Is every sigma-compact topological space is Lindelof space ? I know about $\sigma$-compact and Lindelof space.Say about converse is true or not.
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How to prove that every sigma-compact space is Lindelof space – Mukesh Suthar Mar 16 '21 at 07:48
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1Every open cover of the space has a finite subcover on each compact. Take the countable union and you get a countable subcover. – Mar 16 '21 at 07:55
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If you have a proof please share with me. – Mukesh Suthar Mar 16 '21 at 07:58
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Even every sigma-Lindelöf space is Lindelöf. – bof Mar 16 '21 at 08:19
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Suppose that $X= \bigcup_{n \in \Bbb N} K_n$ where each $K_n$ is compact.
Let $\mathcal{U}$ be any open cover of $X$.
We have for each $n$ that $K_n \subseteq \bigcup \mathcal{U}$ so there is a finite subcover $\mathcal{U}_n \subseteq \mathcal{U}$ such that $K_n \subseteq \bigcup \mathcal{U}_n$ as $K_n$ is compact.
Then $\bigcup_{n \in \Bbb N} \mathcal{U}_n$ is a countable (a countable union of finite sets is countable) subcover of $\mathcal{U}$.
As $\mathcal{U}$ was arbitrary, $X$ is Lindelöf.

Henno Brandsma
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