Math4201 Topology I (Lecture 35)

Countability axioms

Kolmogorov classification

Consider the topological space $X$ .

$X$ is $T_0$ means for every pair of points $x,y\in X$ , $x\neq y$ , there is one of $x$ and $y$ is in an open set $U$ containing $x$ but not $y$ .

$X$ is $T_1$ means for every pair of points $x,y\in X$ , $x\neq y$ , each of them have a open set $U$ and $V$ such that $x\in U$ and $y\in V$ and $x\notin V$ and $y\notin U$ . (singleton sets are closed)

$X$ is $T_2$ means for every pair of points $x,y\in X$ , $x\neq y$ , there exists disjoint open sets $U$ and $V$ such that $x\in U$ and $y\in V$ . (Hausdorff)

$X$ is $T_3$ means that $X$ is regular: for any $x\in X$ and any close set $A\subseteq X$ such that $x\notin A$ , there are disjoint open sets $U,V$ such that $x\in U$ and $A\subseteq V$ .

$X$ is $T_4$ means that $X$ is normal: for any disjoint closed sets, $A,B\subseteq X$ , there are disjoint open sets $U,V$ such that $A\subseteq U$ and $B\subseteq V$ .

Example

Let $\mathbb{R}_{\ell}$ with lower limit topology.

$\mathbb{R}_{\ell}$ is normal since for any disjoint closed sets, $A,B\subseteq \mathbb{R}_{\ell}$ , $x\in A$ and $B$ is closed and doesn’t contain $x$ . Then there exists $\epsilon_x>0$ such that $[x,x+\epsilon_x)\subseteq A$ and does not intersect $B$ .

Therefore, there exists $\delta_y>0$ such that $[y,y+\delta_y)\subseteq B$ and does not intersect $A$ .

Let $U=\bigcup_{x\in A}[x,x+\epsilon_x)$ is open and contains $A$ .

$V=\bigcup_{y\in B}[y,y+\delta_y)$ is open and contains $B$ .

We show that $U$ and $V$ are disjoint.

If $U\cap V\neq \emptyset$ , then there exists $x\in A$ and $Y\in B$ such that $[x,x+\epsilon_x)\cap [y,y+\delta_y)\neq \emptyset$ .

This is a contradiction since $[x,x+\epsilon_x)\subseteq A$ and $[y,y+\delta_y)\subseteq B$ .

Theorem Every metric space is normal

Use the similar proof above.

Proof

Let $A,B\subseteq X$ be closed.

Since $B$ is closed, for any $x\in A$ , there exists $\epsilon_x>0$ such that $B_{\epsilon_x}(x)\subseteq B$ .

Since $A$ is closed, for any $y\in B$ , there exists $\delta_y>0$ such that $A_{\delta_y}(y)\subseteq A$ .

Let $U=\bigcup_{x\in A}B_{\epsilon_x/2}(x)$ and $V=\bigcup_{y\in B}B_{\delta_y/2}(y)$ .

We show that $U$ and $V$ are disjoint.

If $U\cap V\neq \emptyset$ , then there exists $x\in A$ and $Y\in B$ such that $B_{\epsilon_x/2}(x)\cap B_{\delta_y/2}(y)\neq \emptyset$ .

Consider $z\in B_{\epsilon_x/2}(x)\cap B_{\delta_y/2}(y)$ . Then $d(x,z)<\epsilon_x/2$ and $d(y,z)<\delta_y/2$ . Therefore $d(x,y)\leq d(x,z)+d(z,y)<\epsilon_x/2+\delta_y/2$ .

If $\delta_y<\epsilon_x$ , then $d(x,y)<\delta_y/2+\delta_y/2=\delta_y$ . Therefore $x\in B_{\delta_y}(y)\subseteq A$ . This is a contradiction since $U\cap B=\emptyset$ .

If $\epsilon_x<\delta_y$ , then $d(x,y)<\epsilon_x/2+\epsilon_x/2=\epsilon_x$ . Therefore $y\in B_{\epsilon_x}(x)\subseteq B$ . This is a contradiction since $V\cap A=\emptyset$ .

Therefore, $U$ and $V$ are disjoint.

Lemma fo regular topological space

$X$ is regular topological space if and only if for any $x\in X$ and any open neighborhood $U$ of $x$ , there is open neighborhood $V$ of $x$ such that $\overline{V}\subseteq U$ .

Lemma of normal topological space

$X$ is a normal topological space if and only if for any $A\subseteq X$ closed and any open neighborhood $U$ of $A$ , there is open neighborhood $V$ of $A$ such that $\overline{V}\subseteq U$ .

