# Math4201 Topology I (Lecture 36)

## Separation Axioms

### Regular spaces

#### Proposition for $T_1$ spaces

If $X$ is a topological space such that all singleton are closed, then then following holds:

- $X$ is regular if and only if for any point $x\in X$ and any open neighborhood $V$ of $X$ such that $\overline{V}\subseteq U$.
- $X$ is normal if and only if for any closed set $A\subseteq X$, there is an open neighborhood $V$ of $A$ such that $\overline{V}\subseteq U$.

#### 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)

<details>
<summary>Proof</summary>

1. If $X$ is regular and $Y\subseteq X$,then we want to show that $Y$ is regular.

We take a point $y\in Y$ and a closed subset $A$ of $Y$ which doesn't contain $y$.

Observe that $A\subseteq Y$ is closed if and only if $A=\overline{A}\cap Y$ where $\overline{A}$ is the closure of $A$ in $X$.

Notice that $y\in Y$ but $y\notin A$, then $y\notin \overline{A}$ (otherwise $y\in A$).

By regularity of $X$, we can find an open neighborhood $y\in U\subseteq X$ and $\overline{A}\subseteq V\subseteq X$ and $U,V$ are open and disjoint.

This implies that $U\cap Y$ and $V\cap Y$ are open neighborhood of $y$ and $A$ and disjoint from each other.

(Proof for Hausdorff is similar)

---

2. If $\{X_\alpha\}$ is a collection of regular topological spaces, then their product with the product topology is regular.

We take a collection of regular spaces $\{X_\alpha\}_{\alpha\in I}$.

We want to show that their product with the product topology is regular.

Take $\underline{x}=(x_\alpha)_{\alpha\in I}\in \prod_{\alpha\in I}X_\alpha$, any open neighborhood of $x$ contains a basis element of the form 

$$
\prod_{\alpha\in I}U_\alpha
$$

with $x_\alpha\in U_\alpha$ for all $\alpha\in I$, and all but finitely many $U_\alpha$ are equal to $X_\alpha$. (By definition of product topology)

Now for each $\alpha_i$ take an open neighborhood $V_\alpha$ of $x_{\alpha_i}$ such that

1. $\overline{V_\alpha}\subseteq U_\alpha$ (This can be cond by regularity of $X_\alpha$)
2. $V_\alpha=X_\alpha$ if $U_\alpha=X_\alpha$

The product of $\prod_{\alpha\in I}V_\alpha$ is an open neighborhood of $\underline{x}$ and

$$
\overline {\prod_{\alpha\in I}V_\alpha}=\prod_{\alpha\in I}\overline{V_\alpha}\subseteq \prod_{\alpha\in I}U_\alpha
$$

Therefore, $X$ is regular.

(Proof for Hausdorff is similar)


If we replace the regularity with Hausdorffness, then we can take two points $\underline{x},\underline{y}$. Then there exists $\alpha_0$ such that $x_{\alpha_0}\neq y_{\alpha_0}$. We can use this to build disjoint open neighborhoods

$$
x_{\alpha_0}\in U_{\alpha_0}\subseteq X_{\alpha_0},\quad y_{\alpha_0}\in V_{\alpha_0}\subseteq X_{\alpha_0}
$$

of $x_{\alpha_0}$ and $y_{\alpha_0}$.

Then we take $U_\alpha=\begin{cases} U_{\alpha_0} &\alpha= \alpha_0\\ X_{\alpha_0} & \text{otherwise}\end{cases}$ and $V_\alpha=\begin{cases} V_{\alpha_0} &\alpha= \alpha_0\\ X_{\alpha_0} & \text{otherwise}\end{cases}$.

These two sets are disjoint and $\prod_{\alpha\in I}U_{\alpha}$ and $\prod_{\alpha\in I}V_{\alpha}$ are open neighborhoods of $x$ and $y$.

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Recall that this property does not hold for subspace of normal spaces.

### How we construct new normal spaces from existing one

#### Theorem for constructing normal spaces

1. Any compact Hausdorff space is normal
2. Any regular second countable space is normal

<details>
<summary>Proof</summary>

For the first proposition

Earlier we showed that any compact Hausdorff space $X$ is regular, i.e., for any closed subspace $A$ of $X$ and a point $x\in X$ not in $A$. There are open neighborhoods $U_x$ of $A$ and $V_x$ of $x$ such that $U_x\cap V_x=\emptyset$.

