## 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 to spaces which are merely locally compact Hausdorff.

So, without further ado:

**Locally Compact:** Let be a topological space, then is locally compact if given any there exists a neighborhood of such that is a compact subspace of .

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 be a locally compact Hausdorff space and let be any point not in . Define as the topological space whose open sets are 1) the open sets in , 2) the compliment of a compact subspace of , or 3) latex itself. We claim that this space is compact and Hausdorff.

**Proof:**

We first prove that it is, in fact, a topological space. So, let a class of open sets in . We of course may assume WLOG that (otherwise its union is merely ). Also, if contains only sets of the form 1) then since is a topological space clearly is a set of the form 1) and thus open. Also, if consists entirely of sets of the form 2) then and since is a closed subspace of for any we see that is a compact subspace of and so is the compliment of a compact subspace of and thus of the form 3). Lastly, if contains sets of both types we see that is still a closed subspace of for any 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 . If then clearly is a finite subcover. Thus, we may assume that . But, since for some and we must have that is a set of the form 2). It follows that is a compact subspace of . Thus, we have that is an open cover for and since it is compact we are afforded a finite subcover it follows that is the finite subcover we sought.

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

We now prove the obvious:

**Theorem: **Let be the inclusion map. Then, is an embedding.

**Proof: **Clearly is injective. So, let be open in . If then . If is of the form 1) then is open. Lastly, if is of the form 3) then where is a compact subspace of . It follows that and since is Hausdorff we know that is closed and so is open. It is clear that is an open mapping. The conclusion follows.

**Corollary:** where is the initial topology on and is the subspace topology on as being a subspace of .

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

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

**Proof:** Since is locally compact Hausdorff it has an Alexandroff compactification . Now, since is compact Hausdorff it is normal, and thus Tychonoff. But, since a subspace of a Tychonoff space is Tychonoff we see that is a Tychonoff space and since the conclusion follows.

**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 be a Hausdorff space, then is locally compact if and only if for every there exists some compact subspace of such that .

**Proof:** Suppose that is locally compact, then given any we have that there exists some neighborhood of such that is compact. But, .

Conversely, suppose that has the property that given any there exists some compact subspace of such that . So, let be a neighborhood of such that . It follows that and so is a closed subspace of a compact subspace and thus compact. The conclusion follows.

From this we can easily prove the following:

**Theorem:** Let be an open continuous surjection, then is locally compact and is Hausdorff then is locally compact.

**Proof:** Merely note that given any there exists some neighborhood of such that is compact. So, by ‘s openness we have that is a neighborhood of which is contained in which is compact. It follows that . The conclusion follows. .

**Corollary:** If is locally compact and Hausdorff so is for each .

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 |

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 |

[…] Let be a semigroup then if is defined to be the set (where is a formal symbol–compare to Alexandroff Compactification) with the operation by for all and for every . Then, is a […]

Pingback by Interesting Example of a Monoid « Abstract Nonsense | March 25, 2011 |

[…] we consider , which topologically is just the one-point compactification of . Clearly is a chart on . We claim that […]

Pingback by Riemann Surfaces (Pt. III) « Abstract Nonsense | October 2, 2012 |