Abstract Nonsense

Crushing one theorem at a time

Stone-Weierstrass Theorem (Alexandroff compactification of locally compact Hausdorff spaces)


As the title suggest we shall make a brief expedition into the world of locally compact Hausdorff spaces. We do this not for fun and games (although it is fun) but mainly so that we speak of the Alexandroff compactification of such spaces. This concept enables us to weaken our theorems (in a sense) regarding \mathcal{C}[X,\mathbb{R}] to spaces which are merely locally compact Hausdorff.

So, without further ado:

Locally Compact: Let X be a topological space, then X is locally compact if given any x\in X there exists a neighborhood U of X such that \overline{U} is a compact subspace of X.

We are not concerned (in this setting) much with the actual structural details of locally compact Hausdorff spaces besides the fact that they are candidates for a particularly nice compactification. This is summed up in the following theorem.

Theorem (Alexandroff  compactification): Let X be a locally compact Hausdorff space and let \infty be any point not in X. Define X_\infty=X\cup\{\infty\} as the topological space whose open sets are 1) the open sets in X, 2) the compliment of a compact subspace of X, or 3) latex X_\infty itself. We claim that this space is compact and Hausdorff.

Proof:

We first prove that it is, in fact, a topological space. So, let \mathcal{E} a class of open sets in X_\infty. We of course may assume WLOG that X_\infty\notin\mathcal{E} (otherwise its union is merely X_\infty). Also, if \mathcal{E} contains only sets of the form 1) then since X is a topological space clearly \displaystyle \bigcup_{E\in\mathcal{E}}E is a set of the form 1) and thus open. Also, if \mathcal{E} consists entirely of sets of the form 2) then  \displaystyle \bigcup_{E\in\mathcal{E}}E=\left[\bigcap_{E\in\mathcal{E}}E\prime\right]\prime and since \displaystyle \bigcap_{E\in\mathcal{E}}E\prime is a closed subspace of E_0\prime for any E_0\in\mathcal{E} we see that \displaystyle \bigcap_{E\in\mathcal{E}}E\prime is a compact subspace of X and so \displaystyle \bigcup_{E\in\mathcal{E}}E=\left[\bigcap_{E\in\mathcal{E}}E\prime\right]\prime is the compliment of a compact subspace of X and thus of the form 3). Lastly, if \mathcal{E} contains sets of both types we see that \displaystyle \bigcup_{E\in\mathcal{E}}E=\left[\bigcap_{E\in\mathcal{E}}E\prime\right]\prime is still a closed subspace of E_0 for any E_0 of the form 2) and so it is of the form 3).

A similar analysis shows that the topology is closed under finite unions.

We now prove that it’s compact. To see this let \Omega=\left\{G_j\right\}_{j\in\mathcal{J}}. If X_\infty\Omega then clearly \{X_\infty\} is a finite subcover. Thus, we may assume that X_\infty\notin\Omega. But, since \infty\in G_{j_0} for some j\in\mathcal{J} and G_{j_0}\ne X_\infty we must have that G_{j_0} is a set of the form 2). It follows that X_\infty-G_{j_0} is a compact subspace of X. Thus, we have that \left\{G_j\cap X\right\} is an open cover for X_\infty G_{j_0} and since it is compact we are afforded a finite subcover X\cap G_{j_1},\cdots,X\cap G_{j_n} it follows that \left\{G_{j_0},G_{j_1},\cdots,G_{j_n}\right\} is the finite subcover we sought.

It remains to show that X_\infty is Hausdorff. So, let x,y\in X_\infty be distinct. It is clear that if x,y\in X that the open sets afforded by X‘s Hausdorffness work as the open sets for X_\infty‘s Hausdorffness. Thus, we must only show that x and \infty may be separated for any x\in X. But, this follows immediately from X‘s local compactness since there exists some neighborhood U of x such that \overline{U} is compact in X. Clearly U and X_\infty-\overline{U} are disjoint neighborhoods of x and \infty respectively.  \blacksquare

We now prove the obvious:

Theorem: Let \iota:X\hookrightarrow X_\infty be the inclusion map. Then, \iota is an embedding.

Proof: Clearly \iota is injective. So, let U be open in X_\infty. If U=X_\infty then \iota^{-1}(U)=X. If U is of the form 1) then \iota^{-1}(U)=U is open. Lastly, if U is of the form 3) then U=(X-C)\cup\{\infty\} where C is a compact subspace of X. It follows that \iota^{-1}(U)=X-C and since X is Hausdorff we know that C is closed and so X-C is open. It is clear that \iota is an open mapping. The conclusion follows. \blacksquare

Corollary: (X,\mathcal{I})\approx (X,S) where \mathcal{I} is the initial topology on X and S is the subspace topology on X as being a subspace of X_\infty.

The Alexandroff compactification of X is an important tool for proving structural facts about X, not because it adds any knowledge of X but that it eases many proofs.

Theorem: Let X be a locally compact Hausdorff space, then X is Tychonoff.

Proof: Since X is locally compact Hausdorff it has an Alexandroff compactification X_\infty. Now, since X_\infty is compact Hausdorff it is normal, and thus Tychonoff. But, since a subspace of a Tychonoff space is Tychonoff we see that (X,S) is a Tychonoff  space and since (X, \mathcal{I})\approx (X,S) the conclusion follows. \blacksquare

Corollary: Every locally compact Hausdorff space is regular.

Our next theorem is mainly used to make easier the proof of the theorem that immediately follows it.

Theorem: Let X be a Hausdorff space, then X is locally compact if and only if for every x\in X there exists some compact subspace C of X such that x\in C^{\circ}.

Proof: Suppose that X is locally compact, then given any x\in X we have that there exists some neighborhood U of x such that \overline{U} is compact. But, x\in U=U^{\circ}\subseteq\left(\overline{U}\right)^{\circ}.

Conversely, suppose that X has the property that given any x\in X there exists some compact subspace C of X such that x\in C^{\circ}. So, let U be a neighborhood of x such that U\subseteq C^{\circ}\subseteq C. It follows that \overline{U}\subseteq\overline{C}=C and so \overline{U} is a closed subspace of a compact subspace and thus compact. The conclusion follows. \blacksquare

From this we can easily prove the following:

Theorem: Let \varphi:X\to Y be an open continuous surjection, then X is locally compact and Y is Hausdorff then Y is locally compact.

Proof: Merely note that given any \varphi(x)\in Y there exists some neighborhood U of x such that \overline{U} is compact. So, by \varphi‘s openness we have that \varphi(U) is a neighborhood of \varphi(x) which is contained in \varphi\left(\overline{U}\right) which is compact. It follows that \varphi(x)\in\varphi(U)=\varphi(U)^{\circ}\subseteq\varphi\left(\overline{U}\right)^{\circ}. The conclusion follows. \blacksquare.

Corollary: If \displaystyle X=\prod_{j\in\mathcal{J}}X_j is locally compact and Hausdorff so is X_j for each j\in\mathcal{J}.

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March 29, 2010 - Posted by | General Topology, Topology, Uncategorized | , , , ,

4 Comments »

  1. Dear Alex:

    “A similar analysis shows that the topology is closed under finite unions.”

    It should be “under finite intersections”.

    By the way, great blog!

    Best regards,

    Daniel

    Comment by Daniel Lopez Aguayo | October 19, 2010 | Reply

    • Daniel,

      Thank you, I will be sure to change that! And thanks also for the compliment! I always appreciate when people take the time to say something nice.

      Comment by drexel28 | October 19, 2010 | Reply

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