# 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? Commented Jun 22, 2013 at 19:45
• @Daniel: It’s not a compactification: $X$ isn’t a dense subset of it. Commented Jun 22, 2013 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 :) Commented Jun 22, 2013 at 20:03
• @BrianM.Scott very true. Thanks for pointing that out. Commented Jun 22, 2013 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... Commented Jun 22, 2013 at 20:14

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 $$\{q\}\cup (E\setminus F)$$ 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 $$X\setminus K=\{p\}\cup D$$ 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 $X\backslash K=V\cup\left\{ p\right\}$? Commented Jun 22, 2013 at 21:10
• @Serpahimz: Yep; good catch. Thanks. Commented Jun 22, 2013 at 21:12
• Excellent answer, simple and elegant, thanks as ever Brian. Commented Jun 22, 2013 at 21:14
• @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 $(\operatorname{cl}_XK)\setminus K$. Commented Jun 22, 2013 at 22:11
• @PatrickR: Yes, I seem accidentally to have included $D$ in the nbhds of $q$; it should be okay now. Thanks for catching it. I also fixed a typo in the description of $X\setminus K$. Commented Sep 5, 2020 at 16:27

Brian has already shown the uniqueness of one-point compactifications in the Hausdorff case. Here is a simple example of non-uniqueness in the non-Hausdorff case.

Let $$X$$ be a non-compact topological space. Take a point $$\infty$$ not in $$X$$ and form the (non-Hausdorff) topological space $$Y=X\cup\{\infty\}$$ where open sets in $$X$$ remain open in $$Y$$ and the only nbhd of $$\infty$$ is all of $$Y$$.

$$X$$ is embedded in $$Y$$ and is dense in $$Y$$. And $$Y$$ is compact because any open set containing $$\infty$$ is all of $$Y$$. So $$Y$$ is a compactification of $$X$$ in the OP's sense. But in general it will be a strictly weaker topology than the Alexandroff compactification. For example if $$X$$ is $$T_1$$, the Alexandroff compactification would also contain all cofinite sets containing $$\infty$$ as nbhds of $$\infty$$.

One can cook up similar examples by restricting the nbhds of $$\infty$$ to a subcollection of the complements of closed compact subsets of $$X$$. It is also the case that if $$X$$ is not compact, every one-point compactification of $$X$$ is an open embedding. This is shown here. So the topology on $$Y=X\cup\{\infty\}$$ constructed above is the smallest topology that is a one-point compactification of $$X$$. The Alexandroff compactification is the largest topology on $$Y$$ that is a one-point compactification of $$X$$. And any one-point compactification of $$X$$ will have a topology intermediate between these two.

Following Munkres, the uniqueness up to topological equivalence of Alexandrov-compactification follows:

Assuming $$X^{*}:=X \cup \{\infty\},$$ where $$\infty \notin X$$ is proven to be a one-point compactification of X, with the topology stated above. Let $$Y$$ be arbitrary one-point compactification of $$X.$$ Meaning $$Y \setminus X = \{p\}$$ is singleton, Y is Hausdorff compact and $$X$$ is subspace of $$Y.$$ Directly defining the obvious homeomorphism between them: $$f: X^{*} \to Y$$ as $$f(x):=\begin{cases}x, \hspace{0.2cm} x \in X \\ p, \hspace{0.2cm} x \in \{\infty\} \end{cases}.$$ Then $$f$$ is bijective since $$g : Y \to X^{*}$$ defined as $$g(x) = \begin{cases} x, \hspace{0.2cm} x \in X \\ p, \hspace{0.2cm} x \in \{p\} \end{cases}$$ is its inverse map. Furthermore by the symmetry of definitions of $$f$$ and $$g$$ it suffices to show that $$f$$ is open map. I.e. If $$U$$ is open in $$X^{*},$$ then $$f[U]$$ is open in $$Y.$$ Indeed if we do so, the same procedure for $$\{p\}$$ and $$g$$ in the place of $$\{\infty\}$$ and $$f$$ will prove that if $$V$$ is open in $$Y,$$ then $$f^{-1}[V]$$ is open in $$X^{*}.$$ Then both $$f$$ and $$f^{-1}$$ are continuous and so $$f$$ is homemorphism.

Let $$U$$ be open in $$X^{*}.$$ If $$\infty \notin U,$$ then $$f[U]=\operatorname{Id}_{X}[U]=U.$$ Now $$U$$ is open in $$X^{*}$$ and subset of $$X.$$ Combined with the fact that $$X$$ is a subspace of $$X^{*}$$ we get that $$U$$ is open in $$X.$$ Now $$X$$ is open in $$Y.$$ and so $$U$$ is also open in $$Y.$$ As desired.

If $$\infty \in U.$$ We get $$C:=X^{*} \setminus U$$ is closed set in $$X^{*}.$$ Thus $$C$$ is compact subspace of $$X^{*}.$$ Combined with $$C \subset X,$$ follows that $$C$$ is compact subspace of $$X.$$ Then, since $$X$$ is subspace of $$Y,$$ we get that $$C$$ is compact subspace of $$Y,$$ also. Now finally, using the fact that $$C \subset X,$$ the injectivity of $$f$$ (being bijection) thus $$f[C]=C.$$ Thus $$f[X^{*} \setminus U]=X^{*} \setminus U.$$ Now $$f[X^{*} \setminus U]=f[X^{*}]\setminus f[U].$$ Combined gives $$f[U]=Y \setminus C.$$ For the last we used the fact that $$f[X^{*}]=Y$$ from the surjectivity, and the formula $$A \setminus B = C \setminus D \Rightarrow B = A \setminus (C \setminus D),$$ given $$C \setminus D \subset A.$$ At last, since $$Y$$ is Hausdorff and $$C$$ is closed in $$Y$$ we get $$Y \setminus C = f[U]$$ is open in $$Y.$$