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Let $X$ a Lindelöf space and $A \subset X$ a closed subspace, we have seen here

Closed subsets of Lindelöf spaces are Lindelöf

that $A$ need to be Lindelöf. My question is about the reciprocal : if $A$ is a Lindelöf subspace of a Lindelöf space $X$, then $A$ need to be closed??

I guess not, but I cannot find an example, cause everithing I need is about compact sets.

Can you give me some example?

MJD
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Joãonani
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  • Hint. $\mathbb R$ is a Lindelöf space. Every subspace of $\mathbb R$ is a Lindelöf space. Not every subset of $\mathbb R$ is closed. Every countable space is Lindelöf. – bof Oct 20 '21 at 01:02
  • $\Bbb Q$ is even dense in the Lindelöf space $\Bbb R$. Very far from closed. – Henno Brandsma Oct 20 '21 at 06:01

2 Answers2

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No. A space in which every subspace is Lindelöf is called hereditarily Lindelöf. For example, any separable metrizable space. Or any 2nd-countable space. Or any countable space.

Henno Brandsma
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DanielWainfleet
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    Or any separable (or even ccc) ordered space etc. – Henno Brandsma Oct 20 '21 at 05:57
  • @HennoBrandsma . Here is an amusing exercise: Let $X$ be a $k^+$-Souslin line. Let $Y$ be a dense subset of $X$. Let $Y$ have the induced-order topology, which may be strictly weaker than the subspace topology. Show that $Y$ is $k^+$-Souslin. – DanielWainfleet Oct 20 '21 at 06:11
  • so $c(X)=k$ and $d(X)=k^+$ I presume? Trivial as both cellularity and density are hereditary in ordered spaces so in fact $hd(X)=k^+$ etc. – Henno Brandsma Oct 20 '21 at 06:14
  • In an email (yrs ago) from Franklin J. Tall, he made the same error. The topology on Y is NOT defined as the subspace topology but as the induced order topology. – DanielWainfleet Oct 20 '21 at 08:56
  • Follow-up to my last comment. Consider a non-Souslin example $X=\Bbb R\times {0,1,2}$ with the lexicographic order and $Y=\Bbb R\times {1}$. Then $Y$ is dense in $X$. In the subspace topology $Y$ is discrete. But $Y$ is order-isomorphic to $\Bbb R$ so in the induced-order topology $Y$ is separable. – DanielWainfleet Oct 20 '21 at 09:24
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Since a countable union of Lindelöf subspaces is Lindelöf, if $X$ is a Lindelöf space in which every Lindelöf subspace is closed, then $X$ has the property "every countable union of closed sets is closed," equivalently, "every countable intersection of open sets is open." Such spaces are called P-spaces (in the sense of Gillman and Henriksen), so $X$ must be a Lindelöf P-space. A converse holds for Hausdorff spaces: a Lindelöf subspace of a Hausdorff P-space is closed, by a similar argument to the proof that a compact subspace of a Hausdorff space is closed.

For a concrete counterexample to your question, take any Lindelöf space which is not a P-space, e.g., the space $\mathbb Q$ of rational numbers.

bof
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