NoteNextra-origin/content/Math4201/Math4201_L31.md

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:

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 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

1. U\subseteq X and U is open in X. (\infty\notin U)
2. 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.

1. \emptyset\in \mathcal{T} because \emptyset\subseteq X, Y\in \mathcal{T} because Y-Y=\emptyset is compact.
2. 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.

1. 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.

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