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I fully understand the notion of compactness, but today I as wondering why we restrict the coverings to consist of open sets? I understand there is a separate compactness formulation for closed sets using the finite intersection property, but intuitively, why must the covers be open in the standard definition of compactness?

CBBAM
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  • See this post and also this post for 'intuition' about open sets in topology. – user21820 Oct 22 '20 at 20:05
  • Work backwards. Start with an 'intuitive' property of compactness. Reason that the open covering definition gives you this. – Tyrone Oct 22 '20 at 20:07
  • @Tyrone: I'm not sure that is easy. The easiest intuitive property of compactness comes from sequential compactness, which is not as general as the topological notion. – user21820 Oct 22 '20 at 20:09
  • @user21820 The Heine-Borel Theorem is the most intuitive compactness theorem I can think of. – Tyrone Oct 22 '20 at 20:11
  • Consider for instance any topological in which the singletons are closed sets (for example any metric space). Then any set $A$ can be covered by the "closed covering" $\bigcup_{a\in A} {a}$ and thus only finite sets can have finite subcoverings. – Leander Tilsted Kristensen Oct 22 '20 at 20:12
  • @user21820: I’m not sure that sequential compactness is any more intuitive than limit point compactness, but I agree with your main point, and the history of the concept supports that point: it took quite a while for people to realize that open covers having finite subcovers was the ‘right’ definition of compactness. – Brian M. Scott Oct 22 '20 at 20:12
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    @Tyrone: I don’t agree: the Heine-Borel theorem is too obviously dependent on a metric environment to be a reasonable candidate. – Brian M. Scott Oct 22 '20 at 20:14
  • @BrianM.Scott: I realize I never asked myself this question: Does preservation under continuous functions imply the open cover definition in any reasonable sense? If it does, perhaps it can serve as some motivation. In particular, EVT is an instance of this phenomenon. – user21820 Oct 22 '20 at 20:17
  • @BrianM.Scott I would consider a metric environment the best place to get an intuitive feel for many topological properties. Hence the suggestion. – Tyrone Oct 22 '20 at 20:18
  • @user21820: Continuous functions preserve both countable compactness and sequential compactness (though not limit point compactness), so they can’t help too much. – Brian M. Scott Oct 22 '20 at 20:30
  • @Tyrone: That’s a common notion, but I think it a mistaken one: metric spaces are far too nice to be a good place to develop one’s intuition for topological properties. I actually learned topology via a slightly modified Moore method, starting with the definition of a topological space and gradually adding properties that make spaces ‘nicer’, and I still prefer this approach. I recognize, however, that many students are uncomfortable with it and want something familiar as a touchstone. For that purpose I prefer linearly ordered spaces to metric spaces: some familiar examples are readily ... – Brian M. Scott Oct 22 '20 at 20:36
  • ... available, but there is much less temptation to generalize from them to topological spaces in general than there is with metric spaces, because the fact that they are special is much more apparent. – Brian M. Scott Oct 22 '20 at 20:37
  • @BrianM.Scott: Thank you! I still suspect there might be some characterization of compactness in terms of continuous functions, but I don't have a concrete idea in mind. – user21820 Oct 23 '20 at 05:15

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Here's an observation which doesn't answer your whole question (since it doesn't explain why we should expect the open cover definition to be useful) but I think may help clarify the intuition: namely, that the "closed-compact" property is pretty trivial, at least on well-behaved spaces.

Specifically, every infinite space satisfying a certain mild property (namely, the $T_1$ axiom) is non-"closed-compact." This is easy to see: in a $T_1$ space $X$ every singleton is closed and so $\{\{x\}: x\in X\}$ forms a closed cover of $X$ which of course has no proper subcover, hence no finite subcover if $X$ is infinite. Finite spaces meanwhile are boring according to either notion, since all covers of any type (open, closed, or even arbitrary) are finite. So "closed-compactness" can only be interesting on infinite non-$T_1$ spaces. This rather limits the notion's applicability.

Noah Schweber
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