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Suppose (X,d) is a metric space. Does every open cover of X have a minimal subcover with respect to inclusion?

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    Do you mean minimal in the sense of the cardinality, or minimal in the sense of no sub-subcover exists?2012-10-31
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    I meant the sub-subcover sense.2012-10-31

3 Answers 3

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Let $X=\mathbb R$. and $U_n=(-n,n)$. Then $\{U_n\}$ covers $X$ but it doesn't have a minimal sub-cover.

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As other answers have pointed out, there are easy counterexamples. What is true is that if $\mathscr{U}$ is an open cover of a metric space $X$, then $\mathscr{U}$ has an irreducible open refinement: that is, there is an open cover $\mathscr{R}$ of $X$ such that

  1. for each $R\in\mathscr{R}$ there is a $U\in\mathscr{U}$ such that $R\subseteq U$, and
  2. for each $R\in\mathscr{R}$, the family $\mathscr{R}\setminus\{R\}$ no longer covers $X$.

This is a consequence of two well-known theorems. First, every metric space is paracompact, so every open cover of a metric space has a locally finite open refinement. Secondly, every point finite cover of a set (and a fortiori every locally finite cover) has an irreducible subcover.

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Take any $x\in X$, and consider the covering by the sets $\{y\in X:d(x,y) for all positive integers $n$. If $X$ is unbounded (with respect to the metric $d$), then this open cover has no minimal subcover.

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    Thanks for the answer. Now what if X is bounded?2012-10-31
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    @infinitesimal take $\mathbb R$ with the metric $d(x,y) = \min\{1,|x-y|\}$ and consider the cover by the intervals $(-a,a)$. I'd have to think harder if you said that $X$ was totally bounded.2012-10-31
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    And by think harder I mean: The same argument works just fine for the interval $(-1,1)$.2012-10-31
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    @kahen Thanks. I'm now wondering why I was trying so badly to prove the statement, and how I missed these counterexamples.2012-10-31