Question about density proof in Alexandroff compactification

I'm trying to understand a proof about density of a subset X in its one-point compactification Y. We can do this proof by contradiction, suppose we don't have closure of X = Y. this implies that the closure of X = X.

why? anyone can help me?

thanks

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Suppose $cl(X)\not=Y$. We know $X\subseteq cl(X)$ so we get $cl(X)=X$ and $\infty \notin cl(X).$ So by definition of closure, there exists a (wlog, open) neighborhood U of $\infty$ s.t. $U \cap X=\emptyset$. The topology of the extension is defined to be all open subsets of X together with all sets V that contain $\infty$ and such that X\V is closed and compact. $\infty \in U$ so U is of the second kind of open sets, meaning X\U is closed and compact. But remember that $U \cap X=\emptyset$ so X\U=X. We get that X is compact, in contradiction to the assumption (if X was compact, we wouldn't have needed the compactification).
Y is X plus one point. We know that $X \subseteq cl(X)$, or in other words cl(X) is not smaller than X. Also, of course it's contained in our space, Y (it can't contain "elephant" as element). Mathematically, $cl(X) \subseteq Y$. So what cl(X) can be? We're left with two options: X, or Y (because there no options between them - they differ by a point). So if cl(X) isn't Y, it must be X. –  Idan Oct 2 '12 at 13:09
You’re making it much harder than it really is. $Y=X\cup\{p\}$, where $p$ is the new point. To show that $X$ is dense in $Y$, you need only show that every open neighborhood of $p$ has non-empty intersection with $X$. Go back to the definition of the one-point compactification and see why this is true: what are the open neighborhoods of $p$? Why must each of them have non-empty intersection with $X$? It has to do with the fact that $X$ is not compact.