Showing one point compactification is unique up to homeomorphism 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!
 A: 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.
A: 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.)
A: 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 $Y.$
