gt; x$. Then, $B_x$ is open and $\inf B_x = x$ by definition. If $x\ne y$ for $x,y \in \mathbb{R}$, then $\inf B_x \ne \inf B_y$. Then, we can define an injective function $\Phi: \mathbb{R} \to \mathcal{B}$ ($x \mapsto [x, \cdot)$). Since $\Phi$ is injective, $\mathcal{B}$ must be uncountable, i.e., $\mathbb{R}_L$ is not 2nd countable. Lindelof (existence of countable subcover): Take an open cover $O = \{ [a_\alpha, b_\alpha) : a_\alpha, b_\alpha \in \mathbb{R}, a_\alpha &lt; b_\alpha \} = \{ I_\alpha : \alpha \in I \}$. Consider $J_\alpha := (a_\alpha ,b_\alpha)$. Then, $\mathbb{R} = \cup _\alpha I _{\alpha}$. Take any $x\in \mathbb{R} \setminus \cup _\alpha J _{\alpha}$. Then $x= a _\alpha$ for some $\alpha \in I$. By density of rationals, there exists $q _x \in [x, b) \subset \mathbb{Q}$. Define $\Phi: \mathbb{R} \setminus \cup _\alpha J _{\alpha} \to \mathbb{Q}$ $(x \mapsto q _x)$. Take $x,y \in \mathbb{R} \setminus \cup _\alpha J _{\alpha}$ with $x &lt; y$. If $q _x \geq q _y$, i.e., $y \in (x, b _\alpha)$, then $y \in \cup _\alpha J _{\alpha}$, which contradicts to assumption. Thus, $\mathbb{R} \setminus \cup _\alpha J _{\alpha}$ is at most countable. Hence, $q _x &lt; q _y$, i.e., $q _x \ne q _y$. Topologize $\cup _\alpha J _{\alpha}$ as a subspace of $(\mathbb{R}, T _{std})$. Then, $\exists (\alpha _k)$ s.t. $\cup _\alpha J _{\alpha} = \cup _{k=1}^\infty J _{ \alpha _k}$. Therefore, $\mathbb{R} = \left(\mathbb{R} \setminus \cup _{k=1}^\infty J _{ \alpha _k} \right) \bigcup \cup _{k=1}^\infty J _{ \alpha _k}$. Thus, $O’ := \{ [a _ {\alpha _k}, b _ {\alpha _k}), [a _ {r _k}, b _ {r _k}) : a _i, r _i \in \mathbb{N} \}$ is a countable subcover, i.e., $\mathbb{R} _L$ is Lindelof. T4: Take any disjoint closed set $A,B$ in $\mathbb{R} _L$. $\forall x \in A$, $\exists$ basis element s.t. $[x, r _x) \cap B = \emptyset$. Therefore, $\{ [x, r _x) : x \in A \}$ and $\{ [y, r _y) : y \in B \}$ are open covers of $A, B$ respectively. Take $U _A = \cup _{x \in A} [ x, r_x)$ and $U _B = \cup _{y \in B} [ y, r _y)$, which are open. Since $U _A \cap U _B = \emptyset$, $\mathbb{R} _L$ is T4. $\mathbb{R}_L \times \mathbb{R}_L \text{ is} \nonumber$ Not Lindelof: Note that base element of $\mathbb{R}_L \times \mathbb{R}_L$ is $[a, b) \times [c, d)$ and $\{ [x, b) \times [-x, d) : x \in \mathbb{R} \}$ is a open cover of $\mathbb{R}_L \times \mathbb{R}_L$. Consider a set $S:= \{ (x, -x) : x \in \mathbb{R} \} \subset \mathbb{R}_L \times \mathbb{R}_L$, which is a closed set. Then, $(\mathbb{R}_L \times \mathbb{R}_L) \setminus S$ is open and thus $\{ [x, b) \times [-x, d) , (\mathbb{R}_L \times \mathbb{R}_L) \setminus S : x \in \mathbb{R} \}$ is also a open cover. Since we cannot cover the uncoutable choices from $(\mathbb{R}_L \times \mathbb{R}_L) \setminus S$ by countable number, $\mathbb{R}_L \times \mathbb{R}_L)$ is not Lindelof. T3 but not T4: Since $\mathbb{R}_L$ is T3 from its definition, its product $\mathbb{R}_L \times \mathbb{R}_L$ is also T3. Suppose $\mathbb{R}_L$ is T4. Since $S$ is closed and $S$ is discrete, every subsets of $S$ is open and closed. For each closed $A,B$ in $S$, those are closed in $\mathbb{R}_L \times \mathbb{R}_L$. If $A\cap B = \emptyset$, then $\exists$ nbhds $U_A, U_B$ with $U_A \cap U_B = \emptyset$. Note that $\mathbb{Q}$ is dense in $\mathbb{R}_L$, $\mathbb{Q} \times \mathbb{Q}$ is also dense in $\mathbb{R}_L \times \mathbb{R}_L$. Let define $\Phi : 2^S \to 2^{\mathbb{Q}\times \mathbb{Q}}$ s.t. $A (\ne S) \mapsto U_A \cap (\mathbb{Q}\times \mathbb{Q})$. Note that $A$ and $S\setminus A$ are closed, so $\exists U_A, V_A$ nbhds and $\Phi(S) = \mathbb{Q}\times \mathbb{Q}$ and $\Phi(\emptyset) = \emptyset$. Since $\Phi$ is injective, $2^S \cong \Phi( 2^{\mathbb{Q}\times \mathbb{Q}}) \cong 2^N \cong \mathbb{R}$, i.e., $2^S \cong S$, which is a contradiction. Therefore, $\mathbb{R}_L \times \mathbb{R}_L$ is not T4.Random math: 382021-08-13T00:00:00+00:002021-08-13T00:00:00+00:00https://jongyeong.github.io/rm/rm38<p>Source of most content: <a href="https://www.youtube.com/channel/UCB_AbuIVIG8I3K3_T_j0FSw/">Gunhee Cho</a> and Topology by Munkres.</p> <h1 id="countability-axiom">Countability Axiom</h1> <p><code class="language-plaintext warning highlighter-rouge">Motivation.</code> Let $X$ be a <strong>metric space</strong> and $A \subseteq X$ be open. Then,</p> <ol> <li>If $\exists (x_n) \subseteq A$ s.t. $x_n \to x$, then $x \in \overline{A}$. Note that converse holds if $X$ is a MS.</li> <li>$f$ is cts at $x=p$, i.e., $f(\overline{A}) \subseteq \overline{f(A)}$ iff $\forall (p_n)$ which converges to $p$, $f(p_n) \to f(p)$ holds.</li> </ol> <p>Here, we might induce similar results without the assumption of a metric space: <br /> For $x \in \overline{A}$, if $F = \{ V_x \mid V_x : \text{nbhd of }x \}$ is countable, let $\forall n\in \mathbb{N}$, $V_1 \cap \ldots V_n \in F$, $x_n \in V_1 \cap \ldots V_n\cap A$. Then, $\exists (x_n) \subseteq A$ and $x_n \to x$.</p> <h1 id="countable-ts">Countable TS</h1> <p class="info"><code class="language-plaintext info highlighter-rouge">Def.</code> Given T.S. $(X, T)$, $B_X (\subseteq T)$ is a <strong>local base</strong> at $x$ if $\forall$ nbhd $V$ of $x$, $\exists B \in B_X$ s.t. $x\in B \subseteq V$. And we say $X$ is <strong>1st countable</strong> if $\forall x \in X$, $\exists B_x$ a <em>countable</em> local base at $x$.</p> <p>NB1; If $X$ is 1st countable, then $\exists (x_n) \subseteq A$ s.t. $x_n \to x \Leftrightarrow x \in \overline{A}$. <br /> NB2; If $X$ is 1st countable, then $f$ is cts at $x=p \Leftrightarrow \forall (p_n) \to p$, $f(p_n) \to f(p)$. Note that left arrow of NB1,2 can be shown by the def. of 1st countable.</p> <p class="info"><code class="language-plaintext info highlighter-rouge">Def.</code> Given T.S. $(X, T)$, $X$ is <strong>2nd countable</strong> if $\exists$ countable basis $\mathcal{B}$.</p> <p>NB3; if $X$ is T3(regular) and 2nd countable, then $X$ is metrizable. (Uryshon’s metrization thm)<br /> NB4; 2nd countable $\rightarrow$ 1st countable.</p> <blockquote> <p>$\forall x \in X, B_x = \{ B \in \mathcal{B} : x \in B \}$ is countable since $\mathcal{B}$ is countable.</p> </blockquote> <p>NB5; $(\mathbb{R}^w, T_{uniform})$: countable product of real space with uniform topology is 1st countable but not 2nd countable.</p> <h2 id="propositions">Propositions</h2> $\text{Let } X \text{ be a 2nd countable T.S.. Then, the followings hold: } \nonumber$ <ol> <li>For any open cover of $X$, $\exists$ countable subcover of $X$, i.e., $X$ is Lindelof space.</li> <li>$\exists$ countable dense subset $A$ of $X$. Here, <strong>dense</strong> implies $\overline{A} = X$, i.e., $X$ is a <strong>separable space</strong>.</li> </ol> <p>NB6; If $X$ is metrizable, TFAE: 2nd countable, Lindelof, separable.</p> <h3 id="proof">Proof</h3> <ol> <li>Let $\mathcal{B}=\{ B_n : n \in \mathbb{N} \}$ be a countable basis and take any open cover $U = \{ U_\alpha : \alpha \in I\}$ of $X$. For each $n \in \mathbb{N}$, let $U’ = \{ A_n \in U : B_n \subseteq A_n \in U\}$. If we take any $x \in X$, then $\exists U_\alpha$ s.t. $x \in U_\alpha$. Then, $\exists n \in \mathbb{N}$ s.t. $x \in B_n \subseteq U_\alpha$, i.e., $U_\alpha = U_{\alpha, n} \in U’$. Hence, $X = \bigcup_{A \in U’} A$. Note that $U’$ is countable by definition, which concludes the proof.</li> <li>For each $n \in \mathbb{N}$, choose $x_n \in B_n \ne \emptyset$. Let $A := \{ x_n : n \in \mathbb{N}\}$. NTS: $\overline{A} = X$.<br /> Since $\overline{A} \subseteq X$ is clear, take any $x \in X$ and any basis $B_n \ni x$. If $x_n \ne x$, then $x_n \in (B_n \setminus \{x \})$. Thus, $(B_n \setminus \{x \}) \cap X \ne \emptyset$. If $x_n = x$, then $x=x_n \in A$. Therefore, $x \in \overline{A}$.</li> </ol> <hr /> $\text{For T.S. }X, \text{ the followings hold:} \nonumber$ <ol> <li>If $X$ is 1st or 2nd countable, then $Y \subseteq X$ as a subspace also has the same axioms.</li> <li>If $X_i$’s are 1st or 2nd countable for $i \in I$ (all $X_i$ share the same axiom), then $\prod_i X_i$ is also has the same axiom.</li> </ol> <h3 id="proof-1">Proof</h3> <p>Consider the 2nd countable case, then the proof of the 1st countable case would be similar.</p> <ol> <li>Let $\mathcal{B}$ be a countable basis for $X$. Then, $\{B \cap A : B \in \mathcal{B} \}$ is a countable basis for the subspace $A$ of $X$.</li> <li>Let $\mathcal{B}_i$ be a countable basis for the space $X_i$. Then, $\{ \prod_i U_i : U_i \in \mathcal{B}_i \}$ for finitely many $i$ and $U_i = X_i$ for other values of $i$, is a countable basis for $\prod_i X_i$.</li> </ol> <h1 id="topologies-of-function-space">Topologies of function space</h1> <p class="info"><code class="language-plaintext info highlighter-rouge">Def.</code> For a given T.S. $(Y,T)$, let $Y^X = \{ f : X \to Y\}$ be a set of functions. Let $S(x \in Y, U \in T) := \{f \in Y^X : f(x) \in U \}$ be a <strong>subbasis of $Y^X$</strong>. Then, $\mathcal{B} = S(x_1, U_1) \cap \cdots \cap S(x_n, U_n)$ becomes a basis and $T_{\mathcal{B}}$ is called <strong>pointwise-convergence topology</strong> of $Y^X$.</p> <p>NB7; $X_1 \times X_2 = \{ (x_1, x_2) : x_1 \in X_1, x_2 \in X_2\} = \{ f: \{1,2\} \to \cup_{i=1,2} X_i : f(1) \in X_1, f(2) \in X_2\}$, i.e., $\prod_{\alpha \in I}X_\alpha = \{ f \mid f:I \to \cup_{\alpha \in I} X_\alpha, f(\alpha) \in X_{\alpha}, \forall \alpha \in I \}=Y^X$. <br /> NB8; Given $(Y^X, T_{pc}), (f_n) : X \to Y$, $f_n \to f$ in ptwise-convergence is equal to $\forall x \in X, \lim_{n\to \infty} f_n(x) = f(x)$. <br /> NB9; Let $X = I = [0, 1]$ and $f_n(x) = x^n \to f(x) = \begin{cases} 0 &amp;\text{ if } x &lt;1 \\ 1 &amp;\text{ if } x = 1\end{cases}$. Then, for the collection $C(I, \mathbb{R})$ of continuous functions $f: I \to \mathbb{R}$, it holds that $(f_n) \subseteq C(I, \mathbb{R}) \lneq (\mathbb{R}^{[0, 1]}, T_{pc})$. Then, $f_n \to f \not\in C(I, \mathbb{R})$, i.e., $C(I, \mathbb{R})$ is not closed.</p> <p class="info"><code class="language-plaintext info highlighter-rouge">Def.</code> Let $X$ be a set and $Y$ be a metric space with $d$. Then, we say $f_n$ <strong>converges uniformly</strong> to $f$, ${f_n}^\to_\to f$iff $\forall \epsilon &gt; 0, \exists N \in \mathbb{N}$ s.t. $\forall n \geq N, \forall x \in X$, $d(f_n(x), f(x)) &lt; \epsilon$.</p> <p class="info"><code class="language-plaintext info highlighter-rouge">Def.</code> Given T.S. $(X,T)$ and metric space $(Y, d)$, define $B_C(f, \epsilon) := \{ g \in Y^X : \sup_{x\in C } d(f(x), g(x)) &lt; \epsilon\}$ for each compact set $C$ in $X$, $f:X\to Y$ and $\epsilon &gt; 0$. Then, $\mathcal{B} = \{ B_C(f, \epsilon)\}$ is a basis of $Y^X$ and we call $T_{\mathcal{B}}$ a <strong>compact convergence topology</strong> on $Y^X$.</p> <p>NB10; $(f_n) \subseteq (Y^X, T_{cct})$, $f_n \to f$ in cct sense is equal to $\forall$ cpt $C$, $f_n {\mid_{C}}^\to_\to f\mid_C$. <br /> NB11; For metric space $Y$, $T_{cct} = T_{\text{compact open topology}}$.</p> <h2 id="propositions-1">Propositions</h2> $\text{Given } (X,T), (Y, d), \text{ if } (f_n)\subseteq C(X,Y) \text{ s.t. } {f_n}^\to_\to f, \text{ then } f\in C(X,Y), \text{ i.e., } f \text{ is cts.} \nonumber$ <h3 id="proof-2">Proof</h3> <p>Fix $x_0 \in X$ and $\epsilon &gt;0$ and consider a ball $B_{\epsilon}(f(x_0))$. By uniform convergence, for any $x\in X$, $\exists N \in \mathbb{N}$ s.t. $d(f(x), f_N(x) ) &lt; \frac{\epsilon}{3}$. By continuity, $\exists$ nbhd $U$ of $x_0$ s.t. $d(f_N(x), f_N(x_0)) &lt; \frac{\epsilon}{3}$. By uniform convergence, $\exists N \in \mathbb{N}$ s.t. $d(f_N(x_0), f(x_0)) &lt; \frac{\epsilon}{3}$. Then, $d(f(x), f(x_0)) \leq d(f(x), f_N(x) ) + d(f_N(x), f_N(x_0)) + d(f_N(x_0), f(x_0)) &lt; \epsilon$.</p> <hr /> $(\mathbb{R}^w , T_{\text{uniformly convergence topolgy} }) \text{ is 1st countable but not 2nd countable.} \nonumber$ <h3 id="proof-3">Proof</h3> <ul> <li>For a metric space $Y$ and index set $J$, consider a function space $Y^J$. Then, $x,y : J \to Y \, ( \alpha \mapsto x_{\alpha}, y_{\alpha})$. Let $\bar{d}(x_\alpha, y_\alpha) := \max (d(x_\alpha, y_\alpha),1)$ and $\tilde{d}(x_\alpha, y_\alpha) := \sup_{\alpha \in J}\bar{d}(x_\alpha, y_\alpha) \, (\leq 1)$. Then, $(Y^J, \tilde{d})$ is called uniformly convergence topology. <ol> <li>If $\mathbb{R}^w$ is 2nd countable, then $\exists$ countable basis $\mathcal{B}$. Then, for any discrete subspace $D$ of $\mathbb{R}^w$, $D$ must be at most countable since one-point set is open. Precisely, define $\Phi: D \to \mathcal{B}\, (a \mapsto B_a)$ s.t. $B_a \cap D = \{ a \}$. If $a \ne b$ then, $\Phi(a) \ne \Phi(b)$, i.e., $\Phi$ is injective. Thus, $D$ is at most countable.</li> <li>Define $F:= \{ a \in \mathbb{R}^w : \text{each entry of }a \text{ is either } 0 \text{ or }1 \}$ which is uncountable. Then, $\forall a,b \in F, \tilde{d}(a,b) = \begin{cases} 1 \quad a \ne b \\ 0 \quad a = b\end{cases}$, i.e., $F$ is discrete subspace of $\mathbb{R}^w$. This implies that $F$ is at most countable, which is contradicts to the definition of $F$.</li> </ol> </li> </ul>Jongyeong Leelee@ms.k.u-tokyo.ac.jpSource of most content: Gunhee Cho and Topology by Munkres. Countability Axiom Motivation. Let $X$ be a metric space and $A \subseteq X$ be open. Then, If $\exists (x_n) \subseteq A$ s.t. $x_n \to x$, then $x \in \overline{A}$. Note that converse holds if $X$ is a MS. $f$ is cts at $x=p$, i.e., $f(\overline{A}) \subseteq \overline{f(A)}$ iff $\forall (p_n)$ which converges to $p$, $f(p_n) \to f(p)$ holds. Here, we might induce similar results without the assumption of a metric space: For $x \in \overline{A}$, if $F = \{ V_x \mid V_x : \text{nbhd of }x \}$ is countable, let $\forall n\in \mathbb{N}$, $V_1 \cap \ldots V_n \in F$, $x_n \in V_1 \cap \ldots V_n\cap A$. Then, $\exists (x_n) \subseteq A$ and $x_n \to x$. Countable TS Def. Given T.S. $(X, T)$, $B_X (\subseteq T)$ is a local base at $x$ if $\forall$ nbhd $V$ of $x$, $\exists B \in B_X$ s.t. $x\in B \subseteq V$. And we say $X$ is 1st countable if $\forall x \in X$, $\exists B_x$ a countable local base at $x$. NB1; If $X$ is 1st countable, then $\exists (x_n) \subseteq A$ s.t. $x_n \to x \Leftrightarrow x \in \overline{A}$. NB2; If $X$ is 1st countable, then $f$ is cts at $x=p \Leftrightarrow \forall (p_n) \to p$, $f(p_n) \to f(p)$. Note that left arrow of NB1,2 can be shown by the def. of 1st countable. Def. Given T.S. $(X, T)$, $X$ is 2nd countable if $\exists$ countable basis $\mathcal{B}$. NB3; if $X$ is T3(regular) and 2nd countable, then $X$ is metrizable. (Uryshon’s metrization thm) NB4; 2nd countable $\rightarrow$ 1st countable. $\forall x \in X, B_x = \{ B \in \mathcal{B} : x \in B \}$ is countable since $\mathcal{B}$ is countable. NB5; $(\mathbb{R}^w, T_{uniform})$: countable product of real space with uniform topology is 1st countable but not 2nd countable. Propositions $\text{Let } X \text{ be a 2nd countable T.S.. Then, the followings hold: } \nonumber$ For any open cover of $X$, $\exists$ countable subcover of $X$, i.e., $X$ is Lindelof space. $\exists$ countable dense subset $A$ of $X$. Here, dense implies $\overline{A} = X$, i.e., $X$ is a separable space. NB6; If $X$ is metrizable, TFAE: 2nd countable, Lindelof, separable. Proof Let $\mathcal{B}=\{ B_n : n \in \mathbb{N} \}$ be a countable basis and take any open cover $U = \{ U_\alpha : \alpha \in I\}$ of $X$. For each $n \in \mathbb{N}$, let $U’ = \{ A_n \in U : B_n \subseteq A_n \in U\}$. If we take any $x \in X$, then $\exists U_\alpha$ s.t. $x \in U_\alpha$. Then, $\exists n \in \mathbb{N}$ s.t. $x \in B_n \subseteq U_\alpha$, i.e., $U_\alpha = U_{\alpha, n} \in U’$. Hence, $X = \bigcup_{A \in U’} A$. Note that $U’$ is countable by definition, which concludes the proof. For each $n \in \mathbb{N}$, choose $x_n \in B_n \ne \emptyset$. Let $A := \{ x_n : n \in \mathbb{N}\}$. NTS: $\overline{A} = X$. Since $\overline{A} \subseteq X$ is clear, take any $x \in X$ and any basis $B_n \ni x$. If $x_n \ne x$, then $x_n \in (B_n \setminus \{x \})$. Thus, $(B_n \setminus \{x \}) \cap X \ne \emptyset$. If $x_n = x$, then $x=x_n \in A$. Therefore, $x \in \overline{A}$. $\text{For T.S. }X, \text{ the followings hold:} \nonumber$ If $X$ is 1st or 2nd countable, then $Y \subseteq X$ as a subspace also has the same axioms. If $X_i$’s are 1st or 2nd countable for $i \in I$ (all $X_i$ share the same axiom), then $\prod_i X_i$ is also has the same axiom. Proof Consider the 2nd countable case, then the proof of the 1st countable case would be similar. Let $\mathcal{B}$ be a countable basis for $X$. Then, $\{B \cap A : B \in \mathcal{B} \}$ is a countable basis for the subspace $A$ of $X$. Let $\mathcal{B}_i$ be a countable basis for the space $X_i$. Then, $\{ \prod_i U_i : U_i \in \mathcal{B}_i \}$ for finitely many $i$ and $U_i = X_i$ for other values of $i$, is a countable basis for $\prod_i X_i$. Topologies of function space Def. For a given T.S. $(Y,T)$, let $Y^X = \{ f : X \to Y\}$ be a set of functions. Let $S(x \in Y, U \in T) := \{f \in Y^X : f(x) \in U \}$ be a subbasis of $Y^X$. Then, $\mathcal{B} = S(x_1, U_1) \cap \cdots \cap S(x_n, U_n)$ becomes a basis and $T_{\mathcal{B}}$ is called pointwise-convergence topology of $Y^X$. NB7; $X_1 \times X_2 = \{ (x_1, x_2) : x_1 \in X_1, x_2 \in X_2\} = \{ f: \{1,2\} \to \cup_{i=1,2} X_i : f(1) \in X_1, f(2) \in X_2\}$, i.e., $\prod_{\alpha \in I}X_\alpha = \{ f \mid f:I \to \cup_{\alpha \in I} X_\alpha, f(\alpha) \in X_{\alpha}, \forall \alpha \in I \}=Y^X$. NB8; Given $(Y^X, T_{pc}), (f_n) : X \to Y$, $f_n \to f$ in ptwise-convergence is equal to $\forall x \in X, \lim_{n\to \infty} f_n(x) = f(x)$. NB9; Let $X = I = [0, 1]$ and $f_n(x) = x^n \to f(x) = \begin{cases} 0 &amp;\text{ if } x &lt;1 \\ 1 &amp;\text{ if } x = 1\end{cases}$. Then, for the collection $C(I, \mathbb{R})$ of continuous functions $f: I \to \mathbb{R}$, it holds that $(f_n) \subseteq C(I, \mathbb{R}) \lneq (\mathbb{R}^{[0, 1]}, T_{pc})$. Then, $f_n \to f \not\in C(I, \mathbb{R})$, i.e., $C(I, \mathbb{R})$ is not closed. Def. Let $X$ be a set and $Y$ be a metric space with $d$. Then, we say $f_n$ converges uniformly to $f$, ${f_n}^\to_\to f$iff $\forall \epsilon &gt; 0, \exists N \in \mathbb{N}$ s.t. $\forall n \geq N, \forall x \in X$, $d(f_n(x), f(x)) &lt; \epsilon$. Def. Given T.S. $(X,T)$ and metric space $(Y, d)$, define $B_C(f, \epsilon) := \{ g \in Y^X : \sup_{x\in C } d(f(x), g(x)) &lt; \epsilon\}$ for each compact set $C$ in $X$, $f:X\to Y$ and $\epsilon &gt; 0$. Then, $\mathcal{B} = \{ B_C(f, \epsilon)\}$ is a basis of $Y^X$ and we call $T_{\mathcal{B}}$ a compact convergence topology on $Y^X$. NB10; $(f_n) \subseteq (Y^X, T_{cct})$, $f_n \to f$ in cct sense is equal to $\forall$ cpt $C$, $f_n {\mid_{C}}^\to_\to f\mid_C$. NB11; For metric space $Y$, $T_{cct} = T_{\text{compact open topology}}$. Propositions $\text{Given } (X,T), (Y, d), \text{ if } (f_n)\subseteq C(X,Y) \text{ s.t. } {f_n}^\to_\to f, \text{ then } f\in C(X,Y), \text{ i.e., } f \text{ is cts.} \nonumber$ Proof Fix $x_0 \in X$ and $\epsilon &gt;0$ and consider a ball $B_{\epsilon}(f(x_0))$. By uniform convergence, for any $x\in X$, $\exists N \in \mathbb{N}$ s.t. $d(f(x), f_N(x) ) &lt; \frac{\epsilon}{3}$. By continuity, $\exists$ nbhd $U$ of $x_0$ s.t. $d(f_N(x), f_N(x_0)) &lt; \frac{\epsilon}{3}$. By uniform convergence, $\exists N \in \mathbb{N}$ s.t. $d(f_N(x_0), f(x_0)) &lt; \frac{\epsilon}{3}$. Then, $d(f(x), f(x_0)) \leq d(f(x), f_N(x) ) + d(f_N(x), f_N(x_0)) + d(f_N(x_0), f(x_0)) &lt; \epsilon$. $(\mathbb{R}^w , T_{\text{uniformly convergence topolgy} }) \text{ is 1st countable but not 2nd countable.} \nonumber$ Proof For a metric space $Y$ and index set $J$, consider a function space $Y^J$. Then, $x,y : J \to Y \, ( \alpha \mapsto x_{\alpha}, y_{\alpha})$. Let $\bar{d}(x_\alpha, y_\alpha) := \max (d(x_\alpha, y_\alpha),1)$ and $\tilde{d}(x_\alpha, y_\alpha) := \sup_{\alpha \in J}\bar{d}(x_\alpha, y_\alpha) \, (\leq 1)$. Then, $(Y^J, \tilde{d})$ is called uniformly convergence topology. If $\mathbb{R}^w$ is 2nd countable, then $\exists$ countable basis $\mathcal{B}$. Then, for any discrete subspace $D$ of $\mathbb{R}^w$, $D$ must be at most countable since one-point set is open. Precisely, define $\Phi: D \to \mathcal{B}\, (a \mapsto B_a)$ s.t. $B_a \cap D = \{ a \}$. If $a \ne b$ then, $\Phi(a) \ne \Phi(b)$, i.e., $\Phi$ is injective. Thus, $D$ is at most countable. Define $F:= \{ a \in \mathbb{R}^w : \text{each entry of }a \text{ is either } 0 \text{ or }1 \}$ which is uncountable. Then, $\forall a,b \in F, \tilde{d}(a,b) = \begin{cases} 1 \quad a \ne b \\ 0 \quad a = b\end{cases}$, i.e., $F$ is discrete subspace of $\mathbb{R}^w$. This implies that $F$ is at most countable, which is contradicts to the definition of $F$.Random math: 372021-06-11T00:00:00+00:002021-06-11T00:00:00+00:00https://jongyeong.github.io/rm/rm37<p>Source of most content: <a href="https://www.youtube.com/channel/UCB_AbuIVIG8I3K3_T_j0FSw/">Gunhee Cho</a>, <a href="http://pi.math.cornell.edu/%7Ematsumura/math4530/math4530web.html">Lecture note by Matsumura</a> and Topology by Munkres.</p> <h1 id="path-connected">Path-connected</h1> <p class="info"><code class="language-plaintext info highlighter-rouge">Def.</code> Given T.S. $(X, T)$, $X$ is <strong>path-connected</strong> if $\forall p, q \in X$, $\exists$ cts map $f:[a, b] (\subseteq \mathbb{R}) \to X$ s.t. $f(a)=p$ and $f(q)=b$. We call $f$ a <strong>path</strong>.</p> <p>NB1; If $X$ is path-connected, then $X$ is connected.</p> <blockquote> <p>Suppose $X$ is disconnected, i.e., $X = U \mathbin{\unicode{x228D}} V$ for non-empty open sets $U, V$. Since $f$ is continuous and $[a, b]$ is connected (<a href="/rm/rm36#linear-continum">NB6</a>), $f([a, b])=X$ is also connected (<a href="/rm/rm36#proof-2">2</a>) which is a contradiction.</p> </blockquote> <p>NB2; Every ball is path-connected.</p> <blockquote> <p>Let $B_{\epsilon}(0) \subsetneq \mathbb{R}^n$ be a $\epsilon$-ball centered zero. Then, $\forall x,y \in B_{\epsilon}(0) = B$, $\forall t \in [0, 1]$, let $f(t) := tx + (1-t)y$. Since $\Vert x \Vert, \Vert y \Vert \leq \epsilon$, $\Vert f(t) \Vert \leq \epsilon$, i.e., $f(t) \in B$.</p> </blockquote> <p>NB3; The closure of a connected space is connected, but the closure of a path connected space may not be path connected. <br /> NB4; $S^{n-1} = \{ (x_1, \ldots, x_n) | x_1^2 + \ldots + x_n^2 = 1, x_i \in \mathbb{R} \} \subsetneq \mathbb{R}^n$. Consider $f : \mathbb{R}^n \setminus \{0\} \to S^{n-1} (x \mapsto \frac{x}{\Vert x \Vert}).$ <br /> NB5; Example taht is connected but not path-connected. (<a href="https://en.wikipedia.org/wiki/Topologist%27s_sine_curve">Topologist’s sine curve</a>)</p> <blockquote> <p>Consider $I= [0, 1]$ which is connected. Then, $I \times I$ is also connected. If $I \times I$ is path-connected, we can tak $(0,0)$ and $(1,1)$. Then, $\exists$ cts $f: [a, b] \to I \times I$ with $f(a) = (0,0)$ and $f(b)=(1,1)$. By <a href="/rm/rm36#intermediate-value-thm">IVT</a>, $\forall x \in I$, $f^{-1}(x \times (0,1)) \ne \emptyset$. Note that $\forall x \ne y$ in $I$, $f^{-1}(x\times (0,1)) \cap f^{-1}(y \times (0,1))=\emptyset$. Since $f^{-1}(x\times (0,1))$ is open in $[a, b]$, it contains some interval in $f^{-1}(x\times (0,1))$. Then, by the density of $\mathbb{Q}$, choose $q_x$ in that interval. In this way, we can define $\Phi : I \to \mathbb{Q} (x \mapsto q_x)$. Then, $\Phi$ is injective by construction. This implies that $|I| \leq |Q|$, which is a contradiction since $I$ is uncountable.</p> </blockquote> <h1 id="locally-connected">Locally connected</h1> <p class="info"><code class="language-plaintext info highlighter-rouge">Def.</code> Given T.S. $(X, T)$, $X$ is <strong>locally connected</strong> if $\forall x \in X,$ $\forall$ nbhd $U_x$ of $x$, $\exists$ connected nbhd $V_x$ of $x$ s.t. $x \in V_x \subseteq U_x$.</p> <p class="info"><code class="language-plaintext info highlighter-rouge">Def.</code> Given T.S. $(X, T)$, $X$ is <strong>locally path-connected</strong> if $\forall x \in X,$ $\forall$ nbhd $U_x$ of $x$, $\exists$ path-connected nbhd $V_x$ of $x$ s.t. $x \in V_x \subseteq U_x$.</p> <p>NB6; $\mathbb{R}-\{0\}$ is not connected but locaaly connected. <br /> NB7; $\mathbb{Q}$ is not connected and not locally connected.</p>Jongyeong Leelee@ms.k.u-tokyo.ac.jpSource of most content: Gunhee Cho, Lecture note by Matsumura and Topology by Munkres. Path-connected Def. Given T.S. $(X, T)$, $X$ is path-connected if $\forall p, q \in X$, $\exists$ cts map $f:[a, b] (\subseteq \mathbb{R}) \to X$ s.t. $f(a)=p$ and $f(q)=b$. We call $f$ a path. NB1; If $X$ is path-connected, then $X$ is connected. Suppose $X$ is disconnected, i.e., $X = U \mathbin{\unicode{x228D}} V$ for non-empty open sets $U, V$. Since $f$ is continuous and $[a, b]$ is connected (NB6), $f([a, b])=X$ is also connected (2) which is a contradiction. NB2; Every ball is path-connected. Let $B_{\epsilon}(0) \subsetneq \mathbb{R}^n$ be a $\epsilon$-ball centered zero. Then, $\forall x,y \in B_{\epsilon}(0) = B$, $\forall t \in [0, 1]$, let $f(t) := tx + (1-t)y$. Since $\Vert x \Vert, \Vert y \Vert \leq \epsilon$, $\Vert f(t) \Vert \leq \epsilon$, i.e., $f(t) \in B$. NB3; The closure of a connected space is connected, but the closure of a path connected space may not be path connected. NB4; $S^{n-1} = \{ (x_1, \ldots, x_n) | x_1^2 + \ldots + x_n^2 = 1, x_i \in \mathbb{R} \} \subsetneq \mathbb{R}^n$. Consider $f : \mathbb{R}^n \setminus \{0\} \to S^{n-1} (x \mapsto \frac{x}{\Vert x \Vert}).$ NB5; Example taht is connected but not path-connected. (Topologist’s sine curve) Consider $I= [0, 1]$ which is connected. Then, $I \times I$ is also connected. If $I \times I$ is path-connected, we can tak $(0,0)$ and $(1,1)$. Then, $\exists$ cts $f: [a, b] \to I \times I$ with $f(a) = (0,0)$ and $f(b)=(1,1)$. By IVT, $\forall x \in I$, $f^{-1}(x \times (0,1)) \ne \emptyset$. Note that $\forall x \ne y$ in $I$, $f^{-1}(x\times (0,1)) \cap f^{-1}(y \times (0,1))=\emptyset$. Since $f^{-1}(x\times (0,1))$ is open in $[a, b]$, it contains some interval in $f^{-1}(x\times (0,1))$. Then, by the density of $\mathbb{Q}$, choose $q_x$ in that interval. In this way, we can define $\Phi : I \to \mathbb{Q} (x \mapsto q_x)$. Then, $\Phi$ is injective by construction. This implies that $|I| \leq |Q|$, which is a contradiction since $I$ is uncountable. Locally connected Def. Given T.S. $(X, T)$, $X$ is locally connected if $\forall x \in X,$ $\forall$ nbhd $U_x$ of $x$, $\exists$ connected nbhd $V_x$ of $x$ s.t. $x \in V_x \subseteq U_x$. Def. Given T.S. $(X, T)$, $X$ is locally path-connected if $\forall x \in X,$ $\forall$ nbhd $U_x$ of $x$, $\exists$ path-connected nbhd $V_x$ of $x$ s.t. $x \in V_x \subseteq U_x$. NB6; $\mathbb{R}-\{0\}$ is not connected but locaaly connected. NB7; $\mathbb{Q}$ is not connected and not locally connected.Random math: 362021-05-07T00:00:00+00:002021-05-07T00:00:00+00:00https://jongyeong.github.io/rm/rm36<p>Source of most content: <a href="https://www.youtube.com/channel/UCB_AbuIVIG8I3K3_T_j0FSw/">Gunhee Cho</a>, <a href="http://pi.math.cornell.edu/%7Ematsumura/math4530/math4530web.html">Lecture note by Matsumura</a> and Topology by Munkres.</p> <h1 id="connected-space">Connected Space</h1> <p class="info"><code class="language-plaintext info highlighter-rouge">Def.</code> Given T.S. $(X, T)$, $\emptyset \ne E \subseteq X$ is <strong>disconnected</strong> if there exist disjoint sets $A, B$ such that (1) non-empty (2) $E \subseteq A \cup B$ (3) $\bar{A} \cap B = \emptyset = A \cap \bar{B}$.</p> <p>NB1; $E$ is <strong>connected</strong> if $E$ is not disconnected. <br /> NB2; Let $E$ be a disconnected set in $X$. Let $V= \bar{A}^c, W = \bar{B}^c$ be open sets. Then, (3) implies $B \subseteq V$ and $A \subseteq W$. So, $E \subseteq V \cup W$. Let $V’ = E \cap V$ and $W’ = E \cap W$ be open in $E$ (subspace topology). Then, $E = V’ \mathbin{\unicode{x228D}} W’$ holds since $V’ \cap W’ = E \cap V \cap W = E \cap \bar{A}^c \cap \bar{B}^c$ $= E \cap (\bar{A}\cup \bar{B})^c \subseteq E \cap (X\setminus E)^c = \emptyset$. Hence, $E$ can be separated by two proper subset of $E$ which are open and closed. <br /> NB3; $\mathbb{Q}\subseteq \mathbb{R}$ is totally disconnected, i.e., only one point subset is connected.</p> <h2 id="propositions">Propositions</h2> $\text{Let } X= U \mathbin{\unicode{x228D}} V, \text{ where } U,V \in T_X. \text{ If } A\subseteq X \text{ is connected, then } A \subseteq U \text{ or } A \subseteq V.$ <h3 id="proof">Proof</h3> <p>If $A\subseteq A_1 \cup A_2$ for non-empty sets $A_1 = A \cap U$ and $A_2 = A \cap V$, then $A_1$ and $A_2$ are open in $A$. Therefore, $A = A_1 \mathbin{\unicode{x228D}} A_2$ holds. Since $A_1$ and $A_2$ are non-empty open sets in $A$, $A$ is disconnected.</p> <hr /> $\forall \alpha \in I, A_\alpha \subseteq X \text{ (connected), } \cap_{\alpha \in I} A_\alpha \ne \emptyset. \text{ Then, } \cup_{\alpha \in I} A_\alpha \text{ is connected.} \nonumber$ <h3 id="proof-1">Proof</h3> <p>Suppose $\cup_{\alpha \in I} A_\alpha$ is disconnected, i.e., $\cup_{\alpha \in I} A_\alpha = U \mathbin{\unicode{x228D}} V$. Let $p \in \cap_{\alpha \in I} A_\alpha$ and $p\in U$. Since $A_\alpha$ is connected, it must lie entirely in either $U$ or $V$, i.e., $A_\alpha \subset U$ for all $\alpha \in I$. So that $\cup_{\alpha \in I} A_\alpha \subset U$, i.e., $V$ is empty, which is a contradiction.</p> <hr /> $\text{Let } X \text{ be a connected space and } f: X \to Y \text{ be continous. Then, } f(X) \text{ is connected.} \nonumber$ <h3 id="proof-2">Proof</h3> <p>Let $Z= f(X)$ and consider a continous surjective map $g: X \to Z$. Suppose $Z$ is disconnected, i.e., $Z = A \mathbin{\unicode{x228D}} B$ for non-empty open sets $A,B$ in $Z$. Then, $X = g^{-1}(Z) = g^{-1}(A) \mathbin{\unicode{x228D}} g^{-1}(B)$ holds since $g$ is surjective. Since $g$ is continous, $g^{-1}(A)$ and $g^{-1}(B)$ are open in $X$, which concludes the proof.</p> <hr /> $\text{The union of a collection of connected subspaces of } X \text{ that have a point in common is connected.}$ <h3 id="proof-3">Proof</h3> <p>Let $\{ A_\alpha \}$ be a collection of connected subspaces of $X$ and $p$ be a common point. Suppose $A = \cup A_\alpha = U \mathbin{\unicode{x228D}} V$, i.e., $A$ is diconnected. If $p \in U$, $A_\alpha \subset U$ for all $\alpha$ by (1). Thus, $A \subset U$, which contradicts the assumption. Hence, $A$ is connected.</p> <hr /> $\text{Let } X_1, \ldots, X_n \text{ be a connected space. Then, finite Cartesian product } \prod_{i=1}^K X_i \text{ is connected.} \nonumber$ <h3 id="proof-4">Proof</h3> <p>It suffices to show $X_1 \times X_2$ is connected. Note that $X_1 \cong X_1 \times \{ p \}$ and $X_2 \cong \{ p \} \times X_2$ are connected from above proposition. Fix $(a,b) \in X_1 \times X_2$. For each $x \in X_1$, Let $V_x := (\{ x \} \times X_2) \cup (X_1 \times \{ x \})$. Then, $V_x$ is connected since its subspace shares a common point $(x,b)$ by (2). Since $X_1 \times X_2 = \cup_x V_x$ and $(a,b) \in C_x$ holds, by (2), $X_1 \times X_2$ is connected.</p> <hr /> $\text{If } A \subseteq X \text{ is connected, then } \forall B \subseteq X \text{ s.t. } A \subseteq B \subseteq \bar{A}, \, B\text{ is connected.} \nonumber$ <h3 id="proof-5">Proof</h3> <p>If $B = U \mathbin{\unicode{x228D}} V$ for non-empty open sets $U,V$. By (1), $A \subseteq U$ or $A\subseteq V$ and assume $A\subseteq U$ w.l.o.g. Therefore, $\bar{A} \subseteq \bar{U} = U$ holds. Note that $U \mathbin{\unicode{x228D}} V \subseteq \bar{A}$ holds by assumption. This implies that $U\mathbin{\unicode{x228D}} V \subseteq U$, i.e., $V$ is empty set, which is a contradiction.</p> <hr /> <p>NB3; $(\mathbb{R}^{w}, T_{box})$ is disconnected.<br /> NB4; $(\mathbb{R}^{w}, T_{prod})$ is connected.</p> <h1 id="linear-continum">Linear Continum</h1> <p class="info"><code class="language-plaintext info highlighter-rouge">Def.</code> Let $L$ be simply ordered (totally ordrered). $L$ is called <strong>linear continum</strong> if it satisfies (1) $L$ has a least upper bound property, i.e., $\exists \sup E \in L$ and (2) $L$ has a density, i.e., $\forall x, y \in L$ with $x&lt; y$, $\exists z \in L$ s.t. $x &lt; z &lt; y$.</p> <p>NB5; $L$ is connected. <br /> NB6; intervals, rays are connected.</p> <h2 id="proposition">Proposition</h2> $\text{For linear continum with ordered topology } (L, T_{order}) \text{, let } C \subseteq L \text{ be a convex subset. Then, } C \text{ is connected.