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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:

1. X_1\times X_2\times X_3\times \dots\times X_n (in this case, product topology = box topology)
2. 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:

1. X_1\times X_2\times X_3\times \dots\times X_n (in this case, product topology = box topology)
2. 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:

1. Any discrete subspace Y of X is countable
2. There exists a countable subset of X that is dense in X (also called separable spaces)
3. 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.

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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.

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Third part next lecture.