Let $X$ be a topological space such that every open cover has a finite refinement. Then is $X$ compact, or is there a counterexample?
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?
Weakening paracompactness condition
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general-topology
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1In 1, do you mean finite **open** refinement? And in 2, locally finite **open** subcover? – 2010-12-09