}$ <h3 id="proof-6">Proof</h3> <p>Suppose $C$ is disconnected, i.e., $C = U\mathbin{\unicode{x228D}} V$ for non-empty open sets $U,V$. Take $a \in U, \, b \in V$ and assume $a&lt; b$ for the simplicity, i.e., $[a, b] \subset C$ since $C$ is convex. Let $[ a, b] \cap U = U’$ and $[ a, b] \cap V = V’$. Then, $\exists \sup U’ \in L$ and $a \leq \sup U’ \leq b$, i.e., $\sup U’ \in [a, b]$.</p> <ul> <li>Claim: $\sup U’ \not\in [a, b]$. <br /> Note that $[ a, b ] = U’ \cup V’$. Suppose $\sup U’ \in U’$, i.e., $\sup U’ \ne b$. Then, $\exists d \in U’$ s.t. $[ \sup U’ , d ) \subseteq U’$, i.e., $\sup U’ &lt; d \in U’$, which is a contradiction. Hence, $\sup U’ \in V’$, i.e., $\sup U’ \ne a$. Then, $\exists d’$ s.t. $(d’, \sup U’) \subseteq V’$. By definition of linear continum, $\exists z \in V’$ s.t. $z &lt; \sup U’$. Since $\sup U’$ is least upper bound and $z$ is upper bound of $U’$, it is contradiction. Hence, $\sup U’ \not\in U’ \mathbin{\unicode{x228D}} V’ = [a, b]$. Therefore, $C$ is connected.</li> </ul> <h1 id="intermediate-value-thm">Intermediate Value Thm</h1> $\text{Let } X \text{ be connected space, } Y \text{ be simply ordered space and } f:X \to Y \text{ be cts.} \nonumber$ $\text{ Then, } \forall a,b \in X \text{ with } f(a) \ne f(b), \, \forall r \in (f(a), f(b)), \exists c \in (a,b) \text{ s.t. } f(c) =r. \nonumber$ <h3 id="proof-7">Proof</h3> <p>Suppose not, i.e., $\exists r \in (f(a), f(b))$ s.t. $\forall c \in (a,b)$, $f(c)\ne r$. Then, $[ f((a,b)) \cap (-\infty, r) ] \mathbin{\unicode{x228D}} [ f((a,b)) \cap (r, \infty) ] = f((a,b))$. Since $(a,b)$ is connected by (3), image is also connected, which is a contradiction.</p>Jongyeong Leelee@ms.k.u-tokyo.ac.jpSource of most content: Gunhee Cho, Lecture note by Matsumura and Topology by Munkres. Connected Space Def. Given T.S. $(X, T)$, $\emptyset \ne E \subseteq X$ is disconnected if there exist disjoint sets $A, B$ such that (1) non-empty (2) $E \subseteq A \cup B$ (3) $\bar{A} \cap B = \emptyset = A \cap \bar{B}$. NB1; $E$ is connected if $E$ is not disconnected. NB2; Let $E$ be a disconnected set in $X$. Let $V= \bar{A}^c, W = \bar{B}^c$ be open sets. Then, (3) implies $B \subseteq V$ and $A \subseteq W$. So, $E \subseteq V \cup W$. Let $V’ = E \cap V$ and $W’ = E \cap W$ be open in $E$ (subspace topology). Then, $E = V’ \mathbin{\unicode{x228D}} W’$ holds since $V’ \cap W’ = E \cap V \cap W = E \cap \bar{A}^c \cap \bar{B}^c$ $= E \cap (\bar{A}\cup \bar{B})^c \subseteq E \cap (X\setminus E)^c = \emptyset$. Hence, $E$ can be separated by two proper subset of $E$ which are open and closed. NB3; $\mathbb{Q}\subseteq \mathbb{R}$ is totally disconnected, i.e., only one point subset is connected. Propositions $\text{Let } X= U \mathbin{\unicode{x228D}} V, \text{ where } U,V \in T_X. \text{ If } A\subseteq X \text{ is connected, then } A \subseteq U \text{ or } A \subseteq V.$ Proof If $A\subseteq A_1 \cup A_2$ for non-empty sets $A_1 = A \cap U$ and $A_2 = A \cap V$, then $A_1$ and $A_2$ are open in $A$. Therefore, $A = A_1 \mathbin{\unicode{x228D}} A_2$ holds. Since $A_1$ and $A_2$ are non-empty open sets in $A$, $A$ is disconnected. $\forall \alpha \in I, A_\alpha \subseteq X \text{ (connected), } \cap_{\alpha \in I} A_\alpha \ne \emptyset. \text{ Then, } \cup_{\alpha \in I} A_\alpha \text{ is connected.} \nonumber$ Proof Suppose $\cup_{\alpha \in I} A_\alpha$ is disconnected, i.e., $\cup_{\alpha \in I} A_\alpha = U \mathbin{\unicode{x228D}} V$. Let $p \in \cap_{\alpha \in I} A_\alpha$ and $p\in U$. Since $A_\alpha$ is connected, it must lie entirely in either $U$ or $V$, i.e., $A_\alpha \subset U$ for all $\alpha \in I$. So that $\cup_{\alpha \in I} A_\alpha \subset U$, i.e., $V$ is empty, which is a contradiction. $\text{Let } X \text{ be a connected space and } f: X \to Y \text{ be continous. Then, } f(X) \text{ is connected.} \nonumber$ Proof Let $Z= f(X)$ and consider a continous surjective map $g: X \to Z$. Suppose $Z$ is disconnected, i.e., $Z = A \mathbin{\unicode{x228D}} B$ for non-empty open sets $A,B$ in $Z$. Then, $X = g^{-1}(Z) = g^{-1}(A) \mathbin{\unicode{x228D}} g^{-1}(B)$ holds since $g$ is surjective. Since $g$ is continous, $g^{-1}(A)$ and $g^{-1}(B)$ are open in $X$, which concludes the proof. $\text{The union of a collection of connected subspaces of } X \text{ that have a point in common is connected.}$ Proof Let $\{ A_\alpha \}$ be a collection of connected subspaces of $X$ and $p$ be a common point. Suppose $A = \cup A_\alpha = U \mathbin{\unicode{x228D}} V$, i.e., $A$ is diconnected. If $p \in U$, $A_\alpha \subset U$ for all $\alpha$ by (1). Thus, $A \subset U$, which contradicts the assumption. Hence, $A$ is connected. $\text{Let } X_1, \ldots, X_n \text{ be a connected space. Then, finite Cartesian product } \prod_{i=1}^K X_i \text{ is connected.} \nonumber$ Proof It suffices to show $X_1 \times X_2$ is connected. Note that $X_1 \cong X_1 \times \{ p \}$ and $X_2 \cong \{ p \} \times X_2$ are connected from above proposition. Fix $(a,b) \in X_1 \times X_2$. For each $x \in X_1$, Let $V_x := (\{ x \} \times X_2) \cup (X_1 \times \{ x \})$. Then, $V_x$ is connected since its subspace shares a common point $(x,b)$ by (2). Since $X_1 \times X_2 = \cup_x V_x$ and $(a,b) \in C_x$ holds, by (2), $X_1 \times X_2$ is connected. $\text{If } A \subseteq X \text{ is connected, then } \forall B \subseteq X \text{ s.t. } A \subseteq B \subseteq \bar{A}, \, B\text{ is connected.} \nonumber$ Proof If $B = U \mathbin{\unicode{x228D}} V$ for non-empty open sets $U,V$. By (1), $A \subseteq U$ or $A\subseteq V$ and assume $A\subseteq U$ w.l.o.g. Therefore, $\bar{A} \subseteq \bar{U} = U$ holds. Note that $U \mathbin{\unicode{x228D}} V \subseteq \bar{A}$ holds by assumption. This implies that $U\mathbin{\unicode{x228D}} V \subseteq U$, i.e., $V$ is empty set, which is a contradiction. NB3; $(\mathbb{R}^{w}, T_{box})$ is disconnected. NB4; $(\mathbb{R}^{w}, T_{prod})$ is connected. Linear Continum Def. Let $L$ be simply ordered (totally ordrered). $L$ is called linear continum if it satisfies (1) $L$ has a least upper bound property, i.e., $\exists \sup E \in L$ and (2) $L$ has a density, i.e., $\forall x, y \in L$ with $x&lt; y$, $\exists z \in L$ s.t. $x &lt; z &lt; y$. NB5; $L$ is connected. NB6; intervals, rays are connected. Proposition $\text{For linear continum with ordered topology } (L, T_{order}) \text{, let } C \subseteq L \text{ be a convex subset. Then, } C \text{ is connected.}$ Proof Suppose $C$ is disconnected, i.e., $C = U\mathbin{\unicode{x228D}} V$ for non-empty open sets $U,V$. Take $a \in U, \, b \in V$ and assume $a&lt; b$ for the simplicity, i.e., $[a, b] \subset C$ since $C$ is convex. Let $[ a, b] \cap U = U’$ and $[ a, b] \cap V = V’$. Then, $\exists \sup U’ \in L$ and $a \leq \sup U’ \leq b$, i.e., $\sup U’ \in [a, b]$. Claim: $\sup U’ \not\in [a, b]$. Note that $[ a, b ] = U’ \cup V’$. Suppose $\sup U’ \in U’$, i.e., $\sup U’ \ne b$. Then, $\exists d \in U’$ s.t. $[ \sup U’ , d ) \subseteq U’$, i.e., $\sup U’ &lt; d \in U’$, which is a contradiction. Hence, $\sup U’ \in V’$, i.e., $\sup U’ \ne a$. Then, $\exists d’$ s.t. $(d’, \sup U’) \subseteq V’$. By definition of linear continum, $\exists z \in V’$ s.t. $z &lt; \sup U’$. Since $\sup U’$ is least upper bound and $z$ is upper bound of $U’$, it is contradiction. Hence, $\sup U’ \not\in U’ \mathbin{\unicode{x228D}} V’ = [a, b]$. Therefore, $C$ is connected. Intermediate Value Thm $\text{Let } X \text{ be connected space, } Y \text{ be simply ordered space and } f:X \to Y \text{ be cts.} \nonumber$ $\text{ Then, } \forall a,b \in X \text{ with } f(a) \ne f(b), \, \forall r \in (f(a), f(b)), \exists c \in (a,b) \text{ s.t. } f(c) =r. \nonumber$ Proof Suppose not, i.e., $\exists r \in (f(a), f(b))$ s.t. $\forall c \in (a,b)$, $f(c)\ne r$. Then, $[ f((a,b)) \cap (-\infty, r) ] \mathbin{\unicode{x228D}} [ f((a,b)) \cap (r, \infty) ] = f((a,b))$. Since $(a,b)$ is connected by (3), image is also connected, which is a contradiction.Random math: 352021-04-03T00:00:00+00:002021-04-03T00:00:00+00:00https://jongyeong.github.io/rm/rm35<p>Source of most content: <a href="https://www.youtube.com/channel/UCB_AbuIVIG8I3K3_T_j0FSw/">Gunhee Cho</a>, <a href="http://pi.math.cornell.edu/%7Ematsumura/math4530/math4530web.html">Lecture note by Matsumura</a> and Topology by Munkres.</p> <h1 id="limit-point-sequentially-compact">Limit Point, Sequentially Compact</h1> <p class="info"><code class="language-plaintext info highlighter-rouge">Def.</code> Given T.S. $(X, T)$, $X$ is <strong>limit point compact</strong> if for every infinite subset $A \subseteq X$, $A$ contains a limit point.</p> <p class="info"><code class="language-plaintext info highlighter-rouge">Def.</code> Given T.S. $(X, T)$, $X$ is <strong>sequentially compact</strong> if for every seq. $(x _n) _{n=1}^\infty$ in $X$, there exists a convergence subsequence $(x _{n _k}) _{k=1}^\infty$.</p> <h2 id="proposition">Proposition</h2> $\text{Let } (X,T) \text{ be a metrizable space. TFAE}\nonumber$ <blockquote> <ol> <li>$X$ is compact.</li> <li>$X$ is limit point compact</li> <li>$X$ is sequentially compact.</li> </ol> </blockquote> <p>i.e., $\exists d : X\times X \to [0, \infty)$ s.t. the topology induced by the metric $d$ is $T$. <br /> NB1; For arbitrary T.S. $(X,T)$, 1 implies 2.</p> <h3 id="proof">Proof</h3> <ul> <li>$(1\to 2)$: Let $X$ be compact. Suppose $A\subset X$ does not contain any limit point of $A$, i.e., $\bar{A} = A$ (check <a href="/rm/rm23#proof-3">Prop-1</a>). Therefore, $A$ is closed, i.e., $X\setminus A$ is open. On the other hand, by Asm, for each point $a \in A$, $\exists$ nbhd $V_a$ of $a$ s.t. $V_a \cap (A \setminus \{ a\}) = \emptyset$. Take an open cover $\{ V_a, X\setminus A : a \in A\}$ of $X$. Then, since $X$ is compact, there exists finite subcover of $X$. Since $A \cap X-A = \emptyset$, A is covered by finitely many $V_a$’s. Note that $A \cap V_a = \{a \}$ and $A \subset \cup_{finite \, a}V_a$. Therefore, $A$ is a finite set since $A = A \cap (\cup V_a) = \cup ( A \cap V_a) = \cup\{a\}$. This shows that if $A$ does not contain any limit point of $A$, $A$ is a finite set. Its contrapositive is that if $A$ is an infinite subset, then it contains a limit point of $A$, i.e., $A$ is limit point compact.</li> <li>$(2\to 3)$: Let $X$ be limit point compact and take any seq. $(x_n)$ in $X$. If $(x_n)$ is a finite set, which implies $(x_n)$ is all but finitely many constant, then we can choose these constants as subseq. which conludes the proof. Otherwise, since $(x_n)$ is an infinite set, there exists a limit point $a \in X$. This implies that $\forall k \in \mathbb{N}$, $B_{\frac{1}{k}}(a) \cap (x_n)-\{a \} \ne \emptyset$. Then, we can choose a subseq. $(x_{n_k})$ s.t. $(d_{x_{n_k}}, a) &lt; \frac{1}{k}$. Hence, $x_{n_k} \to a$, i.e., $X$ is sequentially compact.</li> <li>$(3 \to 1)$: Let $X$ be sequentially compact and $X$ is also a metric space. <ol> <li>Show that <a href="/rm/rm34#lebesgue-number-lemma">Lebesgue number lemma</a> holds when $X$ is sequentially compact.<br /> Let $\mathcal{A}$ be an open cover of $X$. For each $n \in \mathbb{N}$, let non-empty $C_n$ be subch a set which has a diameter less than $\frac{1}{n}$ and $C_n$ cannot be contained in one open set in $\mathcal{A}$, i.e., assume $C_n$ is a set that does not satisfy Lebesgue number lemma. Take any point $x_n \in C_n$ for all $n\in\mathbb{N}$. So, for a seq. $(x_n)\subseteq X$, there exists a convergence seq. $(x_{n_k})$. Let $x_{n_k} \to a \in X \subset \cup_{A \in \mathcal{A}} A$. Thus, $\exists \epsilon &gt;0$ s.t. $B_\epsilon(a) \subset A$ for some $A \in \mathcal{A}$. By <a href="/rm/rm2#archimedean-property">A.P.</a>, $\exists N \in \mathbb{N}$ s.t. $\frac{1}{N} &lt; \frac{\epsilon}{2}$. Since $x_{n_k} \to a$ for this $\epsilon &gt; 0$, $\exists N \in \mathbb{N}$ s.t. $\forall n_k \geq N$, $x_{n_k} \in B_{\frac{\epsilon}{2}}(a)$. Then, $\forall x \in C_{n_k}, \, \forall n_k \geq N$, $d(x,a) \leq d(x, x_{n_k}) + d(x_{n_k}, a) &lt; \frac{1}{n_k} + \frac{\epsilon}{2} &lt; \epsilon$. Therefore, $C_{n_k} \subset B_{\frac{\epsilon}{2}}(a) \subset A$, which is a contradiction. Hence, Lebesgue number lemma holds for sequentially compact $X$.</li> <li>Show that $\forall \epsilon &gt;0$, $X$ can be covered by finite $\epsilon$-balls. <br /> Suppose not. Then, $\exists \epsilon &gt;0$ s.t. $X$ cannot be covered by finite $\epsilon$-balls. Take $x \in B_{\frac{\epsilon}{2}}(x_1) \subset X$. By Asm, $\exists x_2 \in X \setminus B_\epsilon(x_1)$. In general, by Asm, we can find $x_{n+1} \in X \setminus (\cup_{i=1}^n B_\epsilon(x_i))$. In this way, we can construct a seq. $(x_n)$ s.t. for fixed $n\in \mathbb{N}, \, d(x_n, x_k) \geq \epsilon$, for $1\leq k \leq n-1$. This result implies that this sequence does not have a convergent subsequence, which is a contradiction. Hence, $X$ can be covered by finite $\epsilon$-balls.</li> <li>Let $\mathcal{A}$ be an open cover of $X$. Suppose $X$ is sequentially compact. Then, by step 1, there exists Lebesuge number $\delta&gt;0$ s.t. each open set of diameter less than $\delta$ is contained in one of open sets in $\mathcal{A}$. Take $\epsilon = \delta/3$. Then, by step 2, $X$ can be covered by finite $\epsilon$-balls. Then, each open ball has diameter $\frac{2\delta}{3} &lt; \delta$. Thus, $\epsilon$-ball is contained in $A \in \mathcal{A}$. Hence, we can take a finite subcover of $\mathcal{A}$, i.e., $X$ is compact.</li> </ol> </li> </ul> <h1 id="locally-compact">Locally Compact</h1> <p class="info"><code class="language-plaintext info highlighter-rouge">Def.</code> Given T.S. $(X, T)$, $X$ is <strong>locally compact at $x \in X$</strong> if there exists a compact subspace $C$ in $X$ which conatins some neighborhood $V_x$ of $x$.</p> <p class="info"><code class="language-plaintext info highlighter-rouge">Def.</code> Given T.S. $(X, T)$, $X$ is <strong>locally compact</strong> if $X$ is locally compact at any point.</p> <h2 id="proposition-1">Proposition</h2> $X \text{ is locally compact Hausdorff iff there exists a T.S. } Y \text{ satisfying } \nonumber$ <ol> <li>$X \subset Y$.</li> <li>$Y - X$ consists of a single point.</li> <li>$Y$ is a compact Hausdorff.</li> </ol> $\text{Moreover, if } Y, Y' \text{ satisfy 1-3, there exists a homeomorphism } f: Y \to Y' \text{ s.t. } f\mid_{X}: X \to X \text{, i.e., identity map.} \nonumber$ <h3 id="proof-1">Proof</h3> <ul> <li>(Uniqueness): If $Y, Y’$ satisfy 1-3, write $Y= X \mathbin{\unicode{x228D}} \{ p \}$ and $Y’= X \mathbin{\unicode{x228D}} \{ q \}$. Define $f: Y \to Y’, x \mapsto x$ if $x\in X$ and $p \mapsto q$. Then, $f$ is bijective function. To show $f^{-1}$ is cts, it suffices to show $f$ is an open map, i.e., any image of open set is open. Take any open $V$ in $Y$. <ol> <li>If $p \not\in V$, then $V = V\cap X$ is open in $X$. $f(V)=V=V\cap X$ is open in $X$. Since $X$ is open in $Y’$, $f(V)$ is open in $Y’$.</li> <li>If $p \in V$, consider $p \not\in C=Y-V$ that is closed in $Y$. Since $Y$ is compact, $C$ is also compact in $Y$ (Check <a href="/rm/rm26#properties">Property 1</a>). $f(C)=C$ is a compact subspace of $Y’$ since $X$ is a subspace of $Y’$. Since $Y’$ is Hausdorff and $C$ is closed in $Y’$, $f(V)=Y’-C$ is open in $Y’$.</li> </ol> </li> <li>$(\rightarrow)$ Let $X$ be LC Hausdorff. We construct such a $Y$ as follows: pick any point with symbol $\infty \not\in X$ and $Y := X \mathbin{\unicode{x228D}} \{ \infty\}$. Then, define $T_Y := \{ U, Y-C : U \in T_X, C \text{ is compact in } X \}$. <ol> <li>$T_Y$ is a topology. <br /> (a): $\emptyset \in T_Y$ since $\emptyset \in T_X$. $Y \in T_Y$ since $\emptyset$ is compact in $X$. <br /> (b-1): $\cup_{a} U_a \in T_Y$ is obvious as $U_a \in T_X$. <br /> (b-2): $\cup_{a} Y - C_a = Y - \cap_a C_a$. Since $C_a$ is a compact subspace in Hausdorff, it is closed in $X$. Therfore, $\cap_a C_a$ is closed subspace in compact $C_i$, i.e., compact. Hence, $\cup_{a} Y - C_a \in T_Y$. <br /> (b-3): $U \cup (Y-C) = Y - (C-U)$. Since $C-U$ is closed in compact $C$, it is compact. Therefore, $U \cup (Y-C)\in T_Y$. <br /> (c-1): $U_a \cap U_b \in T_Y$. <br /> (c-2): $(Y-C_a)\cap (Y-C_b) = Y - (C_a \cup C_b)$. Since $C_a \cup C_b$ is compact, $(Y-C_a)\cap (Y-C_b) \in T_Y$. <br /> (c-3): $U_a \cap (Y-C_b) = U_a \cap C_b^c$. Since $C_b^c$ is open in $X$, $U_a \cap (Y-C_b) \in T_X$.