Closed subspace of a Lindelöf space is Lindelöf

In summary: In that case, the compactification is not Lindelof, because the compactification of a discrete space is always Lindelof.In summary, Rao's proof of Proposition 1.5.4 uses the fact that for each open cover of a subspace, there is an open cover of the superspace.
  • #1
Rasalhague
1,387
2
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 [itex](X,\cal{T})[/itex] with a subset [itex]A \subseteq X[/itex], then [itex]\cal{T}_A[/itex] is the subspace topology for [itex]A[/itex].

Let [itex]A[/itex] be a closed subspace of a Lindelöf space [itex](X,\cal{T})[/itex]. Let [itex]C=\left \{ G_\lambda : \lambda \in \Lambda \right \}[/itex] be a [itex]\cal{T}_A[/itex]-open cover of [itex]A[/itex]. Then [itex]G_\lambda = H_\lambda \cap A[/itex] for some [itex]H_\lambda \in \cal{T}[/itex].

Now [itex]\left \{ H_\lambda : \lambda \in \Lambda \right \}[/itex] is a [itex]\cal{T}[/itex]-open cover of [itex]A[/itex]. Hence [itex]\left \{ H_\lambda : \lambda \in \Lambda \right \} \cup (X\setminus A)[/itex] is a [itex]\cal{T}[/itex]-open cover for [itex]X[/itex].

Since [itex]X[/itex] is Lindelöf, we can extract from this cover a countable subcover of [itex]X[/itex], say [itex]\left \{ H_{\lambda_1},H_{\lambda_2},... \right \}[/itex]. Accordingly, [itex]\left \{ G_{\lambda_1},G_{\lambda_2},... \right \}[/itex] is an open subcover of [itex]A[/itex]. Hence [itex](A,\cal{T}_A)[/itex] is Lindelöf.
 
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  • #2
This sentence:

Hence [itex]\left \{ H_\lambda : \lambda \in \Lambda \right \} \cup (X\setminus A)[/itex] is a [itex]\cal{T}[/itex]-open cover for [itex]X[/itex].

says not only that it is a cover, but that the [itex]H_\lambda[/itex] and [itex]X\setminus A[/itex] are open. The latter is of course only true if A is closed.
 
  • #3
Okay, I see. In that case, yes, he has only shown that the statement is true when [itex]A[/itex] is closed. But [itex]\left \{ H_\lambda : \lambda \in \Lambda \right \}\cup X[/itex] is also an open cover for [itex]X[/itex], whether or not [itex]A[/itex] is closed.
 
  • #4
Rasalhague said:
Okay, I see. In that case, yes, he has only shown that the statement is true when [itex]A[/itex] is closed. But [itex]\left \{ H_\lambda : \lambda \in \Lambda \right \}\cup X[/itex] is also an open cover for [itex]X[/itex], whether or not [itex]A[/itex] is closed.

No, the [itex]H_\lambda[/itex] 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 [itex]X=\mathbb{R}[/itex] 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
Ah, I get it! Thanks, micromass.

I realized that [itex]\left \{ H_\lambda : \lambda \in \Lambda \right \}[/itex] is not necessarily an open cover of [itex]X[/itex], but I reasoned that since [itex]X[/itex] is open, [itex]\left \{ H_\lambda : \lambda \in \Lambda \right \}\cup X[/itex] is an open cover of [itex]X[/itex], so we can extract an open subcover [itex]\left \{ H_{\lambda_1},H_{\lambda_2},...,X \right \}[/itex]. What I was forgetting is that the corresponding open cover of [itex]A[/itex], namely [itex]\left \{ H_{\lambda_1},H_{\lambda_2},...,Y \right \}[/itex] is not necessarily a subcover for [itex]C=\left \{ G_\lambda : \lambda \in \Lambda \right \}[/itex] since it's not necessarily the case that [itex]Y\in C[/itex].
 
  • #6
Rasalhague said:
Ah, I get it! Thanks, micromass.

I realized that [itex]\left \{ H_\lambda : \lambda \in \Lambda \right \}[/itex] is not necessarily an open cover of [itex]X[/itex], but I reasoned that since [itex]X[/itex] is open, [itex]\left \{ H_\lambda : \lambda \in \Lambda \right \}\cup X[/itex] is an open cover of [itex]X[/itex], so we can extract an open subcover [itex]\left \{ H_{\lambda_1},H_{\lambda_2},...,X \right \}[/itex].

How does that help?? I can extract {X} as subcover. That doesn't really get me anywhere.
 
  • #7
Yeah, and it doesn't help (show that the statement is true of an arbitrary subspace) because the [itex]\cal{T}_Y[/itex]-open set which in the subspace topology corresponds to [itex]X[/itex] is [itex]Y=X\cap Y[/itex], and - of course - there's no guarantee that [itex]Y[/itex] will belong to an arbitrary [itex]\cal{T}_Y[/itex]-open cover of [itex]Y[/itex].
 
  • #8
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
Rasalhague said:
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.
 

1. What is a closed subspace?

A closed subspace is a subset of a topological space that is also a closed set, meaning that it contains all of its limit points.

2. What is a Lindelöf space?

A Lindelöf space is a topological space in which every open cover has a countable subcover. This means that it can be covered by a countable number of open sets.

3. How does a closed subspace of a Lindelöf space relate to the Lindelöf property?

The statement "a closed subspace of a Lindelöf space is Lindelöf" means that if the original space has the Lindelöf property, then any subspace that is also closed will also have the Lindelöf property. In other words, the Lindelöf property is preserved under taking closed subspaces.

4. Why is this property important in mathematics?

This property is important because it helps to simplify the study of topological spaces. By knowing that closed subspaces of Lindelöf spaces are also Lindelöf, we can focus on studying Lindelöf spaces and use this property to extend our results to closed subspaces. This can save time and effort in proving theorems and can also provide more insight into the structure of topological spaces.

5. What are some examples of topological spaces that have the Lindelöf property?

Some examples of topological spaces that have the Lindelöf property include the real line, the space of rational numbers, and the Cantor set. In general, any metric space is Lindelöf, but there are also non-metric examples such as the Sorgenfrey line and the long line.

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