Now let $B$ be a closed subset of $X$ disjoint from $A$.

For any $x\in B$, we know that we have a disjoint open neighborhood $U_x$ of $A$ and $V_x$ of $x$.

$\{V_n\}_{n\in \mathbb{B}}$ gives an open covering of $B$, $B$ is closed subset of a compact space, therefore, $B$ is compact.

This implies that $\exists x_1,x_2,\ldots,x_n$ such that $B\subseteq \bigcup_{i=1}^n V_{x_i}$.

$U\coloneqq \bigcup_{i=1}^n U_{x_i}$ is an open covering of $A$.

$U_{x_i}\cap V_{x_i}=\emptyset$ implies that $\bigcup_{i=1}^n U_{x_i}\cap \bigcup_{i=1}^n V_{x_i}=\emptyset$.

In summary, $U$ and $V$ are disjoint open neighborhoods of $A$ and $B$ respectively.

---

We want to show that any regular second countable space is normal.

Take $A,B$ be two disjoint closed subsets of $X$. We want to show that we can find disjoint open neighborhoods $U$ of $A$ and $V$ of $B$ such that $U\cap V=\emptyset$.

Step 1:

There is a countable open covering $\{U_i\}_{i\in I}$ of $A$ such that for any $i$, $\overline{U_i}\cap B=\emptyset$.

For any $x\in A$, $X-B$ is an open neighborhood of $x$ and by reformulation of regularity, we can find an open set $U_i'$ such that $\overline{U_i'}\subseteq X-B$.

(We will use the second countability of $X$ to produce countable open coverings)

Let $B$ be a countable basis and let $U_x\in \mathcal{B}$ such that $x\in U_x\subseteq U_x'$.

Note that $\overline{U_x}\subseteq \overline{U_x'}\subseteq X-B$.

and $\{U_x\}_{x\in A}\subseteq \mathcal{B}$ is a countable collection.

So this is a countable open covering of $A$ by relabeling the elements of this open covering we can denote it by $\{U_i\}_{i\in\mathbb{N}}$.

Step 2:

There is a countable open covering $\{V_i\}_{i\in I}$ of $B$ such that for any $i$, $\overline{V_i}\cap A=\emptyset$.

Step 3:

Let's define $\hat{U}_i=U_i-\bigcup_{j=1}^i\overline{V}_j=U_i\cap (X-\bigcup_{j=1}^i \overline{V}_j)$, note that $(X-\bigcup_{j=1}^i \overline{V}_j)$ and $U_i$ is open in $X$, therefore $\hat{U}_i$ is open in $X$.

Since $A\subseteq \bigcup_{i=1}^\infty U_i$ and $\overline{V_j}\cap A=\emptyset$, we have $\bigcup_{i=1}^\infty \hat{U}_i\supseteq A$.

Similarly, we have:

$$
\hat{V}_i=V_i-\bigcup_{j=1}^i\overline{U}_j=V_i\cap (X-\bigcup_{j=1}^i \overline{U}_j)
$$

is also open and $\bigcup_{i=1}^\infty \hat{V}_i\supseteq B$.

Then we claim that these two open neighborhoods $U=\bigcup_{i=1}^\infty \hat{U}_i$ and $V=\bigcup_{i=1}^\infty \hat{V}_i$ are disjoint.

To see this, we proceed by contradiction, suppose $x\in \bigcup_{i=1}^\infty \hat{U}_i\cap \bigcup_{i=1}^\infty \hat{V}_i$, $x\in \bigcup_{i=1}^\infty \hat{U}_i$ and $x\in \bigcup_{i=1}^\infty \hat{V}_i$.

$x\in \bigcup_{i=1}^\infty \hat{U}_i$ implies that $\exists k$ such that $x\in \hat{U}_k$ and $\exists l$ such that $x\in \hat{V}_l$.

Suppose without loss of generality that $k\leq l$.

Then $x\in \hat{U}_k\subseteq U_k$, and $x\in \hat{V}$ implies that $\forall j\leq l:x\notin U_j$. This gives $x\notin U_k$.

This is a contradiction.

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

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