Proof

$\implies$

Let $A$ and $U$ are given as in the statement.

So $A$ and $(X-U)$ are disjoint closed.

Since $X$ is normal and $A\subseteq V\subseteq X$ and $V\cap W=\emptyset$ . $X-U\subseteq W\subseteq X$ . where $W$ is open in $X$ .

And $\overline{V}\subseteq (X-W)\subseteq U$ .

And $A\subseteq V$ .

The proof of reverse direction is similar.

Let $A,B$ be disjoint and closed.

Then $A\subseteq U\coloneqq X-B\subseteq X$ and $X-B$ is open in $X$ .

Apply the assumption to find $A\subseteq V\subseteq X$ and $V$ is open in $X$ and $\overline{V}\subseteq U\coloneqq X-B$ .

Proposition of regular and Hausdorff on subspaces

If $X$ is a regular topological space, and $Y$ is a subspace. Then $Y$ with induced topology is regular. (same holds for Hausdorff)
If $\{X_\alpha\}$ is a collection of regular topological spaces, then their product with the product topology is regular. (same holds for Hausdorff)

Caution

The above does not hold for normal.

Recall that $\mathbb{R}_{\ell}$ with lower limit topology is normal. But $\mathbb{R}_{\ell}\times \mathbb{R}_{\ell}$ with product topology is not normal.

Proof that Sorgenfrey plane is not normal

The goal of this problem is to show that $\mathbb{R}^2_\ell$ (the Sorgenfrey plane) is not normal. Recall that $\mathbb{R}_\ell$ is the real line with the lower limit topology, and $\mathbb{R}_\ell^2$ is equipped with the product topology. Consider the subset

L = \{\, (x,-x) \mid x\in \mathbb{R}_\ell \,\} \subset \mathbb{R}^2_\ell.

Let $A\subset L$ be the set points of the form $(x,-x)$ such that $x$ is rational and $B\subset L$ be the set points of the form $(x,-x)$ such that $x$ is irrational.

Show that the subspace topology on $L$ is the discrete topology. Conclude that $A$ and $B$ are closed subspaces of $\mathbb{R}_\ell^2$

Proof

First we show that $L$ is closed.

Consider $x=(a,b)\in\mathbb{R}^2_\ell-L$ , by definition $a\neq -b$ .

If $a>-b$ , then there exists open neighborhood $U_x=[\frac{\min\{a,b\}}{2},a+1)\times[\frac{\min\{a,b\}}{2},b+1)$ that is disjoint from $L$ (no points of form $(x,-x)$ in our rectangle), therefore $x\in U_x$ .

If $a<-b$ , then there exists open neighborhood $U_x=[a,a+\frac{\min\{a,b\}}{2})\times[b,b+\frac{\min\{a,b\}}{2})$ that is disjoint from $L$ , therefore $x\in U_x$ .

Therefore, $\mathbb{R}^2_\ell-L=\bigcup_{x\in \mathbb{R}_\ell^2-L} U_x$ is open in $\mathbb{R}^2_\ell$ .

So $L$ is closed in $\mathbb{R}^2_\ell$ .

To show $L$ with subspace topology on $\mathbb{R}^2_\ell$ is discrete topology, we need to show that every singleton of $L$ is open in $L$ .

For each $\{(x,-x)\}\in L$ , $[x,x+1)\times [-x,-x+1)$ is open in $\mathbb{R}_\ell^2$ and $\{(x,-x)\}=([x,x+1)\times [-x,-x+1))\cap L$ , therefore $\{(x,-x)\}$ is open in $L$ .

Since $A,B$ are disjoint and $A\cup B=L$ , therefore $A=L-B$ and $B=L-A$ , by definition of discrete topology, $A,B$ are both open therefore the complement of $A,B$ are closed. So $A,B$ are closed in $L$ .

since $L$ is closed in $\mathbb{R}^2_\ell$ , by \textbf{Lemma \ref{closed_set_close_subspace_close}}, $A,B$ is also closed in $\mathbb{R}_\ell^2$ . Therefore $A,B$ are closed subspace of $\mathbb{R}_\ell^2$ .

Let $V$ be an open set of $\mathbb{R}^2_\ell$ containing $B$ . Let $K_n$ consist of all irrational numbers $x\in [0,1]$ such that $[x, x+1/n) \times [-x, -x+1/n)$ is contained in $V$ . Show that $[0,1]$ is the union of the sets $K_n$ and countably many one-point sets.

