NoteNextra-origin/content/Math4201/Math4201_L32.md

Math4201 Topology I (Lecture 30)

Compact and connected spaces

Locally compact

Theorem of one point compactification

X is a locally compact Hausdorff space if and only if there exists topological space Y satisfying the following properties:

1. X is a subspace of Y.
2. Y-X has one point (usually denoted by \infty).
3. Y is compact and Hausdorff.

Y is called one point compactification of X.

Proof for existence of Y (forward direction)

Let's defined the topology of Y as follows:

Let U\subseteq Y is open if and only if either

1. U\subseteq X and U is open in X. (\infty\notin U) (Type 1 open set)
2. \infty \in U and Y-U\subseteq X with subspace topology from X is compact. (Type 2 open set)

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First, we prove that X is a subspace of Y. (That is, every open set U\subseteq X implies that U\cap X is open in X.)

Case 1: U\subseteq X is open in X, then U\cap X=U is open in Y.

Case 2: \infty\in U, then Y-U is a compact subspace of X, since X is Hausdorff. So Y-U is a closed subspace of X.

So X\cap U=X-(Y-U) is open in X.

We also need to show any open U\subseteq X can be written as the intersection of some open in Y and X.

Note that for an open set U\subseteq X, U\cap X is open in X. So U\cap X is open in Y.

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The second part is trivial by observation.

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First we show that Y is Hausdorff.

Let x_1,x_2\in Y, such that x_1\neq x_2.

If one of x, without loss of generality, x_1 is \infty, then by the assumption on X, there is a compact set K containing an open neighborhood U of x_2.

Note that Y-K is an open subspace of Type 2 in Y. In particular, it contains \infty.

This is disjoint from the open neighborhood U of x_2.

If x_1,x_2 are both in X, then by the assumption on X, then by Hausdorff property for X, there are disjoint open neightbors U_1 and U_2 such that x_1\in U_1 and x_2\in U_2. By Type 1 open sets, these are also open and disjoint in Y.

Then we show that Y is compact.

Take an open cover \{U_\alpha\}_{\alpha\in I} of Y.

In particular, there is \alpha_0\in I such that \infty\in U_{\alpha_0}

Note that Y-U_{\alpha_0}\subseteq X with subspace topology from X is compact (by Type 2 set).

So there exists a finite subcover \{U_{\alpha_i}\}_{\alpha_i\in I} such that Y-U_{\alpha_0}\subseteq \bigcup_{i=1}^n U_{\alpha_i}.

So U_{\alpha_0},U_{\alpha_1},\dots,U_{\alpha_n} is a finite cover of Y.

So Y is compact.

Proof for properties from $Y$.(backward direction)

Property 1

X is Hausdorff because it's a subspace of Hausdorff space.

Property 2

By definition

Property 3

X is locally compact.

Let x\in X since Y is Hausdorff, there are disjoint open sets U,V\subseteq Y such that x\in U and \infty\in B.

Let K=Y-V, K is a subset of X since \infty\notin V.

To complete the proof, we need to show that K is compact.

Since V is open in Y, then K is closed in Y. Since Y is compact, then K is compact. (any closed subspace of compact space is compact)

Countability axioms

First countability axiom

Definition for first countability axiom

Let X be a topological space, then X satisfies the first countability axiom if

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 for metric space satisfies the first countability axiom

Any metric space satisfies the first countability axiom.

Take \{B_{\frac{1}{n}}(x)\}_{n=1}^\infty.

Properties for topological spaces that satisfy the first countability axiom

1. If A\subseteq X, then for any x\in \overline{A}, there is a sequence \{x_n\}_{n=1}^\infty\subseteq A such that x_n\to x.
2. If f:X\to Y such that for any sequence \{x_n\}_{n=1}^\infty\subseteq X such that x_n\to x, we have f(x_n)\to f(x) in Y, then f is continuous.

Second countability axiom

Definition for second countability axiom

Let X be a topological space, then X satisfies the second countability axiom if

it has a countable basis.

Clearly any second countable space also satisfies the first countability axiom.

But the converse is not true.

Example for metric space satisfies the second countability axiom

\mathbb{R} satisfies the second countability axiom. Take \{(a,b)|a,b\in\mathbb{Q}\} is a basis for \mathbb{R}.

And \mathbb{Q} is countable.

More generally, \mathbb{R}^n is also countable and satisfies the second countability axiom.

Warning

Not all topological spaces satisfy the second countability axiom is metrizable.