# 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 $B\subseteq V$.

<details>
<summary>Example</summary>

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 $[y,y+\delta_y)\subseteq B$.

</details>

#### Theorem Every metric space is normal

Use the similar proof above.

<details>
<summary>Proof</summary>

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.

</details>

#### 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 $\overline{V}\subseteq U$.

<details>
<summary>Proof</summary>

$\implies$

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 $\overline{V}\subseteq U\coloneqq X-B$.

</details>

#### Proposition of regular and Hausdorff on subspaces

1. If $X$ is a regular topological space, and $Y$ is a subspace. Then $Y$ with induced topology is regular. (same holds for Hausdorff)
2. 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. (In problem set 11)

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

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 $\mathbb{R}_{\ell}$)