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  1. Let $X$ be a topological space such that every open cover has a finite refinement. Then is $X$ compact, or is there a counterexample?

  2. Let $X$ be a topological space such that every open cover has a locally finite subcover. Then is $X$ compact, or is there a counterexample?

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    In 1, do you mean finite open refinement? And in 2, locally finite open subcover? – Nuno Dec 09 '10 at 09:29

2 Answers2

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I quote from Wikipedia's article on paracompact spaces:

Note the similarity between the definitions of compact and paracompact: for paracompact, we replace "subcover" by "open refinement" and "finite" by "locally finite". Both of these changes are significant: if we take the above definition of paracompact and change "open refinement" back to "subcover", or "locally finite" back to "finite", we end up with the compact spaces in both cases.

The page lists Willard's General Topology in its references, so that seems like a good place to go hunting for a proof - or an exercise.

kahen
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    Here is a proof from "Introduction to General Topology" by Joshi. Suppose $X$ is a space with the property that every open cover of it has a locally finite subcover. Let $ \ \mathcal{U} \ $ be a given open cover of $X$. Fix some non-empty $V \in \mathcal{U}$ and let $\mathcal{V} = { G \cup V : G \in \mathcal{U} }$. Then $\mathcal{V}$ is an open cover of $X$. Since every member of $\mathcal{V}$ contains $V$, no subcover of $\mathcal{V}$ can be locally finite unless it is actually finite. But then $\mathcal{U}$ would have a finite subcover. Hence $X$ is compact. – Nuno Dec 09 '10 at 11:06
  • Feel free to add it to your answer if you like. – Nuno Dec 09 '10 at 11:06
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For 1., clearly yes: if a cover has a finite refinement, then the sets it is a refinement of also cover $X$, and form a finite subcover. For 2., see Nuno's comment on the other answer.

Henno Brandsma
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