Closed subspace of a Lindelöf space is Lindelöf

[Rasalhague]

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.

 

[micromass]

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.

 

[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.

 

[micromass]

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.

 

[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 \}. 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.

 

[micromass]

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.

 

[Rasalhague]

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.

 

[Rasalhague]

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.

 

[micromass]

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.