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Let $X$ be Lindelöf and let $A \subseteq X$ be closed. Show that it follows that $A$ is Lindelöf.

That is, we want to show for every open cover of $A$, there is a countable subcover.

Note: (A space is Lindelöf if every open cover of the space has a countable subcover, a weakening of the better-known notion of compactess).

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Hmm... Do you remember how to prove that a closed subset of a compact space is compact? –  Andres Caicedo Nov 26 '12 at 1:54
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Let U be an open cover of A. Since all elements of U are open, they are equal to the intersection of some family of open sets with A. Let this collection of intersections be denoted V. –  Moe Nov 26 '12 at 1:58
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Then consider open cover of X, which consists of X-A and V. Since X is Lindelof, then the union of X-A and V has a countable subcover... –  Moe Nov 26 '12 at 2:00
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Ah, then take away X-A, then V would still be countable, leaving only the desired countable cover of A. –  Moe Nov 26 '12 at 2:01

2 Answers 2

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Using DeMorgan's laws, one can show that a space $X$ is Lindelof iff for every collection $A_\alpha$ of closed subsets of $X$ with the countable intersection property (meaning the intersection of any countable subcollection is nonempty), the total intersection $\cap A_\alpha$ is nonempty. If $A \subset X $ is closed, then every closed set in $A$ is also closed in $X$ so by the previous characterization of Lindelof, $A$ must be Lindelof.

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Very cool, Seth. Thanks. –  Moe Nov 26 '12 at 2:05

Let $C$ be a cover of $A$. Since $A$ is closed, $A^C$, the complement of $A$, is open. Then $C\cup\bigl\{A^C\bigr\}$ is an open cover of $X$. It contains a countable subcover $C'$. Then $C'\setminus\bigl\{A^C\bigr\}$ is countable, covers $A$, and is a subcover of $C$.

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