Math4201 Topology I (Lecture 31)

Compactness

Local compactness

$\mathbb{R}$ is not compact but it has a “lot” of compact subspaces.

An arbitrary point $x\in\mathbb{R}$ then there is a subset $(x-\epsilon,x+\epsilon)U\subseteq \mathbb{R}$ such that $x\in U$ and $U$ is compact.

Definition of local compactness

A space $X$ is locally compact if every point $x\in X$ , there is a compact subspace $K$ of $X$ containing a neighborhood $U$ of $x$ $x\in U\subseteq K$ such that $K$ is compact.

Example

$\mathbb{R}^\omega=\mathbb{R}\times \mathbb{R}\times \mathbb{R}\times \cdots$ with product topology.

where basis is

B=\{(a_1,b_1)\times (a_2,b_2)\times (a_3,b_3)\times \cdots\times \mathbb{R}\times \mathbb{R}\times \cdots \mid a_i,b_i\in \mathbb{R},a_i<b_i\}

all except finitely many of these open intervals are $\mathbb{R}$ .

This space isn’t locally compact.

Consider $\underline{0}=(0,0,0,...)\in \mathbb{R}^\omega$ . If there is a compact subspace $K$ of $\mathbb{R}^\omega$ containing a neighborhood $U$ of $0$ , then it should contain a basis element around $\underline{0}$ .

And $\overline{U}\subseteq K$ .

Since $\overline{U}$ is closed in $K$ , it has to be compact.

But $\overline{U}=[a_1,b_1]\times [a_2,b_2]\times [a_3,b_3]\times \mathbb{R}\times \mathbb{R}\times \mathbb{R}\times \cdots$ which is not compact.

So we can find a open covering

V=\{[a_1,b_1]\times [a_2,b_2]\times [a_3,b_3]\times (-j,j)\times \mathbb{R}\times \mathbb{R}\times \cdots\mid j\in \mathbb{N}\}

which doesn’t have a finite subcover.

Theorem of Homeomorphism over locally compact Hausdorff spaces

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

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

$Y$ is unique in the following sense:

If $Y'$ is another such space, then there is a homeomorphism between $Y$ and $Y'$ $f(x)=x$ for any $x\in X$ .

Proof for existence of Y

Let $Y=X\cup \{\infty\}$ . as a set.

Topology on $Y$ :

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

$U\subseteq X$ and $U$ is open in $X$ . ( $\infty\notin U$ )
$Y-U\subseteq X$ and $Y-U$ with the subspace topology from $X$ is compact. ( $\infty\in U$ )

We need to show that there is a topology on $Y$ that satisfies the definition.

$\emptyset\in \mathcal{T}$ because $\emptyset\subseteq X$ , $Y\in \mathcal{T}$ because $Y-Y=\emptyset$ is compact.
This topology is closed with respect to finite intersections.
Consider $U,U'\in \mathcal{T}$ . Then $U\cap U'$ is open.

Case 1: $\infty\notin U,U'$ , then $U\cap U'$ is open in $X$ .
Case 2: $\infty\in U,U'$ both, then $Y-U$ , $Y-U'$ with subspace topology from $X$ are compact. Note that $Y-(U\cap U')=(Y-U)\cup (Y-U')$ is compact.
Case 3: $\infty\in U$ but not $U'$ , then $Y-U$ with subspace topology from $X$ is compact. So $Y-U\subseteq X$ is compact, and $U'\subseteq X$ is open. And $Y-U\subseteq X$ is closed because $X$ is Hausdorff. and $Y-U\subseteq X$ is compact. So $U\cap U'$ is open in our topology.

Example for such Y

Consider $X=(0,1)$ , we can build $Y=(0,1)\cup \{\infty\}=S^1$ .

Proof for the theorem

First we prove the uniqueness of $f$ .

$Y=X\cup \{\infty\}$ .

$Y'=X\cup \{\infty'\}$ .

the function $f:Y\to Y'$ is defined $f(x)=x$ for $x\in X$ and $f(\infty)=\infty$ .

We show that $f$ is a homeomorphism.

If $f$ is clearly a bijection, we need to show $U\subseteq Y$ is open if and only if $f(U)\subseteq Y'$ is open.

Suppose $U\subseteq Y$ is open.

Case 1, $\infty\notin U$ , so $U\subseteq X$ . (Note $X$ is in $Y$ ) is open. $\{\infty'\}$ is closed in $Y'$ (since $Y'$ is Hausdorff). $f(U)=U\subseteq X$ (Note $X$ is in $Y'$ ) is open. So $U\subseteq X'$ is open.

Case 2, $\infty\in U$ .

Since $U\subseteq Y$ is open, then $Y-U$ is closed. Note that $Y-U$ is closed in $Y$ and $Y$ is Hausdorff. So $Y-U$ is also compact.

Since $\infty\in U$ , then $Y-U\subseteq X$ .

This implies that $f(Y-U)\subseteq X\subset Y'$ is also compact.

Since $Y-U\subseteq Y'$ and $Y'$ is Hausdorff, then $Y-U\subseteq Y'$ is closed.

So $f(U)=U\subseteq Y'$ is open.

Suppose $f(U)\subseteq Y'$ is open.