TITLE: Showing one point compactification is unique up to homeomorphism
QUESTION [22 upvotes]: First for clarity I'll define things as I'm familiar with them:

A compactification of a non-compact topological space $X$
is a compact topological space $Y$
such that $X$
can be densley embedded in $Y$
.
In particular a compacitifaction is said to be a one-point compactification if $\left|Y\backslash X\right|=1$
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:

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

REPLY [2 votes]: 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 only a subset 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.