</li> <li>$X$ is subspace of $Y$, i.e., $T_X = \{ V \cap X : V \in T_Y\}$. <br /> (a): If $V \in T_X$, $V\cap X = V \in T_X$. <br /> (b): Otherwise, $V = Y-C$. Then, $V\cap X = (Y-C)\cap X = X - C \in T_X$. <br /> Hence, $\{ V \cap X : V \in T_Y\} \subseteq T_X$. Since $T_X \subseteq \{ V \cap X : V \in T_Y\}$ is obvious, it conludes the proof.</li> <li>$Y$ is Hausdorff. <br /> Take any $p\ne q$ in $Y$. If $p , q \ne \infty$, since $X$ is Hausdorff, there exist disjoint nbhd of $p,q$. If $q = \emptyset$, we can choose a compact set $C$ in $X$ containing a nbhd $V_p$ of $p$. Then, $V_p$ and $Y-C$ are disjoint neighborhoods of $p$ and $\infty$ in $Y$.</li> <li>$Y$ is compact. <br /> Take any open cover of $Y$. Then, there exists a compact $C$ s.t. $\infty \in Y - C$. Since $C$ is compact in $X$, it will be covered by finitely many open cover of $X$. Since $Y-C$ is open in $Y$, it concludes the proof.</li> </ol> </li> <li>$(\leftarrow)$ Suppose $Y$ satisfying 1-3. Then, $X$ is Hausdorff since it is a subspace of the Hausdorff $Y$. Given $x \in X$, choose disjoint open sets $U$ and $V$ of $Y$ containing $x$ and the single point of $Y-X$, respectively. Then, the set $Y-V$ is closed in $Y$, i.e., it is compact in $Y$. Since $Y-V \subset X$, it is a compact subsapce of $X$ and contains nbhd $U$ of $x$.</li> </ul> <hr /> <p class="info"><code class="language-plaintext info highlighter-rouge">Def.</code> If $Y$ is a compact Hausdorff space and $X$ is a proper subspace of $Y$ whose closure equals $Y$, then $Y$ is said to be a <strong>compactification</strong> of $X$. If $Y-X$ equals to a single point, then $Y$ is called the <strong>one-point compactification</strong> of $X$.</p> <p>NB2; If we take any open nbhd $V$ of $\infty$, then there exists a compact set $C$ s.t. $V = Y - C$ and $C \ne X$ (since $X$ is not compact), i.e., $C \subsetneq X$. Thus, $\exists p \ne \infty, p \in X - C \subset Y - C = V$. Hence, $p \in V \cap Y - \{\infty \}$, i.e., $V \cap Y - \{ \infty\} \ne \emptyset$. This implies that $\infty$ is a limit point of $Y$, i.e., $Y= \bar{X}$.</p> <h2 id="propositions">Propositions</h2> $\text{Let } X \text{ be Hausdorff. Then, } X \text{ is locally compact iff } \forall x \in X, \forall \text{ nbhd } V_x \text{ of } x \nonumber$ $\text{ there is a nbhd } U_x \text{ of } x \text{ s.t. } \bar{U_x} \text{ is compact and } \bar{U_x} \subseteq V_x. \nonumber$ <h3 id="proof-2">Proof</h3> <ul> <li>$(\leftarrow)$: It is obvious by definition since $\bar{U_x}$ is compact by Asm.</li> <li>$(\rightarrow)$: Take any $x \in X$ and any nbhd $V_x$ of $x$. Let $Y$ be the one-point compactification of $X$. Since $V_x$ is open in $Y$, $Y - V_x$ is closed in compact $Y$, $Y-V_x$ is also compact. Note that $x \not\in Y - V_x$, i.e., $x\not\in C$ for a compact subspace $C=Y-V_x$ of Hausdorff $X$. Then, there exists nbhds $A, B$ of $x$ and $C$ s.t. $A \cap B = \emptyset$ from the definition of Hausdorff space. Since $Y$ is also Hausdorff, there is a nbhd $U_x$ of $x$ and $W \supseteq Y-V_x$ s.t. $U_x \cap W =\emptyset$, i.e., $U_x \subset W^c \subseteq (Y - V_x)^c = V_x$. Then, $\bar{U_x} \cap W = \emptyset$. Since $U_x \cap W = \emptyset$, let $p\in U_x’ \cap W$. Since $W$ is nbhd of $p$, $W\cap (U_x -\{p\})\stackrel{\text{limit point}}{\ne} \emptyset$. However, as $W\cap (U_x-\{p\}) \subset W \cap U_x = \emptyset$ holds, it is contradiction. Hence, $\bar{U_x} \cap W = \emptyset$. Therefore, $x \in U_x \subseteq \bar{U_x} \subset W^c \subseteq (Y-V_x)^c = V_x$, i.e., $x \in U_x \subseteq \bar{U_x} \subseteq V_x$. Finally, $\bar{U_x}$ is closed in compact $Y$, i.e., $\bar{U_x}$ is compact in $Y$.</li> </ul> <hr /> $\text{Let } X \text{ be LC Hausdorff. If } A \subseteq X \text{ is open or closed, } A \text{ is LC Hausdorff.}\nonumber$ <h3 id="proof-3">Proof</h3> <ul> <li>open $A$: Let $x \in A \subseteq X$. Then, there exists a compact set $C_x$ s.t. $x\in V_x \subseteq C_x$ for some nbhd $V_x$ by def. of LC. Thus, $x\in V_x \cap A \subseteq C_x \cap A$ holds. Since $C_x$ is compact set in Hausdorff $X$, it is closed. Thus, $C_x \cap A$ is also closed. As closed subset in Hasudorff space is compact, it concludes the proof.</li> <li>closed $A$: For any $x \in A \subseteq X$, take any nbhd $V_x$ of $x$. Since $X$ is LCH, there exists $U_x$ s.t. $\bar{U_x}$ is compact and $x \in U_x \subseteq \bar{U_x} \subseteq V_x := V \cap A$. Thus, $x \in U_x \cap A \subseteq \bar{U_x} \cap A \subseteq V_x$ since $\bar{U_x} \cap A$ is compact in $A$.</li> </ul> <hr /> $X \text{ is LCH iff } X \text{ is open subset of a compact Hausdorff space.} \nonumber$ <h3 id="proof-4">Proof</h3> <p>This is a corollary of above two propositions.</p>Jongyeong Leelee@ms.k.u-tokyo.ac.jpSource of most content: Gunhee Cho, Lecture note by Matsumura and Topology by Munkres. Limit Point, Sequentially Compact Def. Given T.S. $(X, T)$, $X$ is limit point compact if for every infinite subset $A \subseteq X$, $A$ contains a limit point. Def. Given T.S. $(X, T)$, $X$ is sequentially compact if for every seq. $(x _n) _{n=1}^\infty$ in $X$, there exists a convergence subsequence $(x _{n _k}) _{k=1}^\infty$. Proposition $\text{Let } (X,T) \text{ be a metrizable space. TFAE}\nonumber$ $X$ is compact. $X$ is limit point compact $X$ is sequentially compact. i.e., $\exists d : X\times X \to [0, \infty)$ s.t. the topology induced by the metric $d$ is $T$. NB1; For arbitrary T.S. $(X,T)$, 1 implies 2. Proof $(1\to 2)$: Let $X$ be compact. Suppose $A\subset X$ does not contain any limit point of $A$, i.e., $\bar{A} = A$ (check Prop-1). Therefore, $A$ is closed, i.e., $X\setminus A$ is open. On the other hand, by Asm, for each point $a \in A$, $\exists$ nbhd $V_a$ of $a$ s.t. $V_a \cap (A \setminus \{ a\}) = \emptyset$. Take an open cover $\{ V_a, X\setminus A : a \in A\}$ of $X$. Then, since $X$ is compact, there exists finite subcover of $X$. Since $A \cap X-A = \emptyset$, A is covered by finitely many $V_a$’s. Note that $A \cap V_a = \{a \}$ and $A \subset \cup_{finite \, a}V_a$. Therefore, $A$ is a finite set since $A = A \cap (\cup V_a) = \cup ( A \cap V_a) = \cup\{a\}$. This shows that if $A$ does not contain any limit point of $A$, $A$ is a finite set. Its contrapositive is that if $A$ is an infinite subset, then it contains a limit point of $A$, i.e., $A$ is limit point compact. $(2\to 3)$: Let $X$ be limit point compact and take any seq. $(x_n)$ in $X$. If $(x_n)$ is a finite set, which implies $(x_n)$ is all but finitely many constant, then we can choose these constants as subseq. which conludes the proof. Otherwise, since $(x_n)$ is an infinite set, there exists a limit point $a \in X$. This implies that $\forall k \in \mathbb{N}$, $B_{\frac{1}{k}}(a) \cap (x_n)-\{a \} \ne \emptyset$. Then, we can choose a subseq. $(x_{n_k})$ s.t. $(d_{x_{n_k}}, a) &lt; \frac{1}{k}$. Hence, $x_{n_k} \to a$, i.e., $X$ is sequentially compact. $(3 \to 1)$: Let $X$ be sequentially compact and $X$ is also a metric space. Show that Lebesgue number lemma holds when $X$ is sequentially compact. Let $\mathcal{A}$ be an open cover of $X$. For each $n \in \mathbb{N}$, let non-empty $C_n$ be subch a set which has a diameter less than $\frac{1}{n}$ and $C_n$ cannot be contained in one open set in $\mathcal{A}$, i.e., assume $C_n$ is a set that does not satisfy Lebesgue number lemma. Take any point $x_n \in C_n$ for all $n\in\mathbb{N}$. So, for a seq. $(x_n)\subseteq X$, there exists a convergence seq. $(x_{n_k})$. Let $x_{n_k} \to a \in X \subset \cup_{A \in \mathcal{A}} A$. Thus, $\exists \epsilon &gt;0$ s.t. $B_\epsilon(a) \subset A$ for some $A \in \mathcal{A}$. By A.P., $\exists N \in \mathbb{N}$ s.t. $\frac{1}{N} &lt; \frac{\epsilon}{2}$. Since $x_{n_k} \to a$ for this $\epsilon &gt; 0$, $\exists N \in \mathbb{N}$ s.t. $\forall n_k \geq N$, $x_{n_k} \in B_{\frac{\epsilon}{2}}(a)$. Then, $\forall x \in C_{n_k}, \, \forall n_k \geq N$, $d(x,a) \leq d(x, x_{n_k}) + d(x_{n_k}, a) &lt; \frac{1}{n_k} + \frac{\epsilon}{2} &lt; \epsilon$. Therefore, $C_{n_k} \subset B_{\frac{\epsilon}{2}}(a) \subset A$, which is a contradiction. Hence, Lebesgue number lemma holds for sequentially compact $X$. Show that $\forall \epsilon &gt;0$, $X$ can be covered by finite $\epsilon$-balls. Suppose not. Then, $\exists \epsilon &gt;0$ s.t. $X$ cannot be covered by finite $\epsilon$-balls. Take $x \in B_{\frac{\epsilon}{2}}(x_1) \subset X$. By Asm, $\exists x_2 \in X \setminus B_\epsilon(x_1)$. In general, by Asm, we can find $x_{n+1} \in X \setminus (\cup_{i=1}^n B_\epsilon(x_i))$. In this way, we can construct a seq. $(x_n)$ s.t. for fixed $n\in \mathbb{N}, \, d(x_n, x_k) \geq \epsilon$, for $1\leq k \leq n-1$. This result implies that this sequence does not have a convergent subsequence, which is a contradiction. Hence, $X$ can be covered by finite $\epsilon$-balls. Let $\mathcal{A}$ be an open cover of $X$. Suppose $X$ is sequentially compact. Then, by step 1, there exists Lebesuge number $\delta&gt;0$ s.t. each open set of diameter less than $\delta$ is contained in one of open sets in $\mathcal{A}$. Take $\epsilon = \delta/3$. Then, by step 2, $X$ can be covered by finite $\epsilon$-balls. Then, each open ball has diameter $\frac{2\delta}{3} &lt; \delta$. Thus, $\epsilon$-ball is contained in $A \in \mathcal{A}$. Hence, we can take a finite subcover of $\mathcal{A}$, i.e., $X$ is compact. Locally Compact Def. Given T.S. $(X, T)$, $X$ is locally compact at $x \in X$ if there exists a compact subspace $C$ in $X$ which conatins some neighborhood $V_x$ of $x$. Def. Given T.S. $(X, T)$, $X$ is locally compact if $X$ is locally compact at any point. Proposition $X \text{ is locally compact Hausdorff iff there exists a T.S. } Y \text{ satisfying } \nonumber$ $X \subset Y$. $Y - X$ consists of a single point. $Y$ is a compact Hausdorff. $\text{Moreover, if } Y, Y' \text{ satisfy 1-3, there exists a homeomorphism } f: Y \to Y' \text{ s.t. } f\mid_{X}: X \to X \text{, i.e., identity map.} \nonumber$ Proof (Uniqueness): If $Y, Y’$ satisfy 1-3, write $Y= X \mathbin{\unicode{x228D}} \{ p \}$ and $Y’= X \mathbin{\unicode{x228D}} \{ q \}$. Define $f: Y \to Y’, x \mapsto x$ if $x\in X$ and $p \mapsto q$. Then, $f$ is bijective function. To show $f^{-1}$ is cts, it suffices to show $f$ is an open map, i.e., any image of open set is open. Take any open $V$ in $Y$. If $p \not\in V$, then $V = V\cap X$ is open in $X$. $f(V)=V=V\cap X$ is open in $X$. Since $X$ is open in $Y’$, $f(V)$ is open in $Y’$. If $p \in V$, consider $p \not\in C=Y-V$ that is closed in $Y$. Since $Y$ is compact, $C$ is also compact in $Y$ (Check Property 1). $f(C)=C$ is a compact subspace of $Y’$ since $X$ is a subspace of $Y’$. Since $Y’$ is Hausdorff and $C$ is closed in $Y’$, $f(V)=Y’-C$ is open in $Y’$. $(\rightarrow)$ Let $X$ be LC Hausdorff. We construct such a $Y$ as follows: pick any point with symbol $\infty \not\in X$ and $Y := X \mathbin{\unicode{x228D}} \{ \infty\}$. Then, define $T_Y := \{ U, Y-C : U \in T_X, C \text{ is compact in } X \}$. $T_Y$ is a topology. (a): $\emptyset \in T_Y$ since $\emptyset \in T_X$. $Y \in T_Y$ since $\emptyset$ is compact in $X$. (b-1): $\cup_{a} U_a \in T_Y$ is obvious as $U_a \in T_X$. (b-2): $\cup_{a} Y - C_a = Y - \cap_a C_a$. Since $C_a$ is a compact subspace in Hausdorff, it is closed in $X$. Therfore, $\cap_a C_a$ is closed subspace in compact $C_i$, i.e., compact. Hence, $\cup_{a} Y - C_a \in T_Y$. (b-3): $U \cup (Y-C) = Y - (C-U)$. Since $C-U$ is closed in compact $C$, it is compact. Therefore, $U \cup (Y-C)\in T_Y$. (c-1): $U_a \cap U_b \in T_Y$. (c-2): $(Y-C_a)\cap (Y-C_b) = Y - (C_a \cup C_b)$. Since $C_a \cup C_b$ is compact, $(Y-C_a)\cap (Y-C_b) \in T_Y$. (c-3): $U_a \cap (Y-C_b) = U_a \cap C_b^c$. Since $C_b^c$ is open in $X$, $U_a \cap (Y-C_b) \in T_X$. $X$ is subspace of $Y$, i.e., $T_X = \{ V \cap X : V \in T_Y\}$. (a): If $V \in T_X$, $V\cap X = V \in T_X$. (b): Otherwise, $V = Y-C$. Then, $V\cap X = (Y-C)\cap X = X - C \in T_X$. Hence, $\{ V \cap X : V \in T_Y\} \subseteq T_X$. Since $T_X \subseteq \{ V \cap X : V \in T_Y\}$ is obvious, it conludes the proof. $Y$ is Hausdorff. Take any $p\ne q$ in $Y$. If $p , q \ne \infty$, since $X$ is Hausdorff, there exist disjoint nbhd of $p,q$. If $q = \emptyset$, we can choose a compact set $C$ in $X$ containing a nbhd $V_p$ of $p$. Then, $V_p$ and $Y-C$ are disjoint neighborhoods of $p$ and $\infty$ in $Y$. $Y$ is compact. Take any open cover of $Y$. Then, there exists a compact $C$ s.t. $\infty \in Y - C$. Since $C$ is compact in $X$, it will be covered by finitely many open cover of $X$. Since $Y-C$ is open in $Y$, it concludes the proof. $(\leftarrow)$ Suppose $Y$ satisfying 1-3. Then, $X$ is Hausdorff since it is a subspace of the Hausdorff $Y$. Given $x \in X$, choose disjoint open sets $U$ and $V$ of $Y$ containing $x$ and the single point of $Y-X$, respectively. Then, the set $Y-V$ is closed in $Y$, i.e., it is compact in $Y$. Since $Y-V \subset X$, it is a compact subsapce of $X$ and contains nbhd $U$ of $x$. Def. If $Y$ is a compact Hausdorff space and $X$ is a proper subspace of $Y$ whose closure equals $Y$, then $Y$ is said to be a compactification of $X$. If $Y-X$ equals to a single point, then $Y$ is called the one-point compactification of $X$. NB2; If we take any open nbhd $V$ of $\infty$, then there exists a compact set $C$ s.t. $V = Y - C$ and $C \ne X$ (since $X$ is not compact), i.e., $C \subsetneq X$. Thus, $\exists p \ne \infty, p \in X - C \subset Y - C = V$. Hence, $p \in V \cap Y - \{\infty \}$, i.e., $V \cap Y - \{ \infty\} \ne \emptyset$. This implies that $\infty$ is a limit point of $Y$, i.e., $Y= \bar{X}$. Propositions $\text{Let } X \text{ be Hausdorff. Then, } X \text{ is locally compact iff } \forall x \in X, \forall \text{ nbhd } V_x \text{ of } x \nonumber$ $\text{ there is a nbhd } U_x \text{ of } x \text{ s.t. } \bar{U_x} \text{ is compact and } \bar{U_x} \subseteq V_x. \nonumber$ Proof $(\leftarrow)$: It is obvious by definition since $\bar{U_x}$ is compact by Asm. $(\rightarrow)$: Take any $x \in X$ and any nbhd $V_x$ of $x$. Let $Y$ be the one-point compactification of $X$. Since $V_x$ is open in $Y$, $Y - V_x$ is closed in compact $Y$, $Y-V_x$ is also compact. Note that $x \not\in Y - V_x$, i.e., $x\not\in C$ for a compact subspace $C=Y-V_x$ of Hausdorff $X$. Then, there exists nbhds $A, B$ of $x$ and $C$ s.t. $A \cap B = \emptyset$ from the definition of Hausdorff space. Since $Y$ is also Hausdorff, there is a nbhd $U_x$ of $x$ and $W \supseteq Y-V_x$ s.t. $U_x \cap W =\emptyset$, i.e., $U_x \subset W^c \subseteq (Y - V_x)^c = V_x$. Then, $\bar{U_x} \cap W = \emptyset$. Since $U_x \cap W = \emptyset$, let $p\in U_x’ \cap W$. Since $W$ is nbhd of $p$, $W\cap (U_x -\{p\})\stackrel{\text{limit point}}{\ne} \emptyset$. However, as $W\cap (U_x-\{p\}) \subset W \cap U_x = \emptyset$ holds, it is contradiction. Hence, $\bar{U_x} \cap W = \emptyset$. Therefore, $x \in U_x \subseteq \bar{U_x} \subset W^c \subseteq (Y-V_x)^c = V_x$, i.e., $x \in U_x \subseteq \bar{U_x} \subseteq V_x$. Finally, $\bar{U_x}$ is closed in compact $Y$, i.e., $\bar{U_x}$ is compact in $Y$. $\text{Let } X \text{ be LC Hausdorff. If } A \subseteq X \text{ is open or closed, } A \text{ is LC Hausdorff.}\nonumber$ Proof open $A$: Let $x \in A \subseteq X$. Then, there exists a compact set $C_x$ s.t. $x\in V_x \subseteq C_x$ for some nbhd $V_x$ by def. of LC. Thus, $x\in V_x \cap A \subseteq C_x \cap A$ holds. Since $C_x$ is compact set in Hausdorff $X$, it is closed. Thus, $C_x \cap A$ is also closed. As closed subset in Hasudorff space is compact, it concludes the proof. closed $A$: For any $x \in A \subseteq X$, take any nbhd $V_x$ of $x$. Since $X$ is LCH, there exists $U_x$ s.t. $\bar{U_x}$ is compact and $x \in U_x \subseteq \bar{U_x} \subseteq V_x := V \cap A$. Thus, $x \in U_x \cap A \subseteq \bar{U_x} \cap A \subseteq V_x$ since $\bar{U_x} \cap A$ is compact in $A$. $X \text{ is LCH iff } X \text{ is open subset of a compact Hausdorff space.} \nonumber$ Proof This is a corollary of above two propositions.Random math: 342021-03-20T00:00:00+00:002021-03-20T00:00:00+00:00https://jongyeong.github.io/rm/rm34<p>Source of most content: <a href="https://www.youtube.com/channel/UCB_AbuIVIG8I3K3_T_j0FSw/">Gunhee Cho</a> and Topology by Munkres.</p> <h1 id="heine-borel-theorem">Heine-Borel Theorem</h1> $\text{Let } X \text{ be a simply ordered set having the least upper bound property.