Math4201 Topology I (Lecture 33)

Countability axioms

First countability axiom

For any $x\in X$ , there is a countable collection $\{B_n\}_n$ of open neighborhoods of $x$ such that any open neighborhood $U$ of $x$ contains one of $B_n$ .

Example: Metric spaces

Second countability axiom

There exists countable basis for a topology on $X$ .

Example: $\mathbb{R}$ and more generally $\mathbb{R}^n$

Consider the following topology on $\mathbb{R}^\omega$ :

There exists different topology can be defined on $\mathbb{R}^\omega$

Product topology
Box topology
Union topology

Definition of product topology

Let $\{X_\alpha\}_{\alpha\in I}$ be a family of topological spaces:

\prod_{\alpha\in I}X_\alpha=\{f:I\to \bigcup_{\alpha\in I} X_\alpha | \forall \alpha\in I, f(\alpha)\in X_\alpha\}

The product topology defined on the basis that:

The set of following forms:

\mathcal{B}_{prod}=\{\prod_{\alpha\in I}U_\alpha| U_\alpha\subseteq X_\alpha\text{ is open and }U_\alpha=X_\alpha\text{ for all except finitely many }\alpha\}

So $\mathbb{R}^\omega$ with product topology is second countable.

Proof that product topology defined above is second countable

A countable basis for $\mathbb{R}^\omega$ is given by the union of following sets:

B_1=\{(a_1,b_1)\times \mathbb{R}\times \mathbb{R}\times \dots |a_1,b_1\in \mathbb{Q}\}\\ B_2=\{(a_1,b_1)\times(a_2,b_2) \mathbb{R}\times \dots |a_2,b_2\in \mathbb{Q}\}\\ B_n=\{(a_1,b_1)\times(a_2,b_2)\times \dots (a_n,b_n)\times \mathbb{R}\times \dots |a_n,b_n\in \mathbb{Q}\}\\

Each $B_i$ is countable and there exists a bijection from $B_n\cong\mathbb{Q}^{2n}$

And the union of countably many countable sets is also countable.

So $B_1\cup B_2\cup \dots$ is countable. This is also a baiss for the product topology on $\mathbb{R}^\omega$

Lemma of second countable spaces

If $X_1,X_2,X_3,\dots$ are second countable topological spaces, then the following spaces are also second countable:

$X_1\times X_2\times X_3\times \dots\times X_n$ (in this case, product topology = box topology)
$X_1\times X_2\times X_3\times \dots$ with the product topology

Ideas for proof

For $X_1\times X_2\times X_3\times \dots\times X_n$ :

$B_1\times B_2\times B_3\times \dots\times B_n$ is also countable basis for $X_1\times X_2\times X_3\times \dots\times X_n$

Let $\tilde{B}\coloneqq B_1\times B_2\times B_3\times \dots\times B_i\times X_{i+1}\times X_{i+2}\times \dots$ then $\tilde{B}$ is a countable basis for $X_1\times X_2\times X_3\times \dots$

Lemma of first countable spaces

If $X_1,X_2,X_3,\dots$ are first countable topological spaces, then the following spaces are also first countable:

$X_1\times X_2\times X_3\times \dots\times X_n$ (in this case, product topology = box topology)
$X_1\times X_2\times X_3\times \dots$ with the product topology

Ideas for proof

Basically the same as before but now you have analyze the basis for each $x\in X_1\times X_2\times X_3\times \dots$

Definition of box topology

Let $\{X_\alpha\}_{\alpha\in I}$ be a family of topological spaces:

\prod_{\alpha\in I}X_\alpha=\{f:I\to \bigcup_{\alpha\in I} X_\alpha | \forall \alpha\in I, f(\alpha)\in X_\alpha\}

The product topology defined on the basis that:

The set of following forms:

\mathcal{B}_{box}=\{\prod_{\alpha\in I}U_\alpha| U_\alpha\subseteq X_\alpha\text{ is open}\}

Definition of uniform topology

The uniform topology on $X$ is the topology induced by the uniform metric on $X$ .

\rho(x,y)=\sup_{i\in \mathbb{N}}\overline{d}(x_\alpha,y_\alpha)

To get a finite number for $\rho(x,y)$ , we define the bounded metric $\overline{d}$ on $X$ by $\overline{d}(x,y)=\min\{d(x,y),1\}$ where $d$ is the usual metric on $X$ .

where $x=(x_1,x_2,x_3,\dots), y=(y_1,y_2,y_3,\dots)\in X$

In particular, the $\mathbb{R}^\omega$ with the uniform topology is first countable because it’s a metric space.

However, it’s not second countable.

Recall that $Y\subseteq X$ is a discrete subspace if the subspace topology on $Y$ is the discrete topology, i.e. any point of $y$ is open in $Y$ .

Define $A\subseteq \mathbb{R}^\omega$ be defined as follows:

A=\{\underline{x}=(x_1,x_2,x_3,\dots)|x_i\in \{0,1\}\}

Let $\underline{x} and \underline{x}'$ be two distinct elements of $A$ .

\rho(\underline{x},\underline{x}')=\sup_{i\in \mathbb{N}}\overline{d}(\underline{x}_\alpha,\underline{x}'_\alpha)=1

(since there exists at least one entry different in $\underline{x}$ and $\underline{x}'$ )

In particular, $B_1^\rho(\underline{x})\cap A=\{\underline{x}\}$ , so $A$ is a discrete subspace of $\mathbb{R}^\omega$ .

This subspace is also uncountable ( $A$ can create a surjective map to $(0,1)$ using binary representation) which implies that $\mathbb{R}^\omega$ is not second countable.

Proposition of second countable spaces

Let $X$ be a second countable topological space. Then the following holds:

Any discrete subspace $Y$ of $X$ is countable
There exists a countable subset of $X$ that is dense in $X$ (also called separable spaces)
Every open covering of $X$ has countable subcover (That is if $X=\bigcup_{\alpha\in I} U_\alpha$ , then there exists a countable subcover $\{U_{\alpha_1}, ..., U_{\alpha_\infty}\}$ of $X$ ) (also called Lindelof spaces)

Proof

First we prove that any discrete subspace $Y$ of $X$ is countable.

Let $Y$ be a discrete subspace of $X$ . In particular, for any $y\in Y$ we can find an element $B_y$ of the countable basis $\mathcal{B}$ for $Y$ such that $B_y\cap Y=\{y\}$ .

In particular, if $y\neq y'$ , then $B_y\neq B_{y'}$ . Because $\{y\}=B_y\cap Y\neq B_{y'}\cap Y=\{y'\}$ .

This shows that $\{B_y\}_{y\in Y}\subseteq B$ has the same number of elements as $Y$ .

So $Y$ has to be countable.

Next we prove that there exists a countable subset of $X$ that is dense in $X$ .

For each basis element $B\in \mathcal{B}$ , we can pick an element $x\in B$ and let $A$ be the union of all such $x$ .

We claim that $A$ is dense.

To show that $A$ is dense, let $U$ be a non-empty open subset of $X$ .

Take an element $x\in U$ . Note that by definition of basis, there is some element $B\in \mathcal{B}$ such that $x\in B$ . So $x\in B\cap U$ . $U\cap B\neq \emptyset$ , so $A\cap U\neq \emptyset$ .

Since $A\cap U\neq \emptyset$ this shows that $A$ is dense.

Third part next lecture.