If $(X,\tau) $is compact and $\tau'\subseteq \tau$, then $(X,\tau')$ is compact.
I have already read several posts on the subject, but it is still unclear to me. The usual argument is:
"In a coarser space, more sets are compact, essentially because there are fewer open covers to need finite subcovers. That is, if a set is compact in the finer topology then it is compact in the coarser topology." (as found here What does compactness in one topology tell us about compactness in another (coarser or finer) topology?)
But still I am not very convinced, specifically because, if I go to a coarser topology, some open sets are missing with respect to the initial topology, and what if I needed those sets for the extract the fine subcover, what guarantees they aren't needed?
Best Answer
The usual argument is the proof, which is very short and straightforward:
Let $\mathscr{U}\subseteq\tau'$ be a $\tau'$-open cover of $X$. Then $\mathscr{U}\subseteq\tau$, so $\mathscr{U}$ is a $\tau$-open cover of $X$, and there is therefore a finite $\mathscr{R}\subseteq\mathscr{U}$ that covers $X$. $\mathscr{R}\subseteq\mathscr{U}\subseteq\tau'$, so $\mathscr{R}$ is a finite $\tau'$-open subcover of $\mathscr{U}$, and $\langle X,\tau'\rangle$ is therefore compact.
In words, if we start with a $\tau'$-open cover $\mathscr{U}$, it is also automatically a $\tau$-open cover, so it has a finite subfamily that covers the $X$. The members of that subfamily are members of $\mathscr{U}$, so we have the desired finite subcover; no extra sets can possibly be needed, because we’re using only sets that are in the original cover $\mathscr{U}$.
It would be different if we were asking for an open refinement with some particular property instead of for a subcover: then we might actually need some of the sets in $\tau\setminus\tau'$. For example, let $\tau'$ be any non-paracompact topology on $X$, and let $\tau$ be the discrete topology. Then $\tau'\subseteq\tau$, $\langle X,\tau\rangle$ is paracompact, and $\langle X,\tau'\rangle$ is not paracompact.