} \nonumber$ $\text{Then, any closed interval } [a, b] = \{x\in X : a \leq x \leq b \} \text{ is compact}. \nonumber$ <h3 id="proof">Proof</h3> <ol> <li>Take any $a,b\in X$ with $a &lt; b$. Let $\mathcal{A}$ be an open cover of $[a, b]$. (*) If $x \in [a, b)$, then $\exists y \in (x ,b]$ s.t. $[x, y]$ can be covered by at most two elements of $\mathcal{A}$. <ul> <li>(*): Let $x \in [a,b)$. If $\exists y &gt; x$ s.t. $(x, y) = \emptyset$, then $[x, y]$ consists of two points. Thus, it can be covered by at most two elements of $\mathcal{A}$.</li> <li>(*): If not, choose an element $A$ of $\mathcal{A}$ contatining $x$. Since $x \ne b$, $\exists z \in (x, b]$ s.t. $(x, z) \subset A$. Since $x &lt; z, \exists y$ s.t. $x &lt; y &lt; z$. Therefore, $[x, y] \subset A$.</li> </ul> </li> <li>Define $C := \{ y \in (a, b] : [a, y] \text{ can be covered by finite subcover of } \mathcal{A} \}$. Since $a\ne b$, by 1, $C \ne \emptyset$. Moreover, as $y \leq b$, $C$ is bounded aboved. By LUP, $\exists \sup C \in X$. THus, $a &lt; \sup C \leq b$.</li> <li>Since $\sup C \in (a, b], \, \exists A \in \mathcal{A}$ s.t. $\sup C \in A$. Let $z$ be a left end point of $A$. Since $z &lt; \sup C, \, \exists y \in C$ s.t. $z &lt; y \leq \sup C$. Thus, $[a, y]$ can be covered by finite subcover of $\mathcal{A}$ by def. of $C$. Also, $(y , \sup C] \subset A$. Thus, $[a, \sup C] = [a, y] \cup (y , \sup C]$, i.e., $[a, \sup C]$ can be covered by finite subcover of $\mathcal{A}$. Therefore, $\sup C \in C$.</li> <li>If $\sup C \ne b$, i.e., $\sup C &lt; b$, then since $\sup C \in C$, we can take $A\in \mathcal{A}$ containing $\sup C$. Let $z$ be a right end point of $A$. Since $z \in (\sup C, b]$, $[\sup C, z]$ can be covered by finite subcover of $\mathcal{A}$. This means that $z \in C$, contradicting the definition of $\sup C$. Hence, $\sup C = b$.</li> </ol> <hr /> <p>Cor1; Any closed interval of $\mathbb{R}$ is compact. <br /> Cor2; In $\mathbb{R}^n, \, [a_1, b_1] \times \cdots [a_n, b_n]$ is compact.</p> <h2 id="proposition">Proposition</h2> $A \subseteq \mathbb{R}^n \text{ is compact iff } A \text{ is closed an bounded in the euclidean metric } d \text{ or the square metric }\rho. \nonumber$ <h3 id="proof-1">Proof</h3> <ul> <li>$d : \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R} \, (x,y) \mapsto d(x,y) := \sqrt{\sum_{i=1}^n (x_i - y_i)^2}$.</li> <li>$\rho: \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R} \, (x,y) \mapsto \rho(x,y) := \max_{i}{\mid x_i - y_i\mid}$</li> <li>It suffices to consider only $\rho$ since $\rho(x,y) \leq d (x,y) \leq \sqrt{n} \rho(x,y)$ holds, i.e., if $A$ is bounded under $d$ iff it is bounded under $\rho$.</li> <li>$(\Rightarrow)$ Let $A$ be compact. Since $\mathbb{R}^n$ is Hausdorff, $A$ is closed by <a href="/rm/rm26#properties">Prop2</a>. Define $C := \{ B _\rho (O; M) : M \in \mathbb{R} _{+}\}$, the collection of open balls centered at origin with radius $M$. Then, $C$ is an open cover of $A$ since there exists $M$ s.t. $A \subset B _\rho (O; M)$. Since $A$ is compact, $\exists$ finite subcover of $C$. Take $N = \max \{ M _1, \ldots, M _k \}$, then $A \subset B _\rho (0; N)$. Thus, $\rho(x, O) \leq N$ for all $x\in X$, i.e., $A$ is bounded under $\rho$.</li> <li>$(\Leftarrow)$ Suppose $A$ is closed and bounded under $\rho$. Since $A$ is bounded, $\exists N \in \mathbb{R}$ s.t. $\forall x^1, x^2 \in A$, $\rho(x^1, x^2) \leq N$. Fix $a \in A$ and take $\rho(a, O) = M \in \mathbb{R}$. Then, $\forall y \in A, \, \rho(y, O) \leq \rho(y, a) + \rho(a, O) \leq N + M$ holds by triangle inequality. Take $P = M + N &gt; 0$. Then, $A \subset [-P , P]^n$. Since closed interval is compact by <a href="/rm/rm34#heine-borel-theorem">Heine-Borel thm</a> and <a href="/rm/rm33#product-of-compact-spaces">product of compact spaces is compact</a>, $[-P, P]^n$ is compact. The fact that $A$ is closed concludes the proof since <a href="/rm/rm26#properties">closed subspace of compact set is compact</a></li> </ul> <h1 id="extreme-value-theorem">Extreme Value Theorem</h1> $\text{Let }(X,T_X) \text{ be C.S. and } Y \text{ be simply ordered set with } T_{order}. \text{ Let } f: X \to Y \text{ be cts. Then, } \exists \max f , \min f. \nonumber$ <h3 id="proof-2">Proof</h3> <p>Since $X$ is compact and $f$ is continuous, $\text{Im}f =: A$ is compact. Suppose $A$ has no largest element. Define $C := \{ (-\infty, a) : a \in A \}$. Since $A$ is compact, $\exists$ finite subcover of $A$ as $\{ (-\infty, a_1), \ldots, (-\infty, a_n) \}$. Take $M= \max \{ a_1, \ldots, a_n \} = a_k = f(x_k)$. Note that $A \subset (-\infty, M ) = (-\infty, f(x_k))$ holds. This implies $f(x_k) \in A$ and $f(x_k) \in (-\infty , f(x_k))$, which is a contradiction. Similarly, we can show that $\exists \min f$.</p> <h1 id="uniformly-continuous">Uniformly Continuous</h1> <p class="info"><code class="language-plaintext info highlighter-rouge">Def.</code> Given metric spaces $(X, d_X), (Y, d_Y)$, we call $f:X \to Y$ <strong>uniformly continuous</strong> if $\forall \epsilon &gt;0, \exists \delta &gt;0$ s.t. if $d_X (x_1, x_2) &lt; \delta$ for any $x_1, x_2 \in X$, then $d_Y (f(x_1), f(x_2)) &lt; \epsilon$.</p> <p>NB2; $f: \mathbb{R} \to \mathbb{R}$ is not uniformly continuous iff $\exists \epsilon &gt;0$ s.t. $\forall n \in \mathbb{N}, \, \mid x_n - y_n \mid &lt; \frac{1}{n} \implies \mid f(x_n) - f(y_n) \mid \geq \epsilon$ $\Leftrightarrow$ $\exists \epsilon &gt;0$ and a seq. $(x_n), (y_n)$ s.t. $\mid x_n - y_n \mid &lt; \frac{1}{n} \implies \mid f(x_n) - f(y_n) \mid \geq \epsilon$.</p> <p class="info"><code class="language-plaintext info highlighter-rouge">Def.</code> Given metric space $(X, d)$, define diameter $dia A := \sup \{ d(a_1, a_2) : a_1, a_2 \in A \}$. $A$ is <strong>bounded</strong> if $dia A &lt; \infty$.</p> <p class="info"><code class="language-plaintext info highlighter-rouge">Def.</code> Given metric space $(X, d)$, let $A \subset X$. Define <strong>distance from $x$ to $A$</strong> $d_A : X \to \mathbb{R}, \, x \mapsto \inf \{ d(x,a) : a \in A\}$, i.e., $d_A(x) = \inf_{a\in A} d (x,a)$.</p> <p>NB3; $d_A : X \to \mathbb{R}$ is uniformly continuous.</p> <h3 id="proof-nb3">Proof (NB3)</h3> <p>Fix $x \in X$. Then, $d_A (x) = \inf d(x,a) \stackrel{\forall a \in A}{\leq} d(x,a) \stackrel{\forall a \in A, \forall y \in X}{\leq} d(x,y) + d(y,a)$. Thus, $d_A (x) - d(x,y) \leq d(y,a)$ holds for all $a\in A, y\in X$. Fix $y\in X$. Then, $d_A(x) - d(x,y) \leq \inf d(y,a) = d_A (y)$, i.e., $d_A(x) - d_A(y) \leq d(x,y)$. Similarly, $d_A(y) - d_A(x) \leq d(y,x) = d(x,y)$. Therefore, $\mid d_A(x) - d_A(y) \mid \leq d (x,y)$, which concludes the proof.</p> <h2 id="lebesgue-number-lemma">Lebesgue number lemma</h2> $\text{Given metric space } (X, d) \text{ with compact } X. \text{ Let } \mathcal{A} \text{ be an open cover of } X. \nonumber$ $\text{ Then, } \exists \delta &gt; 0 \text{ s.t. } \forall A \subset X \text{ s.t. } dia A &lt; \delta, \exists A_i \in \mathcal{A} \text{ satisfying } A \subset A_i. \nonumber$ <h3 id="proof-3">Proof</h3> <p>Since $X$ is compact, $\exists$ finite subcover of $\mathcal{A}$ called $\{ A _1, \ldots, A _n \}$, i.e., $X \subset \cup _{k=1}^n A _k$. Let $C _k := X \setminus A _k$ for all $1\leq k \leq n$. Define $f: X \to \mathbb{R}, \, x \mapsto f(x) := \sum _{k=1}^n d _{C _k}(x)$.</p> <p>Then, $\forall x \in X$, $\exists$ ball $B$ s.t. $x \in B \subset A _j$ for some $j$ since ball is a basis of topology in metric space. Thus, $f(x) \geq \frac{1}{n}d _{C _j}(x) \geq \frac{1}{n}d _{B^c}(x) \geq \frac{1}{n}radius(B) &gt; 0$ holds. Note that $f$ is continuous and $X$ is compact, we can apply extreme value theorem. Then, $\exists x _0 \in X$ s.t. $0 &lt; f(x _0)= \min \text{Im} f$. Take $\delta = f(x _0) &gt; 0$. For any $A \subseteq X$ with $dia A &lt; \delta$, take $x \in A$. By definition, we have $\delta \leq f(x) = \frac{1}{n}\sum _{k=1}^n d _{C _k}(x) \leq d _{C _j}(x)$ where we take $j$ for maximum. Thus, $\delta \leq d _{C _j}(x)$, i.e., $B _\delta (x) \subset A _j$ holds. Since $dia A &lt; \delta, \, A \subset B _\delta (x)$. Hence, $A \subset B _\delta (x) \subset A _j$.</p> <h2 id="uniformly-continuous-theorem">Uniformly continuous theorem</h2> $\text{Let } (X,d_X), (Y, d_Y) \text{ be metric spaces for compact } X \text{ and } f:X \to Y \text{ be cts. Then, } f \text{ is uniformly continuous.} \nonumber$ <h3 id="proof-4">Proof</h3> <p>Given $\epsilon &gt;0$, consider $C:= \{ B(y; \frac{\epsilon}{2}) : y \in Y \}$ that is an open cover of $Y$. Since $f$ is cts, $\{ f^{-1}(B(y; \frac{\epsilon}{2})) : y \in Y\}$ is open cover of $X$. By Lebesque number lemma, $\exists \epsilon$ s.t. $\forall A \subset X$ with $dia A &lt; \delta$, $\exists A_i \in \{ f^{-1}(B(y; \frac{\epsilon}{2})) : y \in Y\}$ that contains $A$. Let $A = \{ x_1, x_2 \} \subset X$. Then, $dia A &lt; \delta$ holds by definition. So $\exists y \in Y$ s.t. $A \subset f^{-1}(B(y; \frac{\epsilon}{2}))$ by Lemma. This implies that $f(A) = \{ f(x_1), f(x_2) \} \subset B(y; \frac{\epsilon}{2})$. Therefore, $d_Y(y, f(x_1)) &lt; \frac{\epsilon}{2}$ and $d_Y(y, f(x_2)) &lt; \frac{\epsilon}{2}$ holds. Applying triangle inequality concludes the proof.</p>Jongyeong Leelee@ms.k.u-tokyo.ac.jpSource of most content: Gunhee Cho and Topology by Munkres. Heine-Borel Theorem $\text{Let } X \text{ be a simply ordered set having the least upper bound property.} \nonumber$ $\text{Then, any closed interval } [a, b] = \{x\in X : a \leq x \leq b \} \text{ is compact}. \nonumber$ Proof Take any $a,b\in X$ with $a &lt; b$. Let $\mathcal{A}$ be an open cover of $[a, b]$. (*) If $x \in [a, b)$, then $\exists y \in (x ,b]$ s.t. $[x, y]$ can be covered by at most two elements of $\mathcal{A}$. (*): Let $x \in [a,b)$. If $\exists y &gt; x$ s.t. $(x, y) = \emptyset$, then $[x, y]$ consists of two points. Thus, it can be covered by at most two elements of $\mathcal{A}$. (*): If not, choose an element $A$ of $\mathcal{A}$ contatining $x$. Since $x \ne b$, $\exists z \in (x, b]$ s.t. $(x, z) \subset A$. Since $x &lt; z, \exists y$ s.t. $x &lt; y &lt; z$. Therefore, $[x, y] \subset A$. Define $C := \{ y \in (a, b] : [a, y] \text{ can be covered by finite subcover of } \mathcal{A} \}$. Since $a\ne b$, by 1, $C \ne \emptyset$. Moreover, as $y \leq b$, $C$ is bounded aboved. By LUP, $\exists \sup C \in X$. THus, $a &lt; \sup C \leq b$. Since $\sup C \in (a, b], \, \exists A \in \mathcal{A}$ s.t. $\sup C \in A$. Let $z$ be a left end point of $A$. Since $z &lt; \sup C, \, \exists y \in C$ s.t. $z &lt; y \leq \sup C$. Thus, $[a, y]$ can be covered by finite subcover of $\mathcal{A}$ by def. of $C$. Also, $(y , \sup C] \subset A$. Thus, $[a, \sup C] = [a, y] \cup (y , \sup C]$, i.e., $[a, \sup C]$ can be covered by finite subcover of $\mathcal{A}$. Therefore, $\sup C \in C$. If $\sup C \ne b$, i.e., $\sup C &lt; b$, then since $\sup C \in C$, we can take $A\in \mathcal{A}$ containing $\sup C$. Let $z$ be a right end point of $A$. Since $z \in (\sup C, b]$, $[\sup C, z]$ can be covered by finite subcover of $\mathcal{A}$. This means that $z \in C$, contradicting the definition of $\sup C$. Hence, $\sup C = b$. Cor1; Any closed interval of $\mathbb{R}$ is compact. Cor2; In $\mathbb{R}^n, \, [a_1, b_1] \times \cdots [a_n, b_n]$ is compact. Proposition $A \subseteq \mathbb{R}^n \text{ is compact iff } A \text{ is closed an bounded in the euclidean metric } d \text{ or the square metric }\rho. \nonumber$ Proof $d : \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R} \, (x,y) \mapsto d(x,y) := \sqrt{\sum_{i=1}^n (x_i - y_i)^2}$. $\rho: \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R} \, (x,y) \mapsto \rho(x,y) := \max_{i}{\mid x_i - y_i\mid}$ It suffices to consider only $\rho$ since $\rho(x,y) \leq d (x,y) \leq \sqrt{n} \rho(x,y)$ holds, i.e., if $A$ is bounded under $d$ iff it is bounded under $\rho$. $(\Rightarrow)$ Let $A$ be compact. Since $\mathbb{R}^n$ is Hausdorff, $A$ is closed by Prop2. Define $C := \{ B _\rho (O; M) : M \in \mathbb{R} _{+}\}$, the collection of open balls centered at origin with radius $M$. Then, $C$ is an open cover of $A$ since there exists $M$ s.t. $A \subset B _\rho (O; M)$. Since $A$ is compact, $\exists$ finite subcover of $C$. Take $N = \max \{ M _1, \ldots, M _k \}$, then $A \subset B _\rho (0; N)$. Thus, $\rho(x, O) \leq N$ for all $x\in X$, i.e., $A$ is bounded under $\rho$. $(\Leftarrow)$ Suppose $A$ is closed and bounded under $\rho$. Since $A$ is bounded, $\exists N \in \mathbb{R}$ s.t. $\forall x^1, x^2 \in A$, $\rho(x^1, x^2) \leq N$. Fix $a \in A$ and take $\rho(a, O) = M \in \mathbb{R}$. Then, $\forall y \in A, \, \rho(y, O) \leq \rho(y, a) + \rho(a, O) \leq N + M$ holds by triangle inequality. Take $P = M + N &gt; 0$. Then, $A \subset [-P , P]^n$. Since closed interval is compact by Heine-Borel thm and product of compact spaces is compact, $[-P, P]^n$ is compact. The fact that $A$ is closed concludes the proof since closed subspace of compact set is compact Extreme Value Theorem $\text{Let }(X,T_X) \text{ be C.S. and } Y \text{ be simply ordered set with } T_{order}. \text{ Let } f: X \to Y \text{ be cts. Then, } \exists \max f , \min f. \nonumber$ Proof Since $X$ is compact and $f$ is continuous, $\text{Im}f =: A$ is compact. Suppose $A$ has no largest element. Define $C := \{ (-\infty, a) : a \in A \}$. Since $A$ is compact, $\exists$ finite subcover of $A$ as $\{ (-\infty, a_1), \ldots, (-\infty, a_n) \}$. Take $M= \max \{ a_1, \ldots, a_n \} = a_k = f(x_k)$. Note that $A \subset (-\infty, M ) = (-\infty, f(x_k))$ holds. This implies $f(x_k) \in A$ and $f(x_k) \in (-\infty , f(x_k))$, which is a contradiction. Similarly, we can show that $\exists \min f$. Uniformly Continuous Def. Given metric spaces $(X, d_X), (Y, d_Y)$, we call $f:X \to Y$ uniformly continuous if $\forall \epsilon &gt;0, \exists \delta &gt;0$ s.t. if $d_X (x_1, x_2) &lt; \delta$ for any $x_1, x_2 \in X$, then $d_Y (f(x_1), f(x_2)) &lt; \epsilon$. NB2; $f: \mathbb{R} \to \mathbb{R}$ is not uniformly continuous iff $\exists \epsilon &gt;0$ s.t. $\forall n \in \mathbb{N}, \, \mid x_n - y_n \mid &lt; \frac{1}{n} \implies \mid f(x_n) - f(y_n) \mid \geq \epsilon$ $\Leftrightarrow$ $\exists \epsilon &gt;0$ and a seq. $(x_n), (y_n)$ s.t. $\mid x_n - y_n \mid &lt; \frac{1}{n} \implies \mid f(x_n) - f(y_n) \mid \geq \epsilon$. Def. Given metric space $(X, d)$, define diameter $dia A := \sup \{ d(a_1, a_2) : a_1, a_2 \in A \}$. $A$ is bounded if $dia A &lt; \infty$. Def. Given metric space $(X, d)$, let $A \subset X$. Define distance from $x$ to $A$ $d_A : X \to \mathbb{R}, \, x \mapsto \inf \{ d(x,a) : a \in A\}$, i.e., $d_A(x) = \inf_{a\in A} d (x,a)$. NB3; $d_A : X \to \mathbb{R}$ is uniformly continuous. Proof (NB3) Fix $x \in X$. Then, $d_A (x) = \inf d(x,a) \stackrel{\forall a \in A}{\leq} d(x,a) \stackrel{\forall a \in A, \forall y \in X}{\leq} d(x,y) + d(y,a)$. Thus, $d_A (x) - d(x,y) \leq d(y,a)$ holds for all $a\in A, y\in X$. Fix $y\in X$. Then, $d_A(x) - d(x,y) \leq \inf d(y,a) = d_A (y)$, i.e., $d_A(x) - d_A(y) \leq d(x,y)$. Similarly, $d_A(y) - d_A(x) \leq d(y,x) = d(x,y)$. Therefore, $\mid d_A(x) - d_A(y) \mid \leq d (x,y)$, which concludes the proof. Lebesgue number lemma $\text{Given metric space } (X, d) \text{ with compact } X. \text{ Let } \mathcal{A} \text{ be an open cover of } X. \nonumber$ $\text{ Then, } \exists \delta &gt; 0 \text{ s.t. } \forall A \subset X \text{ s.t. } dia A &lt; \delta, \exists A_i \in \mathcal{A} \text{ satisfying } A \subset A_i. \nonumber$ Proof Since $X$ is compact, $\exists$ finite subcover of $\mathcal{A}$ called $\{ A _1, \ldots, A _n \}$, i.e., $X \subset \cup _{k=1}^n A _k$. Let $C _k := X \setminus A _k$ for all $1\leq k \leq n$. Define $f: X \to \mathbb{R}, \, x \mapsto f(x) := \sum _{k=1}^n d _{C _k}(x)$. Then, $\forall x \in X$, $\exists$ ball $B$ s.t. $x \in B \subset A _j$ for some $j$ since ball is a basis of topology in metric space. Thus, $f(x) \geq \frac{1}{n}d _{C _j}(x) \geq \frac{1}{n}d _{B^c}(x) \geq \frac{1}{n}radius(B) &gt; 0$ holds. Note that $f$ is continuous and $X$ is compact, we can apply extreme value theorem. Then, $\exists x _0 \in X$ s.t. $0 &lt; f(x _0)= \min \text{Im} f$. Take $\delta = f(x _0) &gt; 0$. For any $A \subseteq X$ with $dia A &lt; \delta$, take $x \in A$. By definition, we have $\delta \leq f(x) = \frac{1}{n}\sum _{k=1}^n d _{C _k}(x) \leq d _{C _j}(x)$ where we take $j$ for maximum. Thus, $\delta \leq d _{C _j}(x)$, i.e., $B _\delta (x) \subset A _j$ holds. Since $dia A &lt; \delta, \, A \subset B _\delta (x)$. Hence, $A \subset B _\delta (x) \subset A _j$. Uniformly continuous theorem $\text{Let } (X,d_X), (Y, d_Y) \text{ be metric spaces for compact } X \text{ and } f:X \to Y \text{ be cts. Then, } f \text{ is uniformly continuous.