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!