clarification about one point compactification Munkres in his book Topology writes

$X$ is locally compact Hausdorff iff there exists a space $Y$ such that (1) $X$ is a subspace of $Y$,(2) the set $Y-X$ consists of a single point if (3)$Y$ is a compact Hausdorff space.

Then, he also writes:

If $X$ itself should happen to be compact, then the space $Y$ is obtained from $X$ by adjoining a single isolated point.

So, am I correct that if $X$ indeed happens to be compact, the set consisting of the adjoined isolated point, say $\{\infty\}$, is open because $\infty$ is isolated  but it is also closed because $Y$ is Hausdorff?
 A: $\{\infty\}$ is open in $Y$ because its complement is $X$, which is compact and thus closed (because $Y$ is Hausdorff).
$\{\infty\}$ is closed in $Y$ because indeed a Hausdorff space $Y$ is in particular $T_1$, so all singletons are closed.
So if you you consider these 3 facts to be the definition of the one-point compactification, for compact $X$ we have no choice but to add an isolated point (and note that $X$ also needs to be Hausdorff in this case, because $Y$ is and $X$ is a subspace by 1)).
As an aside:
It's more common to say that a compact space does not have any compactifications and demand that for the one-point compactification of a space $X$: 


*

*$Y$ is compact Hausdorff

*$X$ is a dense subspace of $Y$ 

*$Y \setminus X$ consist of a single point $\infty$.
In this case such a space $X$ (for $Y$ to exist) must be Hausdorff (from 1 and 2 again) and locally compact (as $X$ is open in the compact space $Y$, again because singletons are closed in Hausdorff spaces), but not compact (because otherwise $X$ is already closed in $Y$ and cannot be dense anymore). And for such $X$ we can actually construct a $Y$ that obeys these 3 points, and one can show (in both cases) that it is essentially unique as well. 