} \nonumber$ Proof Given $\epsilon &gt;0$, consider $C:= \{ B(y; \frac{\epsilon}{2}) : y \in Y \}$ that is an open cover of $Y$. Since $f$ is cts, $\{ f^{-1}(B(y; \frac{\epsilon}{2})) : y \in Y\}$ is open cover of $X$. By Lebesque number lemma, $\exists \epsilon$ s.t. $\forall A \subset X$ with $dia A &lt; \delta$, $\exists A_i \in \{ f^{-1}(B(y; \frac{\epsilon}{2})) : y \in Y\}$ that contains $A$. Let $A = \{ x_1, x_2 \} \subset X$. Then, $dia A &lt; \delta$ holds by definition. So $\exists y \in Y$ s.t. $A \subset f^{-1}(B(y; \frac{\epsilon}{2}))$ by Lemma. This implies that $f(A) = \{ f(x_1), f(x_2) \} \subset B(y; \frac{\epsilon}{2})$. Therefore, $d_Y(y, f(x_1)) &lt; \frac{\epsilon}{2}$ and $d_Y(y, f(x_2)) &lt; \frac{\epsilon}{2}$ holds. Applying triangle inequality concludes the proof.Random math: 332021-03-19T00:00:00+00:002021-03-19T00:00:00+00:00https://jongyeong.github.io/rm/rm33<p>Source of most content: <a href="https://www.youtube.com/channel/UCB_AbuIVIG8I3K3_T_j0FSw/">Gunhee Cho</a>.</p> <h1 id="product-of-compact-spaces">Product of Compact Spaces</h1> $\text{Let } X_i \text{ be compact space for } i=1,\ldots, n. \text{ Then, } \prod_{i=1}^n X_i \text{ is compact.} \nonumber$ <h2 id="tube-lemma">Tube Lemma</h2> $\text{Suppose } X, Y \text{ be compact. For any } x_0 \in X \text{ and nbhd } U \text{ of } \{x_0 \} \times Y \, \exists \text{ nbhd } W \text{ of } x_0 \in X \text{ s.t. } W \times Y \subset U. \nonumber$ <h3 id="proof-tube">Proof (Tube)</h3> <p>Let $x_0$ be fixed. For each $(x_0, y) \in X \times Y$, choose $U_y \times V_y$, a nbhd of $(x_0, y)$ where $U_y \in T_X$ and $V_y \in T_Y$. Then, $\{U_x \times V_y \subset U: y \in Y \}$ is an open cover of $\{x_0\} \times Y$. If we consider the projection map $p: X\times Y \to Y$ which is onto and continous, $\{ x_0 \} \times Y$ is homeomorphic to $Y$ since $p^{-1}$ is also continous for the fixed $x_0$. Therefore, $\{x_0\}\times Y$ is compact since $Y$ is compact by assumption. Then, we can choose $y_1, \ldots, y_n \in Y$ s.t. $\{x_0 \} \times Y \subset \cup_{k=1}^n (U_{y_k} \times V_{y_k})$. Set $W := \cap_{k=1}^n U_{y_k}$ that contains $x_0$ and is open. Then, $\{ x_0 \} \times Y \subset W \times Y \subset \cup_{k=1}^n (U_{y_k} \times V_{y_k}) \subset U$.</p> <h3 id="proof">Proof</h3> <p>It suffices to show that for compact sets $X,Y$, $X\times Y$ is compact. Let $\{ A_{\alpha^{x}} : \alpha \in J \}$ be an open cover of $\{x \} \times Y$ for the fixed $x \in X$. Since $\{ x \} \times Y$ is compact, there exists finite open subcover s.t. $\{x \}\times Y \subset \cup_{k=1}^n A_{\alpha_k}$. By Tube lemma, there is $W_x$ s.t. $W_x \times Y \subset \cup_{k=1}^n A_{\alpha_k^x}$. Then, $\{ W_{x} : x \in X\}$ will be an open cover of $X$. Since $X$ is compact, $\exists x_1, \ldots, x_m \in X$ s.t. $X \subset \cup_{k=1}^m W_{x_k}$. Therefore, $X\times Y \subset (\cup_{k=1}^m W_{x_k}) \times Y = \subset \cup_{k=1}^m (W_{x_k} \times Y) \stackrel{Tube}{\subset} \cup_{k=1}^m (\cup_{l=1}^n A_{\alpha_{l}^{x_k}})$. Hence, $X\times Y$ is covered by at most finite open sets.</p> <hr /> <p>NB1; In fact, the infinite product of compact spaces is still compact.(<a href="https://en.wikipedia.org/wiki/Tychonoff%27s_theorem">Tychonoff’s theorem</a>). <br /> NB2; Thus $[ 0, 1]^2$ is compact and $[0, 1]^{[0, 1 ]}$ is compact.</p> <h1 id="finite-intersection-property">Finite Intersection Property</h1> <p class="info"><code class="language-plaintext info highlighter-rouge">Def.</code> Let $X$ be a set and $C \subset 2^X$. Then, $C$ has the <strong>finite intersection property (FIP)</strong> if for any $c_1, \ldots, c_n \in C$, $\cap_{k=1}^n C_n \ne \emptyset$.</p> <h2 id="proposition">Proposition</h2> $X \text{ is compact iff any } C \subset 2^X \text{ which has 1. FIP 2. for any } c \in C, c \text{ is closed in }X \text{ satisfies } \cup_{c \in C} c \ne \emptyset. \nonumber$ <h3 id="proof-1">Proof</h3> <ol> <li>$(\Rightarrow)$ If not, then $\exists C$ s.t. $C$ satisfies 1 and 2 but $\cup_{c \in C} c = \emptyset$, which implies $\cup_{c \in C} c^c = X$. Since $X$ is compact and $c^c$ is open, $\bigcup_{finite \, c} c^c = X$ holds. Then, $\bigcap_{finite \, c} c = \emptyset$ holds, i.e., it contradicts to FIP, which concludes the proof.</li> <li>$(\Leftarrow)$ Suppose not. Then, $\exists$ open cover $\{ O _\alpha \} _{\alpha \in J}$ of $X$ s.t. $X = \cup _{\alpha \in J} O _\alpha$ but $\cup _{finite \, \alpha} O _\alpha \subsetneq X$ for any choice of finite indices. Then, $\cap _{finite \, \alpha} O _\alpha^c \ne \emptyset$. Since $O _\alpha^c$ is closed, $\{ O _\alpha^c : \alpha \in J\}$ has FIP. By Asm, $\cup _{\alpha \in J} O _\alpha^c \ne \emptyset$, i.e., $\cup _{\alpha \in J} O _\alpha \ne X$ which is a contradiction.</li> </ol>Jongyeong Leelee@ms.k.u-tokyo.ac.jpSource of most content: Gunhee Cho. Product of Compact Spaces $\text{Let } X_i \text{ be compact space for } i=1,\ldots, n. \text{ Then, } \prod_{i=1}^n X_i \text{ is compact.} \nonumber$ Tube Lemma $\text{Suppose } X, Y \text{ be compact. For any } x_0 \in X \text{ and nbhd } U \text{ of } \{x_0 \} \times Y \, \exists \text{ nbhd } W \text{ of } x_0 \in X \text{ s.t. } W \times Y \subset U. \nonumber$ Proof (Tube) Let $x_0$ be fixed. For each $(x_0, y) \in X \times Y$, choose $U_y \times V_y$, a nbhd of $(x_0, y)$ where $U_y \in T_X$ and $V_y \in T_Y$. Then, $\{U_x \times V_y \subset U: y \in Y \}$ is an open cover of $\{x_0\} \times Y$. If we consider the projection map $p: X\times Y \to Y$ which is onto and continous, $\{ x_0 \} \times Y$ is homeomorphic to $Y$ since $p^{-1}$ is also continous for the fixed $x_0$. Therefore, $\{x_0\}\times Y$ is compact since $Y$ is compact by assumption. Then, we can choose $y_1, \ldots, y_n \in Y$ s.t. $\{x_0 \} \times Y \subset \cup_{k=1}^n (U_{y_k} \times V_{y_k})$. Set $W := \cap_{k=1}^n U_{y_k}$ that contains $x_0$ and is open. Then, $\{ x_0 \} \times Y \subset W \times Y \subset \cup_{k=1}^n (U_{y_k} \times V_{y_k}) \subset U$. Proof It suffices to show that for compact sets $X,Y$, $X\times Y$ is compact. Let $\{ A_{\alpha^{x}} : \alpha \in J \}$ be an open cover of $\{x \} \times Y$ for the fixed $x \in X$. Since $\{ x \} \times Y$ is compact, there exists finite open subcover s.t. $\{x \}\times Y \subset \cup_{k=1}^n A_{\alpha_k}$. By Tube lemma, there is $W_x$ s.t. $W_x \times Y \subset \cup_{k=1}^n A_{\alpha_k^x}$. Then, $\{ W_{x} : x \in X\}$ will be an open cover of $X$. Since $X$ is compact, $\exists x_1, \ldots, x_m \in X$ s.t. $X \subset \cup_{k=1}^m W_{x_k}$. Therefore, $X\times Y \subset (\cup_{k=1}^m W_{x_k}) \times Y = \subset \cup_{k=1}^m (W_{x_k} \times Y) \stackrel{Tube}{\subset} \cup_{k=1}^m (\cup_{l=1}^n A_{\alpha_{l}^{x_k}})$. Hence, $X\times Y$ is covered by at most finite open sets. NB1; In fact, the infinite product of compact spaces is still compact.(Tychonoff’s theorem). NB2; Thus $[ 0, 1]^2$ is compact and $[0, 1]^{[0, 1 ]}$ is compact. Finite Intersection Property Def. Let $X$ be a set and $C \subset 2^X$. Then, $C$ has the finite intersection property (FIP) if for any $c_1, \ldots, c_n \in C$, $\cap_{k=1}^n C_n \ne \emptyset$. Proposition $X \text{ is compact iff any } C \subset 2^X \text{ which has 1. FIP 2. for any } c \in C, c \text{ is closed in }X \text{ satisfies } \cup_{c \in C} c \ne \emptyset. \nonumber$ Proof $(\Rightarrow)$ If not, then $\exists C$ s.t. $C$ satisfies 1 and 2 but $\cup_{c \in C} c = \emptyset$, which implies $\cup_{c \in C} c^c = X$. Since $X$ is compact and $c^c$ is open, $\bigcup_{finite \, c} c^c = X$ holds. Then, $\bigcap_{finite \, c} c = \emptyset$ holds, i.e., it contradicts to FIP, which concludes the proof. $(\Leftarrow)$ Suppose not. Then, $\exists$ open cover $\{ O _\alpha \} _{\alpha \in J}$ of $X$ s.t. $X = \cup _{\alpha \in J} O _\alpha$ but $\cup _{finite \, \alpha} O _\alpha \subsetneq X$ for any choice of finite indices. Then, $\cap _{finite \, \alpha} O _\alpha^c \ne \emptyset$. Since $O _\alpha^c$ is closed, $\{ O _\alpha^c : \alpha \in J\}$ has FIP. By Asm, $\cup _{\alpha \in J} O _\alpha^c \ne \emptyset$, i.e., $\cup _{\alpha \in J} O _\alpha \ne X$ which is a contradiction.Random math: 322021-03-13T00:00:00+00:002021-03-13T00:00:00+00:00https://jongyeong.github.io/rm/rm32<p>Source of most content: <a href="https://www.youtube.com/channel/UCB_AbuIVIG8I3K3_T_j0FSw/">Gunhee Cho</a> and <a href="https://59clc.files.wordpress.com/2011/01/real-and-complex-analysis.pdf">RCA by Rudin</a>.</p> <h1 id="convexity">Convexity</h1> <p class="info"><code class="language-plaintext info highlighter-rouge">Def.</code> For any open set $(a,b) \in [-\infty, \infty]$, $\varphi: (a,b) \to \mathbb{R}$ is <strong>convex</strong> if $\forall x,y \in (a,b)$ with $x &lt; y$ and $\forall \lambda \in [0, 1]$, $\varphi(\lambda x + (1-\lambda) y) \leq \lambda \varphi(x) + (1- \lambda) \varphi(y)$ holds.</p> <p>NB1; If we put $t=\lambda x + (1-\lambda) y$, it holds that $\varphi(t) \leq \lambda \varphi(x) + (1-\lambda)\varphi(y)$. Since $\lambda (x-y) = t-y$ holds, we have $\frac{\varphi(x)-\varphi(y)}{x-y} \leq \frac{\varphi(t)-\varphi(y)}{t-y}$. <br /> NB2; More generally, for $a &lt; s &lt; t &lt; u &lt; b$, it holds that $\frac{\varphi(t)-\varphi(s)}{t-s} \leq \frac{\varphi(u)-\varphi(t)}{u-t}$. <br /> NB3; Thus, a real differentiable function $\varphi$ is convex in $(a,b)$ iff $a&lt; x &lt; y &lt; b$ implies $\varphi’(x) \leq \varphi’(y)$, i.e., the derivative $\varphi’$ is a monotonically increasing function.<br /> NB4; If $\varphi$ is convex on $(a,b)$, then $\varphi$ is continous on $(a,b)$. <br /> NB5; If $f, g$ are convex on $(a,b)$, then $\max (f, g)$ is convex on $(a,b)$.</p> <h2 id="jensens-inequality">Jensen’s Inequality</h2> $\text{Let }\mu \text{ be a measure on } (X, F) \text{ s.t. } \mu(X) =1. \text{ If a real fn } f \in L^1(\mu) \text{ s.t. } \forall x \in X, \, f(x) \in (a,b), \text{ and if } \varphi \text{ is cvx, } \nonumber$ $\text{then } \varphi \left( \int_X f d\mu \right) \leq \int_X \varphi \circ f d\mu. \nonumber$ <h3 id="proof">Proof</h3> <p>Put $t= \int_X f d\mu \in (a,b)$. Let $\beta := \sup \{ \frac{\varphi(s)-\varphi(t)}{s-t} : a &lt; s &lt; t \} \in \mathbb{R}$. By <a href="/rm/rm2#least-upper-bound-property">LUP</a>, it holds that $\forall s \in (a,b),$ $\varphi(s) - \varphi(t) \geq \beta (s-t)$. Since $\forall x \in X, f(x) \in (a,b)$ holds by assumption, $\varphi(f(x)) - \varphi(t) \geq \beta (f(x) -t )$ holds. Since $\varphi$ is contnous by NB4, if we integrate w.r.t. $\mu$, we have $\int_X \varphi \circ f d\mu - \int_X \varphi(t) d \mu \geq \beta \int_X (f(x) -t) d\mu = \beta \int_X f d\mu - \beta\int_X t d \mu \stackrel{t=\int_X f d\mu}{=} 0$. Hence, we have $\int_X \varphi \circ f d\mu \geq \int_X \varphi(t) d \mu = \varphi(t) \int_X 1 d\mu \stackrel{\mu(X) =1}{=} \varphi(t) = \varphi \left( \int_X f d\mu \right)$, which concludes the proof.</p> <hr /> <p>NB6; Let $\varphi(x) = e^x$ (cvx). By Jensen’s Ineq, $\exp(\int_X f) \leq \int_X \exp f$ holds. Here, if we set $f(x) = \log g(x)$, we have $\exp (\int_X \log g) \leq \int_X g$. <br /> NB7; Let $\mu (\{ \frac{1}{n}\}) = \frac{1}{n}$ and $f(x_i) = p_i\in \mathbb{R}$ for finite set $X= \{ x_1, \ldots, x_n \}$. Then, $\int_X f = \frac{1}{n}(p_1 + \cdots + p_n)$ holds. By applying Jensen’s Ineq, we have $\exp ( \frac{1}{n}(p_1 + \cdots + p_n) ) \leq \frac{1}{n}\left[ e^{p_1} + \cdots + e^{p_n} \right]$. Letting $e^{p_i} = y_i$ gives the Arithmetic-geometric inequality. <br /> NB8; If $\mu(x_i) = \alpha_i$ s.t. $\sum \alpha_i = 1$, we have $\Pi_i y_i^{\alpha_i} \leq \sum \alpha_i y_i$.</p> <h1 id="hölder-and-minkowskis-inequalities">Hölder and Minkowski’s Inequalities</h1> <p class="info"><code class="language-plaintext info highlighter-rouge">Def.</code> Let $p , q \in \mathbb{R}_{+}$ s.t. $p+q = pq$, i.e., $\frac{1}{p} + \frac{1}{q} =1$. We say $(p,q)$ <strong>conjugate exponents</strong>.</p> <p>NB9; $(1, \infty)$ and $(\infty , 1)$ are also considered as conjugate exponents.</p> <hr /> $\text{Given M.S. } (X, F, \mu) \text{, conjugate exponents } (p,q) \text{, measurable functions } f,g \in [0, \infty]. \text{ Then, }\nonumber$ $\text{ Hölder's Ineq: } \int_X fg d\mu \leq \left( \int_X f^p d\mu \right)^{1/p} \left( \int_X g^q d\mu \right)^{1/q} \nonumber$ $\text{Minkowski's Ineq: } \left(\int_X (f+g)^p d\mu \right)^{1/p} \leq \left( \int_X f^p d\mu \right)^{1/p} + \left( \int_X g^p d\mu \right)^{1/p} \nonumber$ <h3 id="proof-1">Proof</h3> <ul> <li>(Hölder) : Let $\left( \int_X f^p d\mu \right)^{1/p} = A$ and $\left( \int_X g^q d\mu \right)^{1/q} =B$. If $A=\infty$, it holds trivially and if $A=0$, then $\int_X f^p d\mu =0$ holds. Since $f\geq 0$, this implies that $f^p = 0$ almost everywhere. Hence, $fg=0$ holds almost everywhere, which gives $\int_X fg =0$. Therefore, assume $0 &lt; A, B &lt; \infty$ holds and set $F(x) := \frac{f(x)}{A}$ and $G(x) := \frac{g(x)}{B}$ so that $\int F^p = \frac{A^p}{A^p} =1$ and $\int G^p = \frac{B^p}{B^p} =1$. Since $F^p, G^p \ne 0, \infty$ a.e., we may assume that $0 &lt; F(x), G(x) &lt; \infty$ holds for all $x\in X$. Hence, for each $x \in X$, $\exists! s,t \in \mathbb{R}$ s.t. $F(x)=\exp \frac{s}{p}, \, G(x)=\exp \frac{t}{q}$ for $\frac{1}{p} + \frac{1}{q} = 1$. By convexity of $\exp$, we have $\exp \frac{s}{p} \exp \frac{t}{q} \leq \frac{1}{p}e^s + \frac{1}{q} e^t$, i.e., $FG \leq \frac{1}{p}F^p + \frac{1}{q}G^q$. Thus, $\int_X FG \leq \frac{1}{p} \int F^p + \frac{1}{q} \int G^q = \frac{1}{p} + \frac{1}{q} = 1$, which concludes the proof since this gives $\frac{1}{AB} \int_X fg \leq 1$.</li> <li>(Minkowski) : Since $(p,q)$ is a conjugate exponent, $(p-1)q = p$ and $(q-1)p =q$ holds. As $(f+g)^p = (f+g)(f+g)^{p-1} = f(f+g)^{p-1} + g(f+g)^{p-1}$ holds, we have $\int_X f(f+g)^{p-1} \stackrel{\text{Hölder}}{\leq} \left( \int f^p \right)^{1/p} \left( \int (f+g)^{(p-1)/q} \right)^{1/q} = \left( \int f^p \right)^{1/p} \left( \int (f+g)^{p} \right)^{1/q}$. Thefore, $\int (f+g)^p = \int f(f+g)^{p-1} + g(f+g)^{p-1} \leq \left( \int (f+g)^{p} \right)^{1/q} \left( \left( \int f^p \right)^{1/p} + \left( \int g^p \right)^{1/p}\right)$ holds, i.e., $\left(\int (f+g)^p\right)^{1-1/q} \leq \left( \int f^p \right)^{1/p} + \left( \int g^p \right)^{1/p}$.</li> </ul> <h1 id="l_p-space">$L_p$ space</h1> <p class="info"><code class="language-plaintext info highlighter-rouge">Def.</code> Given a M.S. $(X, F, \mu)$ for $0 &lt; p &lt; \infty$, define <strong>$L_p$ space</strong> $L^p(\mu) := \{ f: X \to \mathbb{C} : \left(\int_X \mid f \mid^p d \mu \right)^{1/p} &lt; \infty \}$ where $f$ is complex measurable function on $X$. We call $\Vert f \Vert_p := \left(\int_X \mid f \mid^p d \mu \right)^{1/p}$ the <strong>$L_p$ norm</strong> of $f$.</p> <p>NB10; When $p=1$, recall that $L^1(\mu)$ space is a collection of Lebesgue integrable measurable functions. <br /> NB11; When $X$ is countable set and $\mu$ is a counting measure, we denote by $\ell^p(\mu) := \{ (x_n) \in \mathbb{C} : \sum_{n=1}^\infty \mid x_n \mid^p &lt; \infty \}$. <br /> NB12; Fix $p \in [1, \infty]$. If $f \in L^p(\mu)$ and $\alpha \in \mathbb{C}$, then $\alpha f \in L^p(\mu)$ holds clearly and $\Vert \alpha f \Vert_{p} = \mid \alpha \mid \Vert f \Vert_{p}$.</p> <p class="info"><code class="language-plaintext info highlighter-rouge">Def.</code> Given a M.S. $(X, F, \mu)$, suppose $g: X \to [0, \infty]$ be measurable function on $X$. Let $S$ be the set of all real $\alpha$ s.t. $\mu(g^{-1}((\alpha, \infty]))=0$. Define $\Vert g \Vert_{\infty} := \inf S$ and call <strong>essential bound (supremum)</strong> of $g$. And define $L^\infty(\mu) := \{ f: X \to \mathbb{C} : \Vert f \Vert_{\infty} &lt; \infty\}$.</p> <p>NB13; If $S=\emptyset$, then let $\inf S = \infty$. <br /> NB14; If $S \ne \emptyset$, $\beta = \inf S \in \mathbb{R}$, then $g^{-1}((\beta, \infty]) = \cup_{n=1}^\infty g^{-1}((\beta+\frac{1}{n}, \infty])$ which is a measure zero set. Thus, $\mu(g^{-1}((\beta, \infty])) =0$, i.e., $\beta \in S$. <br /> NB15; $L^p(\mu)$ is a complete normed linear space (Banach space) for $1\leq p \leq \infty$.