# Showing one point compactification is unique up to homeomorphism

First for clarity I'll define things as I'm familiar with them:

1. A compactification of a non-compact topological space $X$ is a compact topological space $Y$ such that $X$ can be densley embedded in $Y$ .

2. In particular a compacitifaction is said to be a one-point compactification if $\left|Y\backslash X\right|=1$

3. The Alexandroff one-point compactification of a a topological space $\left(X,\mathcal{T}_{X}\right)$ is the set $X^{*}=X\cup\left\{ \infty\right\}$ for some element $\infty\notin X$ given the topology $$\mathcal{T}^{*}:=\mathcal{T}_{X}\cup\left\{ U\subseteq X^{*}\,|\,\infty\in U\,\wedge\, X\backslash U\,\mbox{is compact and closed in }\left(X,\mathcal{T}_{X}\right)\right\}$$ If $\left(X,\mathcal{T}_{X}\right)$ is a Hausdorff space one can omit the requirement that $X\backslash U$ is closed.

It is easy to show that given two choices of elements $\infty_{1},\infty_{2}\notin X$ the one-point compactifications $X\cup\left\{ \infty_{1}\right\}$ and $X\cup\left\{ \infty_{2}\right\}$ with the topology defined as that of the Alexandroff one-point compactification are homeomorphic. What I'm wondering is why isn't there another possible way to define the topology on $X^{*}$ that would also yield a compactification (which is in particular not homeomorphic to the Alexandroff one-point topology)

As far as I see it there are two approaches to answering this question:

1. Show that any topology on $X^{*}$ that yields a compact space in which $X$ is dense is homeomorphic to $\mathcal{T}^{*}$.

2. Show it's not possible to consturct any other topology on $X^{*}$ that results in a compactification.

I'm quite interested in seeing the reasoning to both approaches if possible. Thanks in advance!

• Note that for any finite topological space $X$, the space $X\cup\{\ast\}$ with any topology extending the topology on $X$ will be a compact space. For instance, take $X$ to be a two-element set with a non-trivial topology. The discrete extension of $X$ by one element is not homeomorphic to the disjoint union $X\sqcup\{\ast\}$. Does this example not count however because $X$ is compact? – Dan Rust Jun 22 '13 at 19:45
• @Daniel: It’s not a compactification: $X$ isn’t a dense subset of it. – Brian M. Scott Jun 22 '13 at 19:59
• @BrianM.Scott I was hoping to see a response from you, I've come to think of you as a topological guru. Hoping you might have the time to answer the original question :) – Serpahimz Jun 22 '13 at 20:03
• @BrianM.Scott very true. Thanks for pointing that out. – Dan Rust Jun 22 '13 at 20:03
• Don't you want to assume that $X$ is locally compact at some point? Without this, the "compactification" does not come out as a compact space... – Jakub Konieczny Jun 22 '13 at 20:14

## 1 Answer

You get the uniqueness result if the space is Hausdorff.

Let $\langle X,\tau\rangle$ be a compact space. Suppose that $p\in X$ is in the closure of $Y=X\setminus\{p\}$, and let $\tau_Y$ be the associated subspace topology on $Y$; $\langle X,\tau\rangle$ is then a compactification of $\langle Y,\tau_Y\rangle$.

Suppose that $p\in U\in\tau$, and let $V=U\cap Y$. Then $\varnothing\ne V\in\tau_Y$, so $Y\setminus V$ is closed in $Y$. Moreover, $Y\setminus V=X\setminus U$ is also closed in $X$, which is compact, so $Y\setminus V$ is compact. That is, every open nbhd of $p$ in $X$ is the complement of a compact, closed subset of $Y$. Thus, if $\tau'$ is the topology on $X$ that makes it a copy of the Alexandroff compactification of $Y$, then $\tau\subseteq\tau'$.

Now let $K\subseteq Y$ be compact and closed in $Y$, and let $V=Y\setminus K\in\tau_Y$. If $X\setminus K=V\cup\{p\}\notin\tau$, then $p\in\operatorname{cl}_XK$. If $X$ is Hausdorff, this is impossible: in that case $K$ is a compact subset of the Hausdorff space $X$ and is therefore closed in $X$. Thus, if $X$ is Hausdorff we must have $\tau=\tau'$, and $X$ is (homeomorphic to) the Alexandroff compactification of $Y$.

If $X$ is not Hausdorff, however, we can have $\tau\subsetneqq\tau'$. A simple example is the sequence with two limits. Let $D$ be a countably infinite set, let $p$ and $q$ be distinct points not in $D$, and let $X=D\cup\{p,q\}$. Points of $D$ are isolated. Basic open nbhds of $p$ are the sets of the form $\{p\}\cup(D\setminus F)$ for finite $F\subseteq D$, and basic open nbhds of $q$ are the sets of the form $\{q\}\cup(D\setminus F)$ for finite $F\subseteq D$. Let $Y=D\cup\{q\}$. Then $Y$ is dense in $X$, and $X$ is compact, and $Y$ itself is a closed, compact subset of $Y$ whose complement is not open in $X$.

Improved example (1 June 2015): Let $D$ and $E$ be disjoint countably infinite sets, let $p$ and $q$ be distinct points not in $D\cup E$, let $X=D\cup E\cup\{p,q\}$, and let $Y=D\cup E\cup\{q\}$. Points of $D\cup E$ are isolated. Basic open nbhds of $q$ are the sets of the form $\{q\}\cup D\cup (E\setminus F)$ for finite $F\subseteq E$, and basic open nbhds of $p$ are the sets of the form $\{p\}\cup\big((D\cup E)\setminus F\big)$ for finite $F\subseteq D\cup E$. Then $Y$ is a non-compact dense subspace of the compact space $X$, so $X$ is a (non-Hausdorff) compactification of $Y$. Let $K=\{q\}\cup E$. Then $K$ is a compact closed subset of $Y$, but $X\setminus K=\{p\}\cup E$ is not open in $X$.

(This avoids the question of whether it’s legitimate to look at the Alexandrov compactification of a compact space.)

• I think you might have a small typo, at the end of the first line of the third paragraph I believe it should say $X\backslash K=V\cup\left\{ p\right\}$? – Serpahimz Jun 22 '13 at 21:10
• @Serpahimz: Yep; good catch. Thanks. – Brian M. Scott Jun 22 '13 at 21:12
• Excellent answer, simple and elegant, thanks as ever Brian. – Serpahimz Jun 22 '13 at 21:14
• @Serpahimz: You’re welcome, and thank you. – Brian M. Scott Jun 22 '13 at 21:17
• @Serpahimz: It means that $K$ is closed in $Y$ but not in $X$. That’s possible iff $p$ is a limit point of $K$ in $X$. $V\in\tau_Y$, so there is a $W\in\tau$ such that $V=W\cap Y$. $W$ must be either $V$ or $V\cup\{p\}$, and the latter isn’t in $\tau$, so $V\in\tau$. Thus, each $y\in Y\setminus K$ has $V$ as a $\tau$-open nbhd disjoint from $K$. If $K$ is not $\tau$-closed, that leaves $p$ as the only possible member of the non-empty set $(\operatorname{cl}_XK)\setminus K$. – Brian M. Scott Jun 22 '13 at 22:11