# Maximal compact topology iff compact sets are closed

The theorem says: A compact space $(X,\tau)$ is maximal compact (i.e. no strictly larger topology on $X$ is compact) if and only if every compact subset of $X$ is closed in $\tau$.

For "if", I prove by contraposition. Let $U\subset X$ be a compact subset of $X$ and $U$ is not closed in $\tau$. Define a finer topology $\tau':=\tau\cup\{X\setminus U\}$, and we show that $(X, \tau')$ is compact.

Let $S$ be a subbasis of $(X,\tau')$ and let $S'\subseteq S$ be an open cover of $X$. By Alexander subbase theorem: https://en.wikipedia.org/wiki/Subbase#Alexander_subbase_theorem, it's enough to show that $S'$ has a finite subcover.

Since $X\setminus U\in\tau'$, it is the finite intersection of elements in $S$. But this is not good enough because ideally I want to express $X\setminus U$ (and $U$) as a finite intersection of sets in $S'$, in order obtain a finite subcover of $S'$. How can I get around this?

Let $X\setminus U=\cap_{k\leq n}S_k$ be a finite intersection of sets in $S$. Since $S'$ covers $X$, for each $k\leq n$, we have $S_k\subseteq\cup_{i\in I_k}S_i$, $S_k$ is a subset of some (possibly infinite) union of sets in $S'$. Hence $X\setminus U\subseteq\cup_{k\leq n}\cup_{i\in I_k}S_i$. This proof would work if I can somehow reduce $S_k$ to some finite union of sets in $S'$, but I'm not sure if that is at all possible.

I feel like I need to use the compactness of the original space $(X,\tau)$.

If $(X,\tau)$ is compact and there is some $C\subseteq X$ that is compact but not closed, then $\tau’=\tau \cup \{X\setminus C\}$ is a subbase for a strictly larger topology than $\tau$ that is also compact by Alexander’s subbase theorem:
Let $\mathcal{O}$ be any cover of $X$ by members of $\tau’$. If we only use members of $\tau$ we have a finite subcover because $(X,\tau)$ is compact. So we can assume $X\setminus C \in\mathcal{O}$ But then pick that open set plus the finitely many we need to cover its complement (as $C$ is compact under $\tau$) to get a finite subcover as well.
This argument is the essence of the proof in one direction: Suppose $(X,\tau)$ is maximally compact, then all compact subsets $C$ of $(X,\tau)$ are closed, otherwise we apply the argument to get a strictly larger topology (as the topology generated by $\tau’$ contains $X\setminus C$ which by assumption is not in $\tau$) that is also compact.
For the reverse: if $(X,\tau)$ is compact and suppose all its compact subsets are closed. Now take any topology $\tau’ \supsetneq \tau$. Let $O \in \tau’ \setminus \tau$. Then $X\setminus O$ is not closed in $(X,\tau)$ (or $O$ would be in $\tau$) so it’s not compact under $\tau$. So there is a cover $\mathcal{O}$ of $X\setminus O$ by sets from $\tau$ that has no finite subcover. But then $\mathcal{O} \cup \{O\}$ is a cover of $X$ under $\tau’$ that has no finite subcover as well, so $(X,\tau’)$ is not compact. So $X$ is maximally compact. (This direction does not use the compactness of the original topology..)