# Closed subspace of a Lindelöf space is Lindelöf

by Rasalhague
Tags: lindelöf, space, subspace
P: 1,402
I'm looking at Rao: Topology, Proposition 1.5.4, "A closed subspace of a Lindelöf space is Lindelöf." He gives a proof, which seems clear enough, using the idea that for each open cover of the subspace, there is an open cover of the superspace. But I can't yet see where he uses the fact that the subspace is closed. That's to say, I can't see why the proposition isn't true of an arbitrary subspace.

Here is Rao's proof in full. In his notation, given a topological space $(X,\cal{T})$ with a subset $A \subseteq X$, then $\cal{T}_A$ is the subspace topology for $A$.

 Let $A$ be a closed subspace of a Lindelöf space $(X,\cal{T})$. Let $C=\left \{ G_\lambda : \lambda \in \Lambda \right \}$ be a $\cal{T}_A$-open cover of $A$. Then $G_\lambda = H_\lambda \cap A$ for some $H_\lambda \in \cal{T}$. Now $\left \{ H_\lambda : \lambda \in \Lambda \right \}$ is a $\cal{T}$-open cover of $A$. Hence $\left \{ H_\lambda : \lambda \in \Lambda \right \} \cup (X\setminus A)$ is a $\cal{T}$-open cover for $X$. Since $X$ is Lindelöf, we can extract from this cover a countable subcover of $X$, say $\left \{ H_{\lambda_1},H_{\lambda_2},... \right \}$. Accordingly, $\left \{ G_{\lambda_1},G_{\lambda_2},... \right \}$ is an open subcover of $A$. Hence $(A,\cal{T}_A)$ is Lindelöf.
Mentor
P: 18,346
This sentence:

 Hence $\left \{ H_\lambda : \lambda \in \Lambda \right \} \cup (X\setminus A)$ is a $\cal{T}$-open cover for $X$.
says not only that it is a cover, but that the $H_\lambda$ and $X\setminus A$ are open. The latter is of course only true if A is closed.
 P: 1,402 Okay, I see. In that case, yes, he has only shown that the statement is true when $A$ is closed. But $\left \{ H_\lambda : \lambda \in \Lambda \right \}\cup X$ is also an open cover for $X$, whether or not $A$ is closed.
Mentor
P: 18,346
Closed subspace of a Lindelöf space is Lindelöf

 Quote by Rasalhague Okay, I see. In that case, yes, he has only shown that the statement is true when $A$ is closed. But $\left \{ H_\lambda : \lambda \in \Lambda \right \}\cup X$ is also an open cover for $X$, whether or not $A$ is closed.
No, the $H_\lambda$ only form a cover of A. It is not known that they also cover X. They might cover X, but nothing has been said about that.

For example, take $X=\mathbb{R}$ and A=[0,1].
Let G1=[0,3/4[ and G2=]1/4,1]. Then we might take H1=]-1,3/4[ and H2=]1/4,2[. But these do not cover X.
 P: 1,402 Ah, I get it! Thanks, micromass. I realized that $\left \{ H_\lambda : \lambda \in \Lambda \right \}$ is not necessarily an open cover of $X$, but I reasoned that since $X$ is open, $\left \{ H_\lambda : \lambda \in \Lambda \right \}\cup X$ is an open cover of $X$, so we can extract an open subcover $\left \{ H_{\lambda_1},H_{\lambda_2},...,X \right \}$. What I was forgetting is that the corresponding open cover of $A$, namely $\left \{ H_{\lambda_1},H_{\lambda_2},...,Y \right \}$ is not necessarily a subcover for $C=\left \{ G_\lambda : \lambda \in \Lambda \right \}$ since it's not necessarily the case that $Y\in C$.
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P: 18,346
 Quote by Rasalhague Ah, I get it! Thanks, micromass. I realized that $\left \{ H_\lambda : \lambda \in \Lambda \right \}$ is not necessarily an open cover of $X$, but I reasoned that since $X$ is open, $\left \{ H_\lambda : \lambda \in \Lambda \right \}\cup X$ is an open cover of $X$, so we can extract an open subcover $\left \{ H_{\lambda_1},H_{\lambda_2},...,X \right \}$.
How does that help?? I can extract {X} as subcover. That doesn't really get me anywhere.
 P: 1,402 Yeah, and it doesn't help (show that the statement is true of an arbitrary subspace) because the $\cal{T}_Y$-open set which in the subspace topology corresponds to $X$ is $Y=X\cap Y$, and - of course - there's no guarantee that $Y$ will belong to an arbitrary $\cal{T}_Y$-open cover of $Y$.
 P: 1,402 Now, I wonder if we can find an example of a Lindelöf space with a non-Lindelöf subspace, what I guess would be called a "weakly" (or nonhereditarily) Lindelöf space.
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P: 18,346
 Quote by Rasalhague Now, I wonder if we can find an example of a Lindelöf space with a non-Lindelöf subspace...
Well, it won't be a metric space. All subspaces of a Lindelof metric space are Lindelof, for the simple reason that Lindelof is equivalent to second countable in metric spaces.

But as a counterexample, perhaps take the one-point compactification of an uncountable, discrete space.

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