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Given a metric space $X$, we say $X$ is compact iff there is an open cover $\{U_\alpha\}_{\alpha \in A}$ of $X$ with a finite subcover, that is to say there is a finite subset $A'$ och $A$ such that $\{U_\alpha\}_{\alpha \in A'}$.

Evidently, whether the cover has a finite subcover or not, we have $X = \bigcup_{\alpha \in A} U_\alpha$, because $X$ is contained in this union by definition of a cover, and each set $U_\alpha$ is a subset of $X$ whence there union is a subset of $X$.

Does this mean a compact space in fact has no open cover? At least this has to be the case for $\mathbb{R}^n$.

Also I wonder how we define a subspace $Y$ of a metric space $X$ to be compact. Is it simply the definition above, that is to say for every collection of open subsets of $X$ that contain $Y$, there should be a finite subcollection of sets that contain $Y$, or do we in fact require that the sets that form the cover be subsets of $Y$?

If it is the former, the evidently a proper compact subspace could have an open cover.

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    $\begingroup$ Compact does not mean there EXISTS an open cover with a finite subcover -- it means that EVERY open cover has a finite subcover. $\endgroup$ – Nick Peterson Mar 25 '14 at 17:13
  • $\begingroup$ @NicholasR.Peterson: Yes, I know. But can a compact space in fact have an open cover? $\endgroup$ – Étienne Bézout Mar 25 '14 at 17:15
  • $\begingroup$ Every topological space $X$ has open covers. Take $\mathscr{U} = \{X\}$ for a very simple open cover. $\endgroup$ – Daniel Fischer Mar 25 '14 at 17:17
  • $\begingroup$ @DanielFischer: I'd prefer if we limited the discussion to metric spaces. Surely $\{X \}$ is not a subset of $X$? $\endgroup$ – Étienne Bézout Mar 25 '14 at 17:21
  • $\begingroup$ No, $\{X\}$ is not a subset of $X$. But a cover is a family of subsets of $X$, and $\{X\}$ is a family (with one member) of subsets of $X$. The subset is $X$. $\endgroup$ – Daniel Fischer Mar 25 '14 at 17:25
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Compactness is a property of a space that is independent of what space it is contained in. So $Y$ is a compact subspace of $X$ if and only if $Y$ is a compact set when taken as a stand-alone space.

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  • $\begingroup$ Okay. So does this mean that, in fact, no compact metric space has an open cover? $\endgroup$ – Étienne Bézout Mar 25 '14 at 17:16
  • $\begingroup$ $\{Y\}$ is an open cover of $Y$ always, if the space is $Y$. $\endgroup$ – user2566092 Mar 25 '14 at 17:18
  • $\begingroup$ But is $\{Y\}$ a subset of $Y$?. $\endgroup$ – Étienne Bézout Mar 25 '14 at 17:23

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