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Let $X$ be a metric space and $E\subset X$. Let {$G_i$} be an open cover of $E$

  1. For every open cover {$G_i$}, there exists a finite subcover {$G_{i_n}$} of $E$ such that $G_{i_n} \in${$G_i$}.

  2. For every open cover {$G_i$}, there exists {$M_n$}, a finite family of open sets, such that $E\subset$$\bigcup M_n \subset \bigcup G_i$.

As you know, if 1 is true, then $E$ is compact. I think 1 and 2 both have the same meaning, but can't prove the equivalence. (1→2 is trivial, but 2→1 is not) If 1&2 are not equivalent, please give me some counterexamples..

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You wrote: As you know, if 1 is true, then $E$ is compact. This is not correct. If 1 is true for every open cover then $E$ is compact. –  Martin Sleziak Jul 24 '12 at 9:40
    
@Martin That's what i meant since i take an arbitrary open cover. –  Katlus Jul 24 '12 at 9:42
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up vote 2 down vote accepted

$2$ does not imply $1$. Take for example $E=M=\mathbb{R}$, with the open cover $\{ G_n=(-\infty,n)|n \in \mathbb{N}\}$. This cover has no finite subcover, but we can just take $M=\mathbb{R}$ and $\{ M \}$ is a set satisfying the conclusion of $2$.

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