Let $X$ be a metric space and $E\subset X$. Let {$G_i$} be an open cover of $E$
For every open cover {$G_i$}, there exists a finite subcover {$G_{i_n}$} of $E$ such that $G_{i_n} \in${$G_i$}.
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..