# Math4201 Topology I (Lecture 37)

## Separation Axioms

### Normal spaces

#### Proposition of normal spaces

A topological space $X$ is normal if and only if for all $A\subseteq U$ closed and $U$ is open in $X$, there exists $V$ open such that $A\subseteq V\subseteq \overline{V}\subseteq U$.

#### Urysohn lemma

Let $X$ be a normal space, $A,B$ be two closed and disjoint set in $X$, then there exists continuous function $f:X\to[0,1]$ such that $f(A)=\{0\}$ and $f(B)=\{1\}$.

We say $f$ separates $A$ and $B$.

> [!NOTE]
>
> We could replace $1,0$ to any $a<b$

<details>
<summary>Proof</summary>

Step 1:

Consider the rationals in $[0,1]$. Let $P=[0,1]\cap \mathbb{Q}$. This set is countable.

First we want to prove that there exists a set of open sets $\{U_p\}_{p\in P}$ such that if $p<q$, then $\overline{U_p}\subseteq U_q$. ($U_p$ is open in $X$.)

We prove the claim using countable induction.

Define $U_1=X-B$, since $B$ is closed in $A,B$ are disjoint, $A\subseteq U_1$.

Since $X$ is normal, then there exists $U_0$ such that $A\subseteq U_0\subseteq \overline{U_0}\subseteq X$.

By induction step, for each $p\in P$, we have $U_0,U_1,\cdots,U_{p_n}$ such that if $p<q$, then $\overline{U_p}\subseteq U_q$.

Choose $U_{p_{n+1}}$ as follows:

$\exists p,q\in P_n\coloneqq\{1,0,p_3,\ldots,p_n\}$ and $p_{n+1}\in (p,q)$

$\exists U_{p_{n+1}}$ such that $\overline{U_p}\subseteq U_{p_{n+1}}\subseteq \overline{U_{p_{n+1}}}\subseteq U_q$ by normality of $X$.

This constructs the set satisfying the claim.

Step 2:

We can extend from $P$ to any rationals $\mathbb{Q}$.

We set $\forall p<0$, $U_p=\emptyset$ and $\forall p>1$, $U_p=X$.

Otherwise we use the $p$ in $P$.

Step 3:

$\forall x\in X$, set $\mathbb{Q}(x)=\{p\in \mathbb{Q}:x\in U_p\}\subseteq [0,\infty)$.

This function has a lower bound and $f(x)=\inf\mathbb{Q}(x)$.

Observe that $A\subseteq U_p,\forall p$ and $f(A)=\inf(0,\infty)=\{0\}$.

$\forall b\in B$, $b\in U_p$ if and only if $p>1$, so $f(b)=\inf(1,\infty)=1$.

Suppose $x\in \overline{U_r}$, then $x\in U_s,\forall s<r$, this implies that $f(x)\leq r$.

Suppose $x\notin \overline{U_r}$, then $x\notin U_s,\forall s<r$, this implies that $f(x)\geq r$.

If $x\in \overline{U_q}-U_p$, then $f(x)\in [p,q]$, $\exists p<q$.

To show continuity of $f(x)$.

Let $x_0\in X$ of $f(x_0)\in (c,d)$,

our goal is to show that there exists open $U\subseteq X$ a neighborhood of $x_0\in U$ such that $f(U)\in (c,d)$.

Choose $p,q\in \mathbb{Q}$ such that $c<p<f(x_0)<q<d$.

By our construction, $x_0\in \overline{U_q}-U_p$, and $f(\overline{U_q}-U_p)\subseteq [p,q]\subset (c,d)$.

</details>