A compactification of a space $X$ is an embedding $f:X \to Y$ so that (1) $Y$ is compact, (2) $f(X)$ is dense in $Y$. If furthermore, $Y\setminus f(X)$ is a single point, we say it is a one point compactification.
If we assume $X$ is a compact and Hausdorff space (thus LCH), does there exist a (non-Hausdorff) $Y$, such that there exists $f:X \to Y$ that is a one-point compactification of $X$?
If we require $Y$ to be Hausdorff, this is impossible see here. However, since compact does not imply closed in general, I'm not sure the case when we relax this condition. In particular, even if compactification exists, is it unique up to homeomorphism?
$S^{1}$ is an example, but I cannot think of a compactification. (Edit: I think taking $S_1 \sqcup S_1$ and quotient it out by a space so that every two points on the circle are combined besides 1 point might be an example, but I need to think more carefully, it might not work)
Best Answer
Let $Y$ be the set given by $X$ together with some other point $\infty$. Declare a set to be open if it's either open in $X$ or the whole space. Then the inclusion of $X$ into $Y$ is continuous and the closure of $X$ is all of $Y$ since $\{ \infty \}$ is not open. Moreover, $Y$ is trivially compact, as any open cover includes an open set containing $\infty$ but the only such set is $Y$ itself.
For what it's worth, in my opinion, the moral is that non-$T_1$ spaces are terrible, not anything interesting about compactifications.