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!

Solutions Collecting From Web of "Showing one point compactification is unique up to homeomorphism"

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.)