# Random math: 35

Source 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

$\text{Let } (X,T) \text{ be a metrizable space. TFAE}\nonumber$
1. $X$ is compact.
2. $X$ is limit point compact
3. $X$ 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. $(d_{x_{n_k}}, a) < \frac{1}{k}$. 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.
1. 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, $\exists \epsilon >0$ s.t. $B_\epsilon(a) \subset A$ for some $A \in \mathcal{A}$. By A.P., $\exists N \in \mathbb{N}$ s.t. $\frac{1}{N} < \frac{\epsilon}{2}$. Since $x_{n_k} \to a$ for this $\epsilon > 0$, $\exists N \in \mathbb{N}$ s.t. $\forall n_k \geq N$, $x_{n_k} \in B_{\frac{\epsilon}{2}}(a)$. Then, $\forall x \in C_{n_k}, \, \forall n_k \geq N$, $d(x,a) \leq d(x, x_{n_k}) + d(x_{n_k}, a) < \frac{1}{n_k} + \frac{\epsilon}{2} < \epsilon$. Therefore, $C_{n_k} \subset B_{\frac{\epsilon}{2}}(a) \subset A$, which is a contradiction. Hence, Lebesgue number lemma holds for sequentially compact $X$.
2. Show that $\forall \epsilon >0$, $X$ can be covered by finite $\epsilon$-balls.
Suppose not. Then, $\exists \epsilon >0$ 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.
3. Let $\mathcal{A}$ be an open cover of $X$. Suppose $X$ is sequentially compact. Then, by step 1, there exists Lebesuge number $\delta>0$ 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 $\frac{2\delta}{3} < \delta$. 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

$X \text{ is locally compact Hausdorff iff there exists a T.S. } Y \text{ satisfying } \nonumber$
1. $X \subset Y$.
2. $Y - X$ consists of a single point.
3. $Y$ 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 $Y$.
1. 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 $Y’$.
2. 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 $T_Y := \{ U, Y-C : U \in T_X, C \text{ is compact in } X \}$.
1. $T_Y$ 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 $X$, $U_a \cap (Y-C_b) \in T_X$.
2. $X$ 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.
3. $Y$ 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 $Y$.
4. $Y$ 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.