# Closed subspace of a Lindelöf space is Lindelöf

1. Jul 6, 2012

### 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 $A$.

2. Jul 6, 2012

### micromass

Staff Emeritus
This sentence:

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.

3. Jul 6, 2012

### 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.

4. Jul 6, 2012

### micromass

Staff Emeritus
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.

5. Jul 6, 2012

### 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 $Y\in C$.

6. Jul 6, 2012

### micromass

Staff Emeritus
How does that help?? I can extract {X} as subcover. That doesn't really get me anywhere.

7. Jul 6, 2012

### 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 $Y$.

8. Jul 6, 2012

### 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.

9. Jul 6, 2012

### micromass

Staff Emeritus
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.