NoteNextra-origin/content/Math4201/Math4201_L38.md

Math4201 Topology I (Lecture 38)

Countability and separability

Metrizable spaces

Let \mathbb{R}^\omega be the set of all countable sequences of real numbers.

where the basis is defined

\mathcal{B}=\{\prod_{i=1}^\infty (a_i,b_i)\text{for all except for finitely many}(a_i,b_i)=\mathbb{R}\}

Lemma \mathbb{R}^\omega is metrizable

Consider the metric define on \mathbb{R}^\omega by D(\overline{x},\overline{y})=\sup\{\frac{\overline{d}(x_i,y_i)}{i}\}

where \overline{x}=(x_1,x_2,x_3,\cdots) and \overline{y}=(y_1,y_2,y_3,\cdots), \overline{d}=\min\{|x_i-y_i|,1\}.

Sketch of proof

1. D is a metric. exercise

2. \forall \overline{x}\in \mathbb{R}^\omega, \forall \epsilon >0, \exists basis open set in product topology U\subseteq B_D(\overline{x},\epsilon) containing \overline{x}.

Choose N\geq \frac{1}{\epsilon}, then \forall n\geq N,\frac{\overline{d}(x_n,y_n)}{n}<\frac{1}{N}<\epsilon

We will use the topological space above to prove the following theorem.

Theorem for metrizable spaces

If X is a regular and second countable topological space, then X is metrizable.

Proof

We will show that there exists embedding F:X\to \mathbb{R}^\omega such that F is continuous, injective and if Z=F(X), F:X\to Z is a open map.

Recall that regular and second countable spaces are normal

1. Since X is regular, then 1 point sets in X are closed.
2. X is regular if and only if \forall x\in U\subseteq X, U is open in X. There exists V open in X such that x\in V\subseteq\overline{V}\subseteq U.

Let \{B_n\} be a countable basis for X (by second countability).

Pass to (n,m) such that \overline{B_n}\subseteq B_m.

By Urysohn lemma, there exists continuous function g_{m,n}: X\to [0,1] such that g_{m,n}(\overline{B_n})=\{1\} and g_{m,n}(B_m)=\{0\}.

Therefore, we have \{g_{m,n}\} is a countable set of functions, where \overline{B_n}\subseteq B_m.

We claim that \forall x_0\in U such that U is open in X, there exists k\in \mathbb{N} such that f_k(\{x_0\})>0 and f_k(X-U)=0.

Definition of basis implies that \exists x_0\in B_m\subseteq U

Note that since X is regular, there exists x_0\in B_n\subseteq \overline{B_n}\subseteq B_m.

Choose f_k=g_{m,n}, then f_k(\overline{B_n})=\{1\} and f_k(B_n)=\{0\}. So f_k(x_0)=1 since x_0\in \overline{B_n}.

So F is continuous since each of the f_k is continuous.

F is injective since x\neq y implies that there exists k, f_k=g_{m,n} where x\in \overline{B_n}\subseteq B_m such that f_k(x)\neq f_k(y).

If F(U) is open for all U\subseteq X, U is open in X, then F:X\to Z is homeomorphism.

We want to show that \forall z_0\in F(U), there exists neighborhood W of z_0, z_0\in W\subseteq F(U).

We know that \exists x_0\in F(x_0) such that F(x_0)=z_0.

We choose N such that f_N(\{x_0\})>0 and f_N(X-U)=0, (V\cap Z\subseteq F(U)).

Let V=\pi_N^{-1}((0,\infty)). By construction, V is open in \mathbb{R}^\omega. and V\cap Z is open in Z containing z_0.