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

$D$ is a metric. exercise
$\forall \overline{x}\in \mathbb{R}^\omega$ , $\forall \epsilon >0$ , $\exists$ 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

Since $X$ is regular, then 1 point sets in $X$ are closed.
$X$ 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 $z_0$ .