</p> <h2 id="propositions">Propositions</h2> $\text{Suppose } 1 \leq p \leq \infty \text{ for conjugate exponents } (p,q) \text{ and if } f \in L^p(\mu) \text{ and } g \in L^q (\mu). \nonumber$ $\text{ Then } fg \in L^1(\mu) \text{ and } \Vert fg \Vert_{1} \leq \Vert f \Vert_{p} \Vert g \Vert_{q}. \nonumber$ <h3 id="proof-2">Proof</h3> <p>For $p \in (1, \infty)$, it is simply Hölder’s Ineq. which was shown. By definition, $\mid fg \mid \leq \Vert f \Vert_{\infty} \mid g \mid$ holds a.e. and thus integrating this concludes the proof when $p=\infty$. The same argument holds for $p=1$ and $q=\infty$ holds.</p> <hr /> $\text{Suppose } 1 \leq p \leq \infty, \, f \in L^p(\mu) \text{ and } g \in L^q (\mu), \text{ then } f+g \in L^p(\mu) \text{ and } \Vert f+g \Vert_{p} \leq \Vert f \Vert_{p} \Vert g \Vert_{p}. \nonumber$ <h3 id="proof-3">Proof</h3> <p>For $p \in (1, \infty)$, it follows from Minkowski’s Ineq. which was shown above. For $p=1$ or $p=\infty$, it holds trivially since $\mid f+g \mid \leq \mid f \mid + \mid g \mid$ holds.</p>Jongyeong Leelee@ms.k.u-tokyo.ac.jpSource of most content: Gunhee Cho and RCA by Rudin. Convexity Def. For any open set $(a,b) \in [-\infty, \infty]$, $\varphi: (a,b) \to \mathbb{R}$ is convex if $\forall x,y \in (a,b)$ with $x &lt; y$ and $\forall \lambda \in [0, 1]$, $\varphi(\lambda x + (1-\lambda) y) \leq \lambda \varphi(x) + (1- \lambda) \varphi(y)$ holds. NB1; If we put $t=\lambda x + (1-\lambda) y$, it holds that $\varphi(t) \leq \lambda \varphi(x) + (1-\lambda)\varphi(y)$. Since $\lambda (x-y) = t-y$ holds, we have $\frac{\varphi(x)-\varphi(y)}{x-y} \leq \frac{\varphi(t)-\varphi(y)}{t-y}$. NB2; More generally, for $a &lt; s &lt; t &lt; u &lt; b$, it holds that $\frac{\varphi(t)-\varphi(s)}{t-s} \leq \frac{\varphi(u)-\varphi(t)}{u-t}$. NB3; Thus, a real differentiable function $\varphi$ is convex in $(a,b)$ iff $a&lt; x &lt; y &lt; b$ implies $\varphi’(x) \leq \varphi’(y)$, i.e., the derivative $\varphi’$ is a monotonically increasing function. NB4; If $\varphi$ is convex on $(a,b)$, then $\varphi$ is continous on $(a,b)$. NB5; If $f, g$ are convex on $(a,b)$, then $\max (f, g)$ is convex on $(a,b)$. Jensen’s Inequality $\text{Let }\mu \text{ be a measure on } (X, F) \text{ s.t. } \mu(X) =1. \text{ If a real fn } f \in L^1(\mu) \text{ s.t. } \forall x \in X, \, f(x) \in (a,b), \text{ and if } \varphi \text{ is cvx, } \nonumber$ $\text{then } \varphi \left( \int_X f d\mu \right) \leq \int_X \varphi \circ f d\mu. \nonumber$ Proof Put $t= \int_X f d\mu \in (a,b)$. Let $\beta := \sup \{ \frac{\varphi(s)-\varphi(t)}{s-t} : a &lt; s &lt; t \} \in \mathbb{R}$. By LUP, it holds that $\forall s \in (a,b),$ $\varphi(s) - \varphi(t) \geq \beta (s-t)$. Since $\forall x \in X, f(x) \in (a,b)$ holds by assumption, $\varphi(f(x)) - \varphi(t) \geq \beta (f(x) -t )$ holds. Since $\varphi$ is contnous by NB4, if we integrate w.r.t. $\mu$, we have $\int_X \varphi \circ f d\mu - \int_X \varphi(t) d \mu \geq \beta \int_X (f(x) -t) d\mu = \beta \int_X f d\mu - \beta\int_X t d \mu \stackrel{t=\int_X f d\mu}{=} 0$. Hence, we have $\int_X \varphi \circ f d\mu \geq \int_X \varphi(t) d \mu = \varphi(t) \int_X 1 d\mu \stackrel{\mu(X) =1}{=} \varphi(t) = \varphi \left( \int_X f d\mu \right)$, which concludes the proof. NB6; Let $\varphi(x) = e^x$ (cvx). By Jensen’s Ineq, $\exp(\int_X f) \leq \int_X \exp f$ holds. Here, if we set $f(x) = \log g(x)$, we have $\exp (\int_X \log g) \leq \int_X g$. NB7; Let $\mu (\{ \frac{1}{n}\}) = \frac{1}{n}$ and $f(x_i) = p_i\in \mathbb{R}$ for finite set $X= \{ x_1, \ldots, x_n \}$. Then, $\int_X f = \frac{1}{n}(p_1 + \cdots + p_n)$ holds. By applying Jensen’s Ineq, we have $\exp ( \frac{1}{n}(p_1 + \cdots + p_n) ) \leq \frac{1}{n}\left[ e^{p_1} + \cdots + e^{p_n} \right]$. Letting $e^{p_i} = y_i$ gives the Arithmetic-geometric inequality. NB8; If $\mu(x_i) = \alpha_i$ s.t. $\sum \alpha_i = 1$, we have $\Pi_i y_i^{\alpha_i} \leq \sum \alpha_i y_i$. Hölder and Minkowski’s Inequalities Def. Let $p , q \in \mathbb{R}_{+}$ s.t. $p+q = pq$, i.e., $\frac{1}{p} + \frac{1}{q} =1$. We say $(p,q)$ conjugate exponents. NB9; $(1, \infty)$ and $(\infty , 1)$ are also considered as conjugate exponents. $\text{Given M.S. } (X, F, \mu) \text{, conjugate exponents } (p,q) \text{, measurable functions } f,g \in [0, \infty]. \text{ Then, }\nonumber$ $\text{ Hölder's Ineq: } \int_X fg d\mu \leq \left( \int_X f^p d\mu \right)^{1/p} \left( \int_X g^q d\mu \right)^{1/q} \nonumber$ $\text{Minkowski's Ineq: } \left(\int_X (f+g)^p d\mu \right)^{1/p} \leq \left( \int_X f^p d\mu \right)^{1/p} + \left( \int_X g^p d\mu \right)^{1/p} \nonumber$ Proof (Hölder) : Let $\left( \int_X f^p d\mu \right)^{1/p} = A$ and $\left( \int_X g^q d\mu \right)^{1/q} =B$. If $A=\infty$, it holds trivially and if $A=0$, then $\int_X f^p d\mu =0$ holds. Since $f\geq 0$, this implies that $f^p = 0$ almost everywhere. Hence, $fg=0$ holds almost everywhere, which gives $\int_X fg =0$. Therefore, assume $0 &lt; A, B &lt; \infty$ holds and set $F(x) := \frac{f(x)}{A}$ and $G(x) := \frac{g(x)}{B}$ so that $\int F^p = \frac{A^p}{A^p} =1$ and $\int G^p = \frac{B^p}{B^p} =1$. Since $F^p, G^p \ne 0, \infty$ a.e., we may assume that $0 &lt; F(x), G(x) &lt; \infty$ holds for all $x\in X$. Hence, for each $x \in X$, $\exists! s,t \in \mathbb{R}$ s.t. $F(x)=\exp \frac{s}{p}, \, G(x)=\exp \frac{t}{q}$ for $\frac{1}{p} + \frac{1}{q} = 1$. By convexity of $\exp$, we have $\exp \frac{s}{p} \exp \frac{t}{q} \leq \frac{1}{p}e^s + \frac{1}{q} e^t$, i.e., $FG \leq \frac{1}{p}F^p + \frac{1}{q}G^q$. Thus, $\int_X FG \leq \frac{1}{p} \int F^p + \frac{1}{q} \int G^q = \frac{1}{p} + \frac{1}{q} = 1$, which concludes the proof since this gives $\frac{1}{AB} \int_X fg \leq 1$. (Minkowski) : Since $(p,q)$ is a conjugate exponent, $(p-1)q = p$ and $(q-1)p =q$ holds. As $(f+g)^p = (f+g)(f+g)^{p-1} = f(f+g)^{p-1} + g(f+g)^{p-1}$ holds, we have $\int_X f(f+g)^{p-1} \stackrel{\text{Hölder}}{\leq} \left( \int f^p \right)^{1/p} \left( \int (f+g)^{(p-1)/q} \right)^{1/q} = \left( \int f^p \right)^{1/p} \left( \int (f+g)^{p} \right)^{1/q}$. Thefore, $\int (f+g)^p = \int f(f+g)^{p-1} + g(f+g)^{p-1} \leq \left( \int (f+g)^{p} \right)^{1/q} \left( \left( \int f^p \right)^{1/p} + \left( \int g^p \right)^{1/p}\right)$ holds, i.e., $\left(\int (f+g)^p\right)^{1-1/q} \leq \left( \int f^p \right)^{1/p} + \left( \int g^p \right)^{1/p}$. $L_p$ space Def. Given a M.S. $(X, F, \mu)$ for $0 &lt; p &lt; \infty$, define $L_p$ space $L^p(\mu) := \{ f: X \to \mathbb{C} : \left(\int_X \mid f \mid^p d \mu \right)^{1/p} &lt; \infty \}$ where $f$ is complex measurable function on $X$. We call $\Vert f \Vert_p := \left(\int_X \mid f \mid^p d \mu \right)^{1/p}$ the $L_p$ norm of $f$. NB10; When $p=1$, recall that $L^1(\mu)$ space is a collection of Lebesgue integrable measurable functions. NB11; When $X$ is countable set and $\mu$ is a counting measure, we denote by $\ell^p(\mu) := \{ (x_n) \in \mathbb{C} : \sum_{n=1}^\infty \mid x_n \mid^p &lt; \infty \}$. NB12; Fix $p \in [1, \infty]$. If $f \in L^p(\mu)$ and $\alpha \in \mathbb{C}$, then $\alpha f \in L^p(\mu)$ holds clearly and $\Vert \alpha f \Vert_{p} = \mid \alpha \mid \Vert f \Vert_{p}$. Def. Given a M.S. $(X, F, \mu)$, suppose $g: X \to [0, \infty]$ be measurable function on $X$. Let $S$ be the set of all real $\alpha$ s.t. $\mu(g^{-1}((\alpha, \infty]))=0$. Define $\Vert g \Vert_{\infty} := \inf S$ and call essential bound (supremum) of $g$. And define $L^\infty(\mu) := \{ f: X \to \mathbb{C} : \Vert f \Vert_{\infty} &lt; \infty\}$. NB13; If $S=\emptyset$, then let $\inf S = \infty$. NB14; If $S \ne \emptyset$, $\beta = \inf S \in \mathbb{R}$, then $g^{-1}((\beta, \infty]) = \cup_{n=1}^\infty g^{-1}((\beta+\frac{1}{n}, \infty])$ which is a measure zero set. Thus, $\mu(g^{-1}((\beta, \infty])) =0$, i.e., $\beta \in S$. NB15; $L^p(\mu)$ is a complete normed linear space (Banach space) for $1\leq p \leq \infty$. Propositions $\text{Suppose } 1 \leq p \leq \infty \text{ for conjugate exponents } (p,q) \text{ and if } f \in L^p(\mu) \text{ and } g \in L^q (\mu). \nonumber$ $\text{ Then } fg \in L^1(\mu) \text{ and } \Vert fg \Vert_{1} \leq \Vert f \Vert_{p} \Vert g \Vert_{q}. \nonumber$ Proof For $p \in (1, \infty)$, it is simply Hölder’s Ineq. which was shown. By definition, $\mid fg \mid \leq \Vert f \Vert_{\infty} \mid g \mid$ holds a.e. and thus integrating this concludes the proof when $p=\infty$. The same argument holds for $p=1$ and $q=\infty$ holds. $\text{Suppose } 1 \leq p \leq \infty, \, f \in L^p(\mu) \text{ and } g \in L^q (\mu), \text{ then } f+g \in L^p(\mu) \text{ and } \Vert f+g \Vert_{p} \leq \Vert f \Vert_{p} \Vert g \Vert_{p}. \nonumber$ Proof For $p \in (1, \infty)$, it follows from Minkowski’s Ineq. which was shown above. For $p=1$ or $p=\infty$, it holds trivially since $\mid f+g \mid \leq \mid f \mid + \mid g \mid$ holds.Random math: 312021-03-06T00:00:00+00:002021-03-06T00:00:00+00:00https://jongyeong.github.io/rm/rm31<p>Source of most content: <a href="https://www.youtube.com/channel/UCB_AbuIVIG8I3K3_T_j0FSw/">Gunhee Cho</a> and <a href="https://59clc.files.wordpress.com/2011/01/real-and-complex-analysis.pdf">RCA by Rudin</a>.</p> <h1 id="almost-everywhere">Almost everywhere</h1> <p class="info"><code class="language-plaintext info highlighter-rouge">Def.</code> Given a M.S. $(X, F, \mu)$, we say a statement $P$ holds <strong>almost everywhere (a.e.)</strong> on $X$ if $\exists N \in F$ s.t. $\mu(N)=0$ and $P$ is true for $X\setminus N$.</p> <p>NB1; Give the relation to measurable fns $f,g: X \to [0, \infty]$, $f \sim g$ if $\mu( \{x \in X : f(x)\ne g(x) \})=0$ holds, i.e., $f \sim g$ if $f=g$ a.e.. Then, $\sim$ is an equivalence relation.</p> <blockquote> <ol> <li>$f\sim f$ since $\{x \in X : f(x)\ne g(x) \} = \emptyset$ and $\mu(\emptyset) = 0$.</li> <li>If $f\sim g$, then $g \sim f$.</li> <li>Suppose $f \sim g$ and $g\sim h$. Let $A= \{x \in X : f(x)\ne g(x) \}$ and $B= \{x \in X : g(x)\ne h(x) \}$. Since $\{x \in X : f(x)\ne h(x) \} \subset A\cup B$, $\mu(\{x \in X : f(x)\ne h(x) \}) \leq \mu(A\cup B) = \mu(A)+\mu(B) = 0$. Thus, $f\sim h$.</li> </ol> </blockquote> <p>Let $E= \{ x \in X : f = g\}$. If $f \sim g$, then $\int_X f = \int_{E \cup E^c} f = \int_E f + \int_{E^c}f = \int_E g + \int_{E^c} f \stackrel{\mu(E^c)=0}{=} \int_E g$ $= \int_E g + \int_{E^c} g = \int_X g$. Hence, if $f\sim g$ then $\int_X f = \int_X g$. So, roughly speaking, we can neglect the measure zero set in integration.</p> <h1 id="completion">Completion</h1> $\text{Given M.S. } (X,F,\mu), \text{ define } F^* := \{ E \subseteq X : \exists A,B \in F \text{ s.t. } A \subset E \subset B \text{ and } \mu(B-A) = 0 \} \text{ and } \nonumber$ $\mu^*:= F^ * \to [0, \infty] (E \mapsto \mu(A)) \text{, where } A \subset E \subset B. \text{ Then, } (X, F^ *, \mu^ *) \text{ is a M.S. called the completion of } (X,F,\mu). \nonumber$ <p>NB2; This extended measure $\mu^*$ is called <strong>complete</strong>, since all subsets of measure zero sets are now measurable and $F^ *$ is called the <strong>$\mu$-completion</strong> of $F$. <br /> NB3; Every measure can be completed, so, whenever it is convenient, we may assume that any given measure is complete. <br /> NB4; For M.S. $(X, F, \mu)$, let $E \subseteq F, \, f: E \to (Y, T)$. We say $f$ is measruable function if $\forall V \in T, f^{-1}(V) \cap E \in F$ and $\mu (E^c) = 0$.</p> <h3 id="proof">Proof</h3> <ul> <li>Well-definedness of $\mu^ *$: Let $E \in F^ *$ and $A_1 \subseteq E \subseteq B_1$ and $A_2 \subseteq E \subseteq B_2$, i.e., $\mu(B_1 - A_1) = \mu(B_2 - A_2) = 0$. Then, $A_1 - A_2 \subset E - A_2 \subset B_2 - A_2$ holds, which gives $\mu(A_1 - A_2)=0$. Similarly, $\mu(A_2 - A_1)=0$ holds. By using countable additivity of measure, we have $\mu (A_1) = \mu(A_1 \cap A_2) + \mu(A_1 \cap A_2^c)$ and $\mu (A_2) = \mu(A_1 \cap A_2) + \mu(A_2 \cap A_1^c)$, which gives $\mu (A_1) = \mu(A_2)$.</li> <li>$F^*$ is $\sigma$-algebra: <ol> <li>Since $F \subseteq F^ *, \, X \in F^ *$.</li> <li>If $A \subseteq E \subseteq B$ then $B^c \subseteq E^c \subseteq A^c$. Since $\mu(A^c - B^c) = \mu(B-A)$, if $E \in F^ *$ then $E^c \in F^ *$ holds.</li> <li>Let $\{ E _i \} _{i=1}^\infty \subseteq F^ *$. $\forall i, \, \exists A _i, B _i \in F$ s.t. $A _i \subseteq E _i \subseteq B _i \implies \cup _{i=1}^\infty A _i \subseteq \cup _{i=1}^\infty E _i \subseteq \cup _{i=1}^\infty B _i \in F$. Hence, $\mu( \cup _{i=1}^\infty B _i - \cup _{i=1}^\infty A _i) = \mu( \cup _{i=1}^\infty (B _i - \cup _{j=1}^\infty A _j)) \leq \mu( \cup _{i=1}^\infty (B _i-A _i)) \stackrel{( * )}{\leq}\sum _{i=1}^\infty \mu(B _i - A _i) =0$, i.e., $\cup _{i=1}^\infty E _i \in F ^*$.</li> </ol> </li> <li>Countable additivity: Let $\{ E _i \} _{i=1}^\infty \subseteq F^ *$ be mutually disjoint subsets. By deifnition, $\forall i, \, \exists A _i, B _i \in F$ s.t. $A _i \subseteq E _i \subseteq B _i, \, \mu(B _i - A _i) =0$. Then, $\mu^ * (\cup _{i=1}^\infty E _i) = \mu (\cup _{i=1}^\infty A _i) = \sum _i \mu (A _i) = \sum _i \mu^ * (E _i)$.</li> <li>$(*)$: Let $F_i = E_i - \cup_{j=1}^{i-1} E_i$. Then, $F_i$’s are mutually disjoint and $\cup_{i=1}^\infty F_i = \cup_{i=1}^\infty E_i$. Hence, $\mu(\cup_{i=1}^\infty E_i) = \mu(\cup_{i=1}^\infty F_i) = \sum_i \mu(F_i) \stackrel{monotonicity}{\leq} \sum_i \mu(E_i)$.</li> </ul> <h2 id="proposition">Proposition</h2> $\text{Let } \{ f_n \}_{n=1}^\infty \text{ be measurable fns defined a.e. on } X \text{ to } \mathbb{C} \text{ s.t. } \sum_{n=1}^\infty \int_X |f_n| d\mu &lt; \infty. \nonumber$ $\text{Then, } f(x) := \sum_{n=1}^\infty f_n(x) \text{ is defined a.e. and } \int_X f d\mu = \int_X \sum_{n=1}^\infty f_n d\mu = \sum_{n=1}^\infty \int_X f_n d\mu. \nonumber$ <h3 id="proof-1">Proof</h3> <p>Let $s_n$ be a set which defines $f_n : s_n \to \mathbb{C}$, i.e., $\forall n \, \mu(s_n^c) =0$. Let $S := \cap_{n=1}^\infty s_n$. Then, $\mu (s^c) = \mu (\cup_{n=1}^\infty s_n^c) \leq \sum_{n} \mu(s_n ^c)=0$. Define $\varphi: S \to [0, \infty] \, (x \mapsto \sum_{i=1}^\infty \mid f_n (x) \mid)$. Then, $\int _X \varphi d\mu = \int _S \varphi d\mu \stackrel{(p)}{=}$ $\sum _{i=1}^\infty \int _S \mid f _n \mid d\mu = \sum _{i=1}^\infty \int _X \mid f _n \mid d\mu &lt; \infty$, i.e., $\varphi \in L^1(\mu)$. Let $E = \{ x \in S : \varphi(x) &lt; \infty \}$. Then, $\mu (E^c) = 0$. If $x\in E$, it holds that $\mid f(x) \mid = \mid \sum _{n=1}^\infty f _n(x) \mid \leq \sum _{n=1}^\infty \mid f _n (x) \mid = \varphi(x) &lt; \infty$. This implies that $f$ is defined on $E$, i.e., $f$ is defined a.e. on $X$ so that $f \in L^1(\mu)$. Next, define $g _n(x):= \sum _{k=1}^n f _k(x)$ on $E$. Then, $g _n \to f$ on $E$ (a.e. on $X$). By <a href="/rm/rm30#lebesgues-dct-ldct">LDCT</a>, we have $\lim _n \int _E g _n d\mu = \int _E \lim _n g _n d\mu = \int _E f d\mu$. This concludes the proof since $\mu(E^c) = 0$. <a href="/rm/rm29#proof-3">(p)</a></p>Jongyeong Leelee@ms.k.u-tokyo.ac.jpSource of most content: Gunhee Cho and RCA by Rudin. Almost everywhere Def. Given a M.S. $(X, F, \mu)$, we say a statement $P$ holds almost everywhere (a.e.) on $X$ if $\exists N \in F$ s.t. $\mu(N)=0$ and $P$ is true for $X\setminus N$. NB1; Give the relation to measurable fns $f,g: X \to [0, \infty]$, $f \sim g$ if $\mu( \{x \in X : f(x)\ne g(x) \})=0$ holds, i.e., $f \sim g$ if $f=g$ a.e.. Then, $\sim$ is an equivalence relation. $f\sim f$ since $\{x \in X : f(x)\ne g(x) \} = \emptyset$ and $\mu(\emptyset) = 0$. If $f\sim g$, then $g \sim f$. Suppose $f \sim g$ and $g\sim h$. Let $A= \{x \in X : f(x)\ne g(x) \}$ and $B= \{x \in X : g(x)\ne h(x) \}$. Since $\{x \in X : f(x)\ne h(x) \} \subset A\cup B$, $\mu(\{x \in X : f(x)\ne h(x) \}) \leq \mu(A\cup B) = \mu(A)+\mu(B) = 0$. Thus, $f\sim h$. Let $E= \{ x \in X : f = g\}$. If $f \sim g$, then $\int_X f = \int_{E \cup E^c} f = \int_E f + \int_{E^c}f = \int_E g + \int_{E^c} f \stackrel{\mu(E^c)=0}{=} \int_E g$ $= \int_E g + \int_{E^c} g = \int_X g$. Hence, if $f\sim g$ then $\int_X f = \int_X g$. So, roughly speaking, we can neglect the measure zero set in integration. Completion $\text{Given M.S. } (X,F,\mu), \text{ define } F^* := \{ E \subseteq X : \exists A,B \in F \text{ s.t. } A \subset E \subset B \text{ and } \mu(B-A) = 0 \} \text{ and } \nonumber$ $\mu^*:= F^ * \to [0, \infty] (E \mapsto \mu(A)) \text{, where } A \subset E \subset B. \text{ Then, } (X, F^ *, \mu^ *) \text{ is a M.S. called the completion of } (X,F,\mu). \nonumber$ NB2; This extended measure $\mu^*$ is called complete, since all subsets of measure zero sets are now measurable and $F^ *$ is called the $\mu$-completion of $F$. NB3; Every measure can be completed, so, whenever it is convenient, we may assume that any given measure is complete. NB4; For M.S. $(X, F, \mu)$, let $E \subseteq F, \, f: E \to (Y, T)$. We say $f$ is measruable function if $\forall V \in T, f^{-1}(V) \cap E \in F$ and $\mu (E^c) = 0$. Proof Well-definedness of $\mu^ *$: Let $E \in F^ *$ and $A_1 \subseteq E \subseteq B_1$ and $A_2 \subseteq E \subseteq B_2$, i.e., $\mu(B_1 - A_1) = \mu(B_2 - A_2) = 0$. Then, $A_1 - A_2 \subset E - A_2 \subset B_2 - A_2$ holds, which gives $\mu(A_1 - A_2)=0$. Similarly, $\mu(A_2 - A_1)=0$ holds. By using countable additivity of measure, we have $\mu (A_1) = \mu(A_1 \cap A_2) + \mu(A_1 \cap A_2^c)$ and $\mu (A_2) = \mu(A_1 \cap A_2) + \mu(A_2 \cap A_1^c)$, which gives $\mu (A_1) = \mu(A_2)$. $F^*$ is $\sigma$-algebra: Since $F \subseteq F^ *, \, X \in F^ *$. If $A \subseteq E \subseteq B$ then $B^c \subseteq E^c \subseteq A^c$. Since $\mu(A^c - B^c) = \mu(B-A)$, if $E \in F^ *$ then $E^c \in F^ *$ holds. Let $\{ E _i \} _{i=1}^\infty \subseteq F^ *$. $\forall i, \, \exists A _i, B _i \in F$ s.t. $A _i \subseteq E _i \subseteq B _i \implies \cup _{i=1}^\infty A _i \subseteq \cup _{i=1}^\infty E _i \subseteq \cup _{i=1}^\infty B _i \in F$. Hence, $\mu( \cup _{i=1}^\infty B _i - \cup _{i=1}^\infty A _i) = \mu( \cup _{i=1}^\infty (B _i - \cup _{j=1}^\infty A _j)) \leq \mu( \cup _{i=1}^\infty (B _i-A _i)) \stackrel{( * )}{\leq}\sum _{i=1}^\infty \mu(B _i - A _i) =0$, i.e., $\cup _{i=1}^\infty E _i \in F ^*$. Countable additivity: Let $\{ E _i \} _{i=1}^\infty \subseteq F^ *$ be mutually disjoint subsets. By deifnition, $\forall i, \, \exists A _i, B _i \in F$ s.t. $A _i \subseteq E _i \subseteq B _i, \, \mu(B _i - A _i) =0$. Then, $\mu^ * (\cup _{i=1}^\infty E _i) = \mu (\cup _{i=1}^\infty A _i) = \sum _i \mu (A _i) = \sum _i \mu^ * (E _i)$. $(*)$: Let $F_i = E_i - \cup_{j=1}^{i-1} E_i$. Then, $F_i$’s are mutually disjoint and $\cup_{i=1}^\infty F_i = \cup_{i=1}^\infty E_i$. Hence, $\mu(\cup_{i=1}^\infty E_i) = \mu(\cup_{i=1}^\infty F_i) = \sum_i \mu(F_i) \stackrel{monotonicity}{\leq} \sum_i \mu(E_i)$. Proposition $\text{Let } \{ f_n \}_{n=1}^\infty \text{ be measurable fns defined a.e. on } X \text{ to } \mathbb{C} \text{ s.t. } \sum_{n=1}^\infty \int_X |f_n| d\mu &lt; \infty. \nonumber$ $\text{Then, } f(x) := \sum_{n=1}^\infty f_n(x) \text{ is defined a.e. and } \int_X f d\mu = \int_X \sum_{n=1}^\infty f_n d\mu = \sum_{n=1}^\infty \int_X f_n d\mu. \nonumber$ Proof Let $s_n$ be a set which defines $f_n : s_n \to \mathbb{C}$, i.e., $\forall n \, \mu(s_n^c) =0$. Let $S := \cap_{n=1}^\infty s_n$. Then, $\mu (s^c) = \mu (\cup_{n=1}^\infty s_n^c) \leq \sum_{n} \mu(s_n ^c)=0$. Define $\varphi: S \to [0, \infty] \, (x \mapsto \sum_{i=1}^\infty \mid f_n (x) \mid)$. Then, $\int _X \varphi d\mu = \int _S \varphi d\mu \stackrel{(p)}{=}$ $\sum _{i=1}^\infty \int _S \mid f _n \mid d\mu = \sum _{i=1}^\infty \int _X \mid f _n \mid d\mu &lt; \infty$, i.e., $\varphi \in L^1(\mu)$. Let $E = \{ x \in S : \varphi(x) &lt; \infty \}$. Then, $\mu (E^c) = 0$. If $x\in E$, it holds that $\mid f(x) \mid = \mid \sum _{n=1}^\infty f _n(x) \mid \leq \sum _{n=1}^\infty \mid f _n (x) \mid = \varphi(x) &lt; \infty$. This implies that $f$ is defined on $E$, i.e., $f$ is defined a.e. on $X$ so that $f \in L^1(\mu)$. Next, define $g _n(x):= \sum _{k=1}^n f _k(x)$ on $E$. Then, $g _n \to f$ on $E$ (a.e. on $X$). By LDCT, we have $\lim _n \int _E g _n d\mu = \int _E \lim _n g _n d\mu = \int _E f d\mu$. This concludes the proof since $\mu(E^c) = 0$. (p)Random math: 302021-03-05T00:00:00+00:002021-03-05T00:00:00+00:00https://jongyeong.github.io/rm/rm30<p>Source of most content: <a href="https://www.youtube.com/channel/UCB_AbuIVIG8I3K3_T_j0FSw/">Gunhee Cho</a> and <a href="https://59clc.files.wordpress.com/2011/01/real-and-complex-analysis.pdf">RCA by Rudin</a>.</p> <h1 id="converse-of-radon-nikodym-theorem">Converse of Radon-Nikodym theorem</h1> $\text{Given a M.S. } (X, F, \mu), \text{ let } f: X \to [0, \infty] \text{ be measurable function. Define } \varphi: F \to [0, \infty] (E\in F \mapsto \int_E f \mathrm{d}\mu). \nonumber$ $\text{Then, } \varphi \text{ is a measure. Moreover, given a measurable function } g : X \to [0, \infty], \int_X g \mathrm{d}\varphi = \int_X g f \mathrm{d}\mu. \nonumber$ <h3 id="proof">Proof</h3> <ul> <li>Let $E_1, E_2, \ldots$ be disjoint members of $F$ whose union is $E$. Then, $f(x) \chi_{E}(x) = \sum_{k=1}^{\infty} f(x) \chi_{E_k}(x)$ holds. Therefore, $\varphi (E) = \int_{E} f \mathrm{d}\mu = \int_{X} f \chi_E \mathrm{d}\mu= \int_{X} \sum_{k=1}^{\infty} f(x) \chi_{E_k}(x) \mathrm{d}\mu \stackrel{f \geq 0}{=} \sum_{k=1}^{\infty} \int_{X} f(x) \chi_{E_k}(x) \mathrm{d}\mu = \sum_{k=1}^{\infty} \varphi (E_{k})$ holds. This implies that $\varphi$ satisfies contunable additivity. Also, as $\varphi(\emptyset) = \int_{\emptyset} f\mathrm{d}\mu =0$, $\varphi$ is a measure.</li> <li>NTS: $\int_{X}g \mathrm{d}\varphi = \int_{X} gf \mathrm{d}\mu$. <ol> <li>When $g=\chi_E$ for $E\in F$: $\int_{X} \chi_E \mathrm{d}\varphi = \int_{E} 1 \mathrm{d}\varphi = \varphi(E) = \int_{E} f \mathrm{d}\mu = \int_{X} \chi_{E} f \mathrm{d}\mu$.</li> <li>When $g$ is simple: $\int_{X}g \mathrm{d}\varphi = \int_{X} \sum_{i} c_{i} \chi_{A_{i}} \mathrm{d}\varphi = \sum_{i} c_{i} \int_{X} \chi_{A_{i}} \mathrm{d}\varphi \stackrel{1}{=} \sum_{i} c_{i} \int_{X} \chi_{A_{i}} f \mathrm{d}\varphi =\int_{X} gf \mathrm{d}\mu$.</li> <li>When $g$ is non-negative measurable: By <a href="/rm/rm28#simple-approximation-theorem">simple approximation thm</a>, there exists a monotonic increaseing seq. of non-negative measurable simple functions $\{ s _{i} \} _{i=1}^{\infty}$ . Then, by 2, $\int _{X} s _i \mathrm{d}\varphi = \int _{X} s _i f \mathrm{d}\varphi$ holds for all $i$. By MCT, $\int _{X}g \mathrm{d}\varphi = \int _{X} gf \mathrm{d}\mu$ holds.</li> </ol> </li> </ul> <p>NB1; From this theorem, we can see the integral as a measure.</p> <h1 id="l1-space">$L^1$ space</h1> <p class="info"><code class="language-plaintext info highlighter-rouge">Def.</code> Given a M.S. $(X, F, \mu)$, <strong>$L^1$ space</strong> $L^1(X, F, \mu) = L^1 (\mu) := \{ f: X \to \mathbb{C} : \int_X | f | \mathrm{d}\mu &lt; \infty \}$ is a collection of Lebesgue integrable functions.</p> <p>NB2; If $f\in L^1(\mu): X \to \mathbb{C}$, i.e., $f = u + iv$ for real measurable functions $u,v$ on $X$, then $\int_{E} f \mathrm{d}\mu = \int_{E} u^+ \mathrm{d}\mu + \int_{E} u^- \mathrm{d}\mu + i \int_{E} v^+ \mathrm{d}\mu + i \int_{E} v^- \mathrm{d}\mu$. Note that $\int_{E} u^+ \mathrm{d}\mu \leq \int_{E} \mid u \mid \mathrm{d}\mu \leq \int_{E} \mid f \mid \mathrm{d}\mu &lt; \infty$.</p> <h2 id="proposition">Proposition</h2> $\text{Suppose } f,g \in L^1(\mu), \alpha, \beta \in \mathbb{C}. \text{ Then, } \alpha f + \beta g \in L^1(\mu) \text{ and } \int_{X} \alpha f + \beta g \mathrm{d}\mu = \alpha \int_{X} f \mathrm{d}\mu + \beta \int_{X} g \mathrm{d}\mu. \nonumber$ <h3 id="proof-1">Proof</h3> <ul> <li>$\int_{X} \mid \alpha f + \beta g \mid \mathrm{d}\mu \leq \int_{X} \mid \alpha \mid \, \mid f \mid + \mid \beta \mid \, \mid g \mid \mathrm{d}\mu = \mid \alpha \mid \int_{X} \mid f \mid \mathrm{d}\mu + \mid \beta \mid \int_{X} \mid g \mid \mathrm{d}\mu &lt; \infty$.</li> <li> <ol> <li>Let $h = f+g$. Then, we have $h^+ +f^- + g^- = f^+ + g^+ + g^+ + h^-$. From <a href="/rm/rm29#proof-3">previous proposition</a>, it holds that $\int h^+ + \int f^- + \int g^- = \int f^+ + \int g^+ + \int h^-$. Since $f$ and $g$ are Lebesgue intergrable, $\int f+g = \int f +\int g$ holds.</li> <li>NTS: $\int \alpha f \mathrm{d}\mu = \alpha \int f \mathrm{d}\mu$ for $\alpha \in \mathbb{C}$. Since it holds for $\alpha \in \mathbb{R}$, it suffices to show when $\alpha = i$. If $f = u + iv$, then $\int if = \int iu -v = \int -v + i \int u = i(\int u + i \int v) = i \int f$.</li> </ol> </li> </ul> <hr /> $\text{If } f \in L^1(\mu), \text{ then it holds that } \mid \int_X f \mathrm{d}\mu \mid \leq \int_X \mid f \mid \mathrm{d}\mu. \nonumber$ <h3 id="proof-2">Proof</h3> <p>Let $z= \int_{X} f \mathrm{d}\mu$ and $\angle z = \theta$. Take $\alpha = e^{i\theta}$, then $\alpha z = \mid z \mid$. Then, $\mid \int_X f \mathrm{d}\mu \mid = \mid z \mid = \alpha z = \int_X Re(f) \mathrm{d}\mu \leq \int_X \mid f \mid \mathrm{d}\mu$ holds.</p> <h1 id="lebesgues-dct-ldct">Lebesgue’s DCT (LDCT)</h1> $\text{Supppose } \{ f_n \} \text{ is a seq. of complex measurable fns on } X \text{ s.t. } f(x) = \lim_{n\to \infty} f_n(x) \text{ exists } \forall x \in X. \nonumber$ $\text{If } \exists g \in L^1(\mu) \text{ s.t. } \mid f_n(x) \mid \leq g(x), \text{ then } f\in L^1(\mu), \lim_{n\to \infty} \int_X \mid f_n - f \mid \mathrm{d}\mu = 0 \text{ and } \lim_{n\to \infty} \int_X f_n \mathrm{d}\mu = \int_X f \mathrm{d}\mu. \nonumber$ <h3 id="proof-3">Proof</h3> <p>First, $\mid f_n \mid \leq g$ gives $\mid f \mid \leq g$, i.e., $f \in L^1 (\mu)$. Since $\mid f_n - f \mid \leq 2g$ and $2g - \mid f_n -f \mid$ is measurable, applying <a href="/rm/rm29#fatous-lemma">Fatou’s Lemma</a>, we have $\int_X \liminf 2g- \mid f_n - f\mid \mathrm{d}\mu = \int_X 2g \mathrm{d}\mu \leq \liminf \int_X 2g- \mid f_n - f\mid \mathrm{d}\mu$ $= \int_X 2g \mathrm{d}\mu + \liminf_{n\to \infty} \left( -\int_X \mid f_n - f \mid \mathrm{d}\mu \right) = \int_X 2g \mathrm{d}\mu - \limsup \int_X \mid f_n - f \mid \mathrm{d}\mu$. It implies that $\limsup \int_X \mid f_n - f \mid \leq 0$, i.e., $\limsup \int_X \mid f_n - f \mid \leq =0$. Since $\liminf \int_X \mid f_n - f \mid \geq 0$, $\lim_{n\to \infty} \int_X \mid f_n - f \mid \mathrm{d}\mu = 0$ holds. Applying above proposition concludes the proof.</p> <hr /> <p>NB1; Define $f_n(x) = \frac{f(x+1/n)-f(x)}{1/n}$. Then, $f_n(x) \to f’(x)$ if $f$ is differentiable. If there exists $g\in L^1(\mu)$ s.t. $\forall n, \, \mid f_n \mid \leq g$, then $\frac{d}{dx}\int_X f \mathrm{d}\mu =\lim_n \int_X f_n \mathrm{d}\mu \stackrel{LDCT}{=} \int_X \lim_n f_n \mathrm{d}\mu = \int_X f’ \mathrm{d}\mu = \int_X \frac{d}{dx}f \mathrm{d}\mu$ holds. This shows that we can switch the derivative and integration.</p>Jongyeong Leelee@ms.k.u-tokyo.ac.jpSource of most content: Gunhee Cho and RCA by Rudin. Converse of Radon-Nikodym theorem $\text{Given a M.S. } (X, F, \mu), \text{ let } f: X \to [0, \infty] \text{ be measurable function. Define } \varphi: F \to [0, \infty] (E\in F \mapsto \int_E f \mathrm{d}\mu). \nonumber$ $\text{Then, } \varphi \text{ is a measure. Moreover, given a measurable function } g : X \to [0, \infty], \int_X g \mathrm{d}\varphi = \int_X g f \mathrm{d}\mu. \nonumber$ Proof Let $E_1, E_2, \ldots$ be disjoint members of $F$ whose union is $E$. Then, $f(x) \chi_{E}(x) = \sum_{k=1}^{\infty} f(x) \chi_{E_k}(x)$ holds. Therefore, $\varphi (E) = \int_{E} f \mathrm{d}\mu = \int_{X} f \chi_E \mathrm{d}\mu= \int_{X} \sum_{k=1}^{\infty} f(x) \chi_{E_k}(x) \mathrm{d}\mu \stackrel{f \geq 0}{=} \sum_{k=1}^{\infty} \int_{X} f(x) \chi_{E_k}(x) \mathrm{d}\mu = \sum_{k=1}^{\infty} \varphi (E_{k})$ holds. This implies that $\varphi$ satisfies contunable additivity. Also, as $\varphi(\emptyset) = \int_{\emptyset} f\mathrm{d}\mu =0$, $\varphi$ is a measure. NTS: $\int_{X}g \mathrm{d}\varphi = \int_{X} gf \mathrm{d}\mu$. When $g=\chi_E$ for $E\in F$: $\int_{X} \chi_E \mathrm{d}\varphi = \int_{E} 1 \mathrm{d}\varphi = \varphi(E) = \int_{E} f \mathrm{d}\mu = \int_{X} \chi_{E} f \mathrm{d}\mu$. When $g$ is simple: $\int_{X}g \mathrm{d}\varphi = \int_{X} \sum_{i} c_{i} \chi_{A_{i}} \mathrm{d}\varphi = \sum_{i} c_{i} \int_{X} \chi_{A_{i}} \mathrm{d}\varphi \stackrel{1}{=} \sum_{i} c_{i} \int_{X} \chi_{A_{i}} f \mathrm{d}\varphi =\int_{X} gf \mathrm{d}\mu$. When $g$ is non-negative measurable: By simple approximation thm, there exists a monotonic increaseing seq. of non-negative measurable simple functions $\{ s _{i} \} _{i=1}^{\infty}$ . Then, by 2, $\int _{X} s _i \mathrm{d}\varphi = \int _{X} s _i f \mathrm{d}\varphi$ holds for all $i$. By MCT, $\int _{X}g \mathrm{d}\varphi = \int _{X} gf \mathrm{d}\mu$ holds. NB1; From this theorem, we can see the integral as a measure. $L^1$ space Def. Given a M.S. $(X, F, \mu)$, $L^1$ space $L^1(X, F, \mu) = L^1 (\mu) := \{ f: X \to \mathbb{C} : \int_X | f | \mathrm{d}\mu &lt; \infty \}$ is a collection of Lebesgue integrable functions. NB2; If $f\in L^1(\mu): X \to \mathbb{C}$, i.e., $f = u + iv$ for real measurable functions $u,v$ on $X$, then $\int_{E} f \mathrm{d}\mu = \int_{E} u^+ \mathrm{d}\mu + \int_{E} u^- \mathrm{d}\mu + i \int_{E} v^+ \mathrm{d}\mu + i \int_{E} v^- \mathrm{d}\mu$. Note that $\int_{E} u^+ \mathrm{d}\mu \leq \int_{E} \mid u \mid \mathrm{d}\mu \leq \int_{E} \mid f \mid \mathrm{d}\mu &lt; \infty$. Proposition $\text{Suppose } f,g \in L^1(\mu), \alpha, \beta \in \mathbb{C}. \text{ Then, } \alpha f + \beta g \in L^1(\mu) \text{ and } \int_{X} \alpha f + \beta g \mathrm{d}\mu = \alpha \int_{X} f \mathrm{d}\mu + \beta \int_{X} g \mathrm{d}\mu. \nonumber$ Proof $\int_{X} \mid \alpha f + \beta g \mid \mathrm{d}\mu \leq \int_{X} \mid \alpha \mid \, \mid f \mid + \mid \beta \mid \, \mid g \mid \mathrm{d}\mu = \mid \alpha \mid \int_{X} \mid f \mid \mathrm{d}\mu + \mid \beta \mid \int_{X} \mid g \mid \mathrm{d}\mu &lt; \infty$. Let $h = f+g$. Then, we have $h^+ +f^- + g^- = f^+ + g^+ + g^+ + h^-$. From previous proposition, it holds that $\int h^+ + \int f^- + \int g^- = \int f^+ + \int g^+ + \int h^-$. Since $f$ and $g$ are Lebesgue intergrable, $\int f+g = \int f +\int g$ holds. NTS: $\int \alpha f \mathrm{d}\mu = \alpha \int f \mathrm{d}\mu$ for $\alpha \in \mathbb{C}$. Since it holds for $\alpha \in \mathbb{R}$, it suffices to show when $\alpha = i$. If $f = u + iv$, then $\int if = \int iu -v = \int -v + i \int u = i(\int u + i \int v) = i \int f$. $\text{If } f \in L^1(\mu), \text{ then it holds that } \mid \int_X f \mathrm{d}\mu \mid \leq \int_X \mid f \mid \mathrm{d}\mu. \nonumber$ Proof Let $z= \int_{X} f \mathrm{d}\mu$ and $\angle z = \theta$. Take $\alpha = e^{i\theta}$, then $\alpha z = \mid z \mid$. Then, $\mid \int_X f \mathrm{d}\mu \mid = \mid z \mid = \alpha z = \int_X Re(f) \mathrm{d}\mu \leq \int_X \mid f \mid \mathrm{d}\mu$ holds. Lebesgue’s DCT (LDCT) $\text{Supppose } \{ f_n \} \text{ is a seq. of complex measurable fns on } X \text{ s.t. } f(x) = \lim_{n\to \infty} f_n(x) \text{ exists } \forall x \in X. \nonumber$ $\text{If } \exists g \in L^1(\mu) \text{ s.t. } \mid f_n(x) \mid \leq g(x), \text{ then } f\in L^1(\mu), \lim_{n\to \infty} \int_X \mid f_n - f \mid \mathrm{d}\mu = 0 \text{ and } \lim_{n\to \infty} \int_X f_n \mathrm{d}\mu = \int_X f \mathrm{d}\mu. \nonumber$ Proof First, $\mid f_n \mid \leq g$ gives $\mid f \mid \leq g$, i.e., $f \in L^1 (\mu)$. Since $\mid f_n - f \mid \leq 2g$ and $2g - \mid f_n -f \mid$ is measurable, applying Fatou’s Lemma, we have $\int_X \liminf 2g- \mid f_n - f\mid \mathrm{d}\mu = \int_X 2g \mathrm{d}\mu \leq \liminf \int_X 2g- \mid f_n - f\mid \mathrm{d}\mu$ $= \int_X 2g \mathrm{d}\mu + \liminf_{n\to \infty} \left( -\int_X \mid f_n - f \mid \mathrm{d}\mu \right) = \int_X 2g \mathrm{d}\mu - \limsup \int_X \mid f_n - f \mid \mathrm{d}\mu$. It implies that $\limsup \int_X \mid f_n - f \mid \leq 0$, i.e., $\limsup \int_X \mid f_n - f \mid \leq =0$. Since $\liminf \int_X \mid f_n - f \mid \geq 0$, $\lim_{n\to \infty} \int_X \mid f_n - f \mid \mathrm{d}\mu = 0$ holds. Applying above proposition concludes the proof. NB1; Define $f_n(x) = \frac{f(x+1/n)-f(x)}{1/n}$. Then, $f_n(x) \to f’(x)$ if $f$ is differentiable. If there exists $g\in L^1(\mu)$ s.t. $\forall n, \, \mid f_n \mid \leq g$, then $\frac{d}{dx}\int_X f \mathrm{d}\mu =\lim_n \int_X f_n \mathrm{d}\mu \stackrel{LDCT}{=} \int_X \lim_n f_n \mathrm{d}\mu = \int_X f’ \mathrm{d}\mu = \int_X \frac{d}{dx}f \mathrm{d}\mu$ holds. This shows that we can switch the derivative and integration.