NoteNextra-origin/content/Math4201/Math4201_L37.md

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

Proof

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).