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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?

some_math_guy
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    Just a guess about intuition: the set is compact if every open cover has a finite subcover, so it's not like an "existence" result where you have a particular cover and are throwing away some of the sets, which is how it sounds like you are thinking about it ("what if I needed those sets"). –  Jul 07 '20 at 18:54

3 Answers3

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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.

Brian M. Scott
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  • Are you considering $\tau'\subseteq \tau$ or$\tau\subseteq \tau'$ ? – some_math_guy Jul 07 '20 at 19:16
  • @J.C.VegaO: $\tau'\subseteq\tau$, as in the question. – Brian M. Scott Jul 07 '20 at 19:16
  • Then why does it start with a $\tau'$-open cover $\mathscr{U}$, shouldn't it start with a with a $\tau$-open cover $\mathscr{U}$ since we have the hypothesis of $\tau$ being compact and then show it is a cover of the coarser one? (which is the source of my confusion, because, some of the sets may be missing in the coarser one that we needed to extract the fnite subcover) – some_math_guy Jul 07 '20 at 19:23
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    @J.C.VegaO: No, it absolutely should not start with a $\tau$-open cover: we’re trying to prove that $\langle X,\color{red}{\tau'}\rangle$ is compact, so we must of course start with a $\tau'$-open cover. – Brian M. Scott Jul 07 '20 at 19:23
  • In the second paragraph, it says "$\mathscr{R}$ is a finite $\tau'$-open subcover of $\mathscr{U}$", shouldn't it read "$\mathscr{R}$ is a finite $\tau'$-open subcover of $X$, because we are covering$ X$, right?, not $\mathscr{U}$ – some_math_guy Jul 07 '20 at 19:43
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    @J.C.VegaO: Both are standard usage. $\mathscr{R}$ covers $X$, so it’s reasonable to call it a subcover of $X$, but it’s a subset of the cover $\mathscr{U}$, so it’s equally reasonable to call it a subcover of $\mathscr{U}$. The question is whether one wants to emphasize the space being covered or the original cover of which we’re finding a subcover. – Brian M. Scott Jul 07 '20 at 19:50
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It might be helpful to think about what happens if the set were somehow compact in the finer topology, but not in the coarser topology. If $X$ is compact in $\tau$ but not compact in $\tau'$, then there is an open cover $\mathcal{U}$ of $X$ in $\tau'$ which admits no finite subcover. However, since $\tau' \subseteq \tau$ we also have that $\mathcal{U}$ is an open cover of $X$ in $\tau$, which is a problem because this means $X$ is also not compact in $\tau.$

DMcMor
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An open cover with respect to the coarser topology is also an open cover with respect to the finer one, this should be clear. Due to compactness with respect to the finer topology, we can find a finite open subcover with respect to the finer topology. This subcover only contains elements which are open with respect to the finer topology, which are open with respect to the coarser topology as well. Meaning that the finite subcover we found is open with respect to the coarser topology, too.

Vercassivelaunos
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