# Abstract Nonsense

## 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.

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 $j\in\mathcal{J}$.

March 29, 2010 -

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

2. […] 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 | Reply

3. […] 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 | Reply