- #1
Rasalhague
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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].
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.