Proof

Since $B$ is open in $L$, for each $b\in B$, by definition of basis in $\mathbb{R}_\ell^2$, and $B$ is open, there exists $b\in ([b,b+\epsilon)\times [-b,-b+\delta))\cap L\subseteq V$ and $0<\epsilon,\delta$, so there exists $n_b$ such that $\frac{1}{n_b}<\min\{\epsilon,\delta\}$ such that $b\in ([b,b+\frac{1}{n_b})\times [-b,-b+\frac{1}{n_b}))\cap L\subseteq V$.

Therefore $\bigcup_{n=1}^\infty K_n$ covers irrational points in $[0,1]$

Note that $B=L-A$ where $A$ is rational points therefore countable.

So $[0,1]$ is the union of the sets $K_n$ and countably many one-point sets.

Use Problem 5-3 to show that some set $\overline{K_n}$ contains an open interval $(a,b)$ of $\mathbb{R}$ . (You don’t need to prove Problem 5.3, if it is not your choice of #5.)

Lemma

Let $X$ be a compact Hausdorff space; let $\{A_n\}$ be a countable collection of closed sets of $X$ . If each sets $A_n$ has empty interior in $X$ , then the union $\bigcup_{n=1}^\infty A_n$ has empty interior in $X$ .

Proof

We proceed by contradiction, note that $[0,1]$ is a compact Hausdorff space since it's closed and bounded.

And $\{\overline{K_n}\}_{n=0}^\infty$ is a countable collection of closed sets of $[0,1]$ .

Suppose for the sake of contradiction, $\overline{K_n}$ has empty interior in $X$ for all $n\in \mathbb{N}$ , by \textbf{Lemma \ref{countable_closed_sets_empty_interior}}, then $\bigcup_{n=1}^\infty \overline{K_n}$ has empty interior in $[0,1]$ , where $\Q\cap[0,1]$ are countably union of singletons, therefore has empty interior in $[0,1]$ .

Therefore $\bigcup_{n=1}^\infty \overline{K_n}$ has empty interior in $[0,1]$ , since $\bigcup_{n=1}^\infty K_n\subseteq \bigcup_{n=1}^\infty \overline{K_n}$ , $\bigcup_{n=1}^\infty K_n$ also has empty interior in $[0,1]$ by definition of subspace of $[0,1]$ , therefore $\bigcup_{n=1}^\infty K_n\cup (\Q\cap[0,1])$ has empty interior in $[0,1]$ . This contradicts that $\bigcup_{n=1}^\infty K_n\cup (\Q\cap[0,1])$ covers $[0,1]$ and should at least have interior $(0.1,0.9)$ .

Show that $V$ contains the open parallelogram consisting of all points of the form

x\times (-x+\epsilon)\quad\text{ for which }\quad a<x<b\text{ and }0<\epsilon<\frac{1}{n}.

Proof

Since $V$ is open, by previous problem we know that there exists $n$ such that $\overline{K_n}$ contains the open interval $(a,b)$.

If $x\in K_n$ , $\forall a<x<b$ , by definition of $K_n$ $[x,x+\frac{1}{n})\times [-x,-x+\frac{1}{n})\subseteq V$ .

If $x$ is a limit point of $K_n$ , since $V$ is open, there exists $0<\epsilon<\frac{1}{n}$ such that $[x,x+\epsilon)\times [-x,-x+\epsilon)\subseteq V$ .

This gives our desired open parallelogram.

Show that if $q$ is a rational number with $a<q<b$ , then the point $q\times (-q)$ of $\mathbb{R}_\ell^2$ is a limit point of $V$ . Conclude that there are no disjoint open neighborhoods $U$ of $A$ and $V$ of $B$ .

Proof

Consider all the open neighborhood of $q\times (-q)$ in $\mathbb{R}_\ell^2$, for all $\delta>0$, $[q,q+\delta)\times (-q,-q+\delta)$ will intersect with some $x\times [-x,-x+\epsilon)\subseteq V$ such that $0<\epsilon<\frac{1}{n}<\delta$.

Therefore, any open set containing $q\times (-q)\in A$ will intersect with $V$ , it is impossible to build disjoint open neighborhoods $U$ of $A$ and $V$ of $B$ .

This shows that $\mathbb{R}_{\ell}$ is not metrizable. Otherwise $\mathbb{R}_{\ell}\times \mathbb{R}_{\ell}$ would be metrizable. Which could implies that $\mathbb{R}_{\ell}$ is normal.

Theorem of metrizability (Urysohn metirzation theorem)

If $X$ is normal and second countable, then $X$ is metrizable.

Note

Every metrizable topological space is normal.
Every metrizable space is first countable.
But there are some metrizable space that is not second countable.

Note that if $X$ is normal and first countable, then it is not necessarily metrizable. (Example $\mathbb{R}_{\ell}$ )