# Random math: 39

Source 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

$\text{Given a T1 T.S. } (X,T), \nonumber$
1. $X$ is regular iff $\forall x \in X$, $\forall$ nbhd $U_x$ of $x$, $\exists$ nbhd $V$ of $x$ s.t. $x\in \bar{V} \subseteq U_x$.
2. $X$ 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 $A$.

$\text{If } X \text{ is regular (or T2), then } \nonumber$
1. Subspace $Y$ of $X$ is regular (or T2).
2. $\prod_i X_i$ 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

1. $(X, T_{cofinite})$ is T1 but not T2.
2. $\mathbb{R} _k := (\mathbb{R}, T _{k})$ 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.

$\mathbb{R}_L := (\mathbb{R}, T_{\text{lower limit top}}), \text{ where } \mathcal{B} = \{ [a,b) : a,b \in \mathbb{R}, a< b \}. \nonumber$

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 $b > x$. 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 $O = \{ [a_\alpha, b_\alpha) : a_\alpha, b_\alpha \in \mathbb{R}, a_\alpha < b_\alpha \} = \{ I_\alpha : \alpha \in I \}$. 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 $x < y$. 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, $q _x < q _y$, 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.

$\mathbb{R}_L \times \mathbb{R}_L \text{ is} \nonumber$
1. 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.
2. 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.