Let $(X,\tau)$ be a compact Hausdorff space, and define $Y=X\setminus\lbrace x\rbrace$ for some (fixed) $x\in X$. Let $\sigma$ denote the subspace topology on $Y$ induced by $\tau$. We say that a space is locally compact if each point in the space has a compact neighbourhood.

In Exercise 7.8.15 of these notes it is asked to show that $Y$ is locally compact Hausdorff, and that its one-point compactification $Y_\infty$ is homeomorphic to $X$.

To prove the first statement we use the following proposition:

Proposition. If $X$ is locally compact Hausdorff and $Y\subseteq X$ is open, then $Y$ is locally compact.

First, since $X$ is Hausdorff we know $\lbrace x\rbrace$ is closed, so $Y$ is open in $X$. Since $X$ is compact, it is also locally compact, and so using the above proposition we know that $Y$ is locally compact. Any subspace of a Hausdorff space is also Hausdorff, so the first result follows.

I am unsure on showing $Y_\infty \cong X$. The one-point compactification $(Y_\infty,\sigma_\infty)$ is defined as follows: $Y_\infty = Y\cup\lbrace \infty\rbrace$ for some $\infty\notin Y$, and $\sigma_\infty = \sigma \cup \lbrace Y_\infty\setminus K:K\subseteq Y \text{ closed and compact} \rbrace$. A compactification usually comes with a continuous embedding $\iota:Y\to Y_\infty$ such that $\iota(Y)$ is dense in $Y_\infty$. I am supposing that in this case the map is simply given by the inclusion mapping $\iota: Y\hookrightarrow Y_\infty:y\mapsto y$.

My guess for a homeomorphism would be the following map: $f:X\to Y_\infty$, where $Y\ni y\mapsto y$ and $x\mapsto \infty$. This is clearly a bijection. It will therefore suffice to show that $f$ is continuous. It will automatically follow that $f$ is closed (and thus a homeomorphism) because $X$ is compact and $Y_\infty$ is Hausdorff (because $Y$ is Hausdorff and locally compact). However that's where I'm having some trouble.

In light of the continuity of $\iota$, which is simply the restriction of $f$ to $Y$, I think it suffices to show that $f$ is continuous at $x$. So let $V$ be an open neighbourhood of $\infty$. Then we can find a (closed) compact set $K\subseteq Y$ such that $V=Y_\infty\setminus K$. Since $K$ does not contain $\infty$ we have $V=\lbrace \infty \rbrace \cup Y\setminus K$. In other words, we can find an open set $W\in\sigma$ such that $V=\lbrace\infty\rbrace \cup W$. Looking at the open sets of $Y$, we find that they are exactly the open sets of $X$ punctured at $x$: $\sigma = \lbrace U\cap (X\setminus\lbrace x\rbrace) :U\in\tau\rbrace = \lbrace U\setminus\lbrace x\rbrace :U\in \tau\rbrace$. Thus we can write $V=\lbrace \infty\rbrace \cup U\setminus\lbrace x\rbrace$ for some $U\in\tau$. This shows that any open neighbourhood of $\infty$ is an open neighbourhood of $x$, where we swap $x$ out with $\infty$. This is exactly what the function $f$ does, so $f(U)=V$. This in particular shows that $f$ is continuous at $x$. The result follows.

Is this a valid proof of the result? At the least it is a bit sloppy, so any tips on how to clean it up are also more than welcome (e.g. if there is a more succinct argument, I would like to see it). Thanks so much in advance!


2 Answers 2


Your proof works as far as I can tell. Note however that the set $U$ you construct can be explicitly written down. In fact, you have $U = X \setminus{K}$ as can be checked by applying $f$. You would also have obtained this by calculating $$f^{-1}(Y_\infty\setminus{K}) = f^{-1}(\{\infty\} \cup Y\setminus{K}) = \{x\} \cup Y\setminus{f^{-1}(K)} = X\setminus{K}.$$ This is essentially what you did, this way of writing it down is just a bit shorter.

And when you say, you "think" that checking continuity at $x$ is enough, you should convince yourself (via a proof) that it is indeed quite easy to show that $f$ is continuous at any other point.


Yes, your argument is correct, though you really ought to finish the proof that $f$ is continuous. This can be done very easily by noting that if $U$ is open in $Y_\infty$, and $\infty\notin U$, then $f^{-1}[U]=U$ is open in $Y$ and hence in $X$, since $Y$ is open in $X$. You do need here the fact that $Y$ is open in $X$.

For a slightly different approach see this answer; it also includes an example showing that you really do need the space to be Hausdorff.

  • $\begingroup$ Thank you Professor Scott! Just to make sure I get the details: since $X$ is Hausdorff, the singleton $\lbrace x\rbrace$ is closed. And so, $Y=X\setminus\lbrace x\rbrace$ is open in $X$. Since $U$ is open in $Y$ we can find a set $V$ which is open in $X$ such that $U=V\cap Y$. Then we easily find that $X\setminus U = (X\setminus V)\cup (X\setminus Y)$, which is closed in $X$, and so $U$ is open. Together with what I did in the OP this shows that $f$ is continuous on all of $X$, am I right? I will check out your linked answer as well! $\endgroup$
    – Nesta
    Aug 18, 2016 at 17:23
  • 1
    $\begingroup$ @Nesta: You’re right, but you don’t have to work that hard: $V\cap Y$ is the intersection of two open sets in $X$, so it’s automatically open in $X$. $\endgroup$ Aug 18, 2016 at 17:30
  • $\begingroup$ Ah of course, I was confused for a second. Thanks again! $\endgroup$
    – Nesta
    Aug 18, 2016 at 17:34
  • $\begingroup$ @Nesta: You’re welcome! $\endgroup$ Aug 18, 2016 at 17:41

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