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Closed subspace of a Lindelöf space is Lindelöf 
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#1
Jul612, 11:36 AM

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 [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
Jul612, 12:11 PM

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This sentence:



#3
Jul612, 01:05 PM

P: 1,402

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
Jul612, 01:08 PM

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Closed subspace of a Lindelöf space is Lindelöf
For example, take [itex]X=\mathbb{R}[/itex] and A=[0,1]. Let G_{1}=[0,3/4[ and G_{2}=]1/4,1]. Then we might take H_{1}=]1,3/4[ and H_{2}=]1/4,2[. But these do not cover X. 


#5
Jul612, 01:52 PM

P: 1,402

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
Jul612, 01:55 PM

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#7
Jul612, 02:11 PM

P: 1,402

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
Jul612, 02:17 PM

P: 1,402

Now, I wonder if we can find an example of a Lindelöf space with a nonLindelöf subspace, what I guess would be called a "weakly" (or nonhereditarily) Lindelöf space.



#9
Jul612, 02:21 PM

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P: 18,278

But as a counterexample, perhaps take the onepoint compactification of an uncountable, discrete space. 


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