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Here's a given definition while I am reading the Topology, Munkres.

Definition. A space $X$ is said to be compact if every open covering $A$ of $X$ contains a finite subcollection that also covers $X$.

Then my question here, when we check whether the open covering $A of X contains a finite sub collection that also covers X, that sub-collection must be proper or not?

I think it must be proper, or not that I haven't known that sub means equal to proper seeing the next example :

The interval $(0, 1]$ is not compact; the open covering $A = \{(1/n, 1] | n ∈ Z^+\}$ contains no finite subcollection covering (0, 1].

Beverlie
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  • No, but the requirement is only meaningful when the original cover is infinite. – saulspatz Apr 25 '18 at 03:18
  • @saulspatz Ah, I've just confused with finiteness with countability. thx – Beverlie Apr 25 '18 at 03:20
  • There's absolutely no need for the sub cover to be proper. But if the original cover is infinite, a finite subcover must be proper. If a set is such that it has no infinite open covers, it is compact. But this is very moot. – fleablood Apr 25 '18 at 05:35

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