On one-point compactification and closed subsets Is it true that of we have a closed subset C of a space X, then C will also be closed in the one-point compactification of X as well?
 A: I will assume that $X$ is not compact, so that $X^*$, the one-point compactification of $X$, is not equal to $X$, and that $p$ is the new point (sometimes called the point at infinity). By definition the open nbhds of $p$ in $X^*$ are the sets $X^*\setminus K$ such that $K$ is a compact subset of $X$. 
Let $C$ be any closed, non-compact subset of $X$; there is at least one such set, namely, $X$ itself. Let $K$ be any compact subset of $X$. If $C\subseteq K$, then $C$, being a closed subset of a compact set, would be compact, so $C\nsubseteq K$, and therefore $C\cap(X^*\setminus K)\ne\varnothing$. Thus, no open nbhd of $p$ is disjoint from $C$, so $p\in\operatorname{cl}_{X^*}C$, and $C$ is not closed in $X^*$.
It follows that if $C$ is a closed subset of $X$, then $C$ is closed in $X^*$ if and only if $C$ is compact.
A: I know it is a bit late but I hope that my explanation is nice and clear.
Let $X^* := X \cup \{\infty\}$ be the Alexandroff extension of $X$.
Then by definition the open sets of $X^*$ are either an open set $U$ of $X$,
or $X^* \setminus C$ for some compact set $C$ in $X$.
It follows then that the closed sets of $X^*$ are exactly the compact subset of $X$ or they are of the form $(X \setminus U) \cup \{\infty\}$ for some open set $U$ in $X$.
Now choose any closed subset $V$ of $X$.
As $V$ cannot contain our new point $\infty$ then $V$ is closed in $X^*$ if and only if it is also compact.
We can see from this result that if $X$ was already compact then every closed set in $X$ would also be compact in $X$, and hence closed in $X^*$ also.
