## T_2 and compactness

Any compact $T_2$ (= Hausdorff) space $(X, \tau)$  is on the brink of being compact and being $T_2$:

(1) If we endow $X$ with a topology $\tau'$ distinct from $\tau$ such that $\tau' \supseteq \tau$, then $(X, \tau')$ is not compact any more; and
(2) if we endow $X$ with a topology $\tau'$ distinct from $\tau$ such that $\tau' \subseteq \tau$, we lose the the Hausdorff property.

More concisely, these statements say: compact $T_2$ spaces are maximal compact and minimal $T_2$. They are both quite straightforward to prove.

Which leads to the question about the converse of these statements, i.e.

(1) Is every maximal compact space $T_2$?
(2) Is every minimal $T_2$ space compact?

As for the first question, the answer is No. One example is the Alexandroff compactification $\mathbb{Q}^*$ of the rationals with the Euclidean topology. Recall that the Alexandroff compactification of a Hausdorff space $(X, \tau)$ is formed by setting $X^* = X \cup \{\infty\}$ where $\infty \notin X$ and endowing $X^*$ with the topology generated by $\tau \cup \{X^* \setminus F: F \textrm{ is compact in }X\}$.

The key to showing that $\mathbb{Q}^*$ is maximal compact is the following
Lemma. A space is maximal compact if and only if every compact subset is closed.

Moreover, we cannot separate any $q \in \mathbb{Q}$ from $\infty \in \mathbb{Q}^*$ by disjoint neighborhoods, so $\mathbb{Q}^*$ is not Hausdorff.

The lemma above characterizes maximal compact spaces. On the other hand I lack a tool for tackling minimal Hausdorffness. So I can answer Question (1) above, but for its ”dual” question whether minimal $T_2$ implies compactness, I don’t have a clue. Any hints are appreciated!