Im a bit confused with the proof of the fact that K is compact iff K is bounded and closed. More with the - K is compact => is bounded and closed. In my proof i use the fact that If K is compact then for every open cover of K there is a finite subcover also containing K, but if we assume that G = { U1, U2, ... , Un } is that subcover, is it true that for every i Ui is finite set?
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No. $U_i$ is an open set (so usually far fro mfinite), more specifically, one of the open sets that occur in the given open cover of $K$. – Hagen von Eitzen May 09 '17 at 14:36
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Note: the usual word for what you call "limited" is "bounded". – Ethan Bolker May 09 '17 at 14:39
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Thanks! I'll keep that in mind. – Zarrie May 09 '17 at 14:42
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No. Take it with a simple example. The set $[0,1]$ is a compact subspace of $\mathbb{R}$. An open cover of it is $\{ [0,1/2[,\ ]1/3, 1] \}$, for the elements of this collection are open in $[0,1]$. None of them is finite.
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2@Zarrie: No, not necessarily. For example, $\left{(-\infty,3/4),(1/4,+\infty)\right}$ is a finite open cover of $[0,1]$ as a subset of $\mathbb{R}$, but neither element of this cover is bounded. – zipirovich May 09 '17 at 15:45