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Math4201 Topology I (Lecture 34)

Countability axioms

Second countability axiom

Example of spaces that is first countable but not second countable

Let \mathbb{R}_l be \mathbb{R} with the lower limit topology generated by the basis:

\mathcal{B}=\{[a,b)\mid a,b\in \mathbb{R},a<b\}

And \mathbb{R}_l is first countable since for each x\in \mathbb{R}, there is a countable collection \{[x,\frac{1}{n}+x)\}_{n\in \mathbb{N}} of open intervals in \mathbb{R}_l such that any open interval U of x contains one of [x,\frac{1}{n}+x)\}

However, \mathbb{R}_l is not second contable:

If \mathbb{R}_l is second countable, then for any real number x, there is an element U of \mathcal{B} contains x and is contained in [x,x+1).

Any such open sets is of the form [x,x+\epsilon)\cap A with \epsilon>0 and any element of A being larger than \min(U_x)=x.

In summary, for any x\in \mathbb{R}, there is an element U_x\in \mathcal{B} with (U_x)=x. In particular, if x\neq y, then U_x\neq U_y. SO there is an injective map f:\mathbb{R}\rightarrow \mathcal{B} sending x to U_x. This implies that \mathbb{B} is uncountable.

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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Then we prove that every open covering of X has countable subcover.

Let \{U_\alpha\}_{\alpha\in I} be an open covering of X. Let \mathcal{B} be a countable basis for X.

For any basis element B of X. If B is in U_\alpha for some \alpha\in I, then pick U_\alpha as an element of our subcover.

This way we get countably many open sets $U_\alpha$'s because \mathcal{B} is countable.

We also claim that the chosen $U_\alpha$'s given an open covering of X.

For any x\in X, there is U_\alpha (possibly not one of the chosen ones) such that x\in U_\alpha. There is B_x\in \mathcal{B} such that x\in B_x\subseteq U_\alpha.

In particular, there is a chosen U_\alpha such that B_x\subseteq U_\alpha. This implies that there is a chosen U_\alpha, containing x.

Separation Axioms

Our goal is to find conditions that if some space is second countable, and xxx, then it's metrizable.

Kolmogorov classification

Let X be a topological space:

• X is T_0 if for any pair of points x,y\in X, x\neq y, there is an open set U containing x but not y. (equivalent to say that any singleton set is closed)
• X is Hausdorff if for any pair of distinct x,y\in X, there are disjoint open sets U and V such that x\in U and y\in V.

Definition of regular spaces

A T_0 space is regular if for any x\in X and any close set A\subseteq X such that x\notin A, there are disjoint open sets U,V such that x\in U and A\subseteq V.

Definition of normal spaes

A T_0 space is normal if for any disjoint closed sets, A,B\subseteq X, there are disjoint open sets U,V such that A\subseteq U and B\subseteq V.

Let \mathbb{R}_K be the topology on \mathbb{R} generated by the basis:

\mathcal{B}=\{(a,b)\mid a,b\in \mathbb{R},a<b\}\cup \{(a,b)-K\mid a,b\in \mathbb{R},a<b\}

where K=\coloneqq \{\frac{1}{n}\mid n\in \mathbb{N}\}.

This is finer than the standard topology on \mathbb{R}.

Since (-1,1)-K is not open in \mathbb{R}, but it is open in \mathbb{R}_K.

\mathbb{R}_K is Hausdorff, since for any x\neq y, there is an open set (x-\epsilon,x+\epsilon) and (y-\epsilon,y+\epsilon) such that (x-\epsilon,x+\epsilon)\cap (y-\epsilon,y+\epsilon)=\emptyset.

However, this space is not regular.

consider x=0 and A=K is a closed set.

Any open neighborhood U of x and V of A are not disjoint.

Otherwise, if there exists such U and V, we can assume U is a basis element. If U=(a,b) with 0\in U then U\cap K\neq \emptyset.

Which is a contradiction.

So U=(a,b)-K. Suppose \frac{1}{n}\in (a,b).

Then V contains an open interval of the form (\frac{1}{n}-\epsilon,\frac{1}{n}+\epsilon). But (a,b)-K\cap (\frac{1}{n}-\epsilon,\frac{1}{n}+\epsilon)\neq \emptyset.

Which is a contradiction.

Let X be a regular space. Take x\in X and an open neighborhood U of x. So A\coloneqq X-U is a closed set disjoint from U.

By regularity assumption, there is an open neighborhood W_1 of x and W_2 of A that are disjoint.

In particular, X-W_2 is a closed set contained in X-A.

In particular, X-W_2 is a closed set contained in X-A=U which also contains W_1. This implies that the closure of W_1 is contained in U.

Lemma of regular spaces

If X is a regular space, and x\in X, and U is an open neighborhood of x, then there is an open neighborhood V of x such that \overline{V}\subseteq U.

Continue next lecture.