Suppose $X$ is compact. $A$,$B$ open sets which cover $X$. Can $X$ be covered by compact sets $C$,$D$ such that $C \subseteq A$ and $D \subseteq B$?
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No. Two complementary half-spaces in euclidean space cover the space but the whole space cannot be covered by two compact sets. – Eduardo Longa Oct 15 '18 at 15:59
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2@EduardoLonga The Euclidean space is not compact. – José Carlos Santos Oct 15 '18 at 15:59
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Oh yes, I did not see that was required – Eduardo Longa Oct 15 '18 at 16:00
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I guess if we add the hypothesis of Hausdorff, then it is true. – Eduardo Longa Oct 15 '18 at 16:02
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@EduardoLonga Proof for the Hausdorff case? I can do it for normal spaces. – davik Oct 15 '18 at 16:17
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1Have a look at this: https://math.stackexchange.com/questions/1329866/compact-hausdorff-spaces-are-normal – Eduardo Longa Oct 15 '18 at 16:19
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1For any normal $X$ we can find closed $C, D$ as required. This shows that the answer is "yes" for compact Hausdorff $X$. – Paul Frost Oct 15 '18 at 16:19
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@PaulFrost thanks forgot compact Hausdorff implies normal... – davik Oct 15 '18 at 16:20
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Let $X = \alpha(\mathbb{Q})$ denote the Alexandroff compactification of $\mathbb{Q}$. It is a compact non-Hausdorff $T_1$-space which is moreover a KC-space (which means that all compact subsets are closed). See the answers to https://math.stackexchange.com/q/2793610.
Now assume that $X$ satisfies the "compact refinement" property of your question. Then $X$ would be normal since it is a KC-space. But among compact $T_1$-spaces, Hausdorff is equivalent to normal. This is a contradiction.
Paul Frost
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Marvelous, let me clarify the proof because I was a little confused reading it. Suppose it does satisfy my requirement, let $a$,$b$ be distinct points and take the two open sets the complements of the points. Apply refinement and taking complements gives disjoint open neighborhoods. This implies Hausdorff which is contradiction. – davik Oct 15 '18 at 20:17
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I've asked a new question which asks whether a counterexample still exists if I add the condition of being locally compact, see https://math.stackexchange.com/q/2965117/116530. If you have any idea, that would be great – davik Oct 21 '18 at 22:40
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We need normality of $X$ (so $X$ Hausdorff is sufficient too), because the fact that every two-set open cover has a two-set closed refinement is equivalent to normality. (easy proof by taking complements.)
Henno Brandsma
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This isn't correct, consider some finite X with a topology which is not discrete – davik Oct 15 '18 at 16:34
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@davik A space $X$ is a normal if for any disjoint closed sets $E$ and $F$ there are open neighbourhoods $U$ of $E$ and $V$ of $F$ that are also disjoint. A normal $T_1$-space is called $T_4$. However, notation is not standardized in the literature, and many authors require a normal space to be Hausdorff (or equivalently $T_1$). Among compact $T_1$-spaces, Hausdorff is equivalent to normal. – Paul Frost Oct 15 '18 at 16:51
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@PaulFrost thanks, but this answer is still not correct. Take cofinite topology on an infinite space. Then every subset is compact. – davik Oct 15 '18 at 16:58
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You are right, a two-set open cover may have a two-set compact but non-closed refinement. – Paul Frost Oct 15 '18 at 17:02
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2It's really equivalent to normality in a KC space (and thus to Hausdorffness in that case). – Henno Brandsma Oct 15 '18 at 20:59