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Closed subspace of a Lindelöf space is Lindelöf

  1. Jul 6, 2012 #1
    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].

  2. jcsd
  3. Jul 6, 2012 #2
    This sentence:

    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.
  4. Jul 6, 2012 #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.
  5. Jul 6, 2012 #4
    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.
  6. Jul 6, 2012 #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].
  7. Jul 6, 2012 #6
    How does that help?? I can extract {X} as subcover. That doesn't really get me anywhere.
  8. Jul 6, 2012 #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].
  9. Jul 6, 2012 #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.
  10. Jul 6, 2012 #9
    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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