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 $\operatorname{cl}(X) = Y$.
This implies that $\operatorname{cl}(X) = X$. 
Why? Can anyone help me?
Thanks
 A: 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.
A: Suppose $\operatorname{cl}(X)\not=Y$. We know $X\subseteq \operatorname{cl}(X)$ so we get $\operatorname{cl}(X)=X$ and $\infty \notin \operatorname{cl}(X).$ 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 $X\setminus V$ is closed and compact. $\infty \in U$ so $U$ is of the second kind of open sets, meaning $X \setminus U$ is closed and compact. But remember that $U \cap X=\emptyset$ so $X\setminus U=X$. We get that $X$ is compact, in contradiction to the assumption (if $X$ was compact, we wouldn't have needed the compactification).
