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As similarly described in a question represented here before:

Let $\langle \mathcal{U}_n: n \in \mathbb{N} \rangle$ be a sequence of $\omega$-covers of $X$, and suppose that $X$ is Lindelöf. Can we always find a sequence $\langle F_n: n \in \mathbb{N} \rangle$ with each $F_n \in \mathcal{U}_n$ such that $\cup F_n$ is an open cover of $X$?

Thank you!

Note: The property that $X$ is Lindelöf was not an assumption in the referred question.

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  • But, as far as I know, $\mathbb R$ is not lindeloff.. am I wrong about that? – topsi Feb 05 '14 at 15:52
  • Of course $\mathbb{R}$ is Lindelöf: every second-countable space is Lindelöf. – user642796 Feb 05 '14 at 16:01
  • But, if we cover each open interval $(n,n+1)$, by open segments of length $\frac{1}{2^{2^n}}$, isn't the union of any countable set of intervals going to be with finite length? what am I missing here – topsi Feb 05 '14 at 16:22
  • @Shir: That won't cover $\mathbb{R}$. In particular, no integers will be contained in any of these sets. – user642796 Feb 05 '14 at 16:32
  • @Shir: Let's put it another way: for each $n \in \mathbb{Z}$ the closed interval $[n,n+1]$ is compact, and $\mathbb{R} = \bigcup_{n \in \mathbb{Z}} [n,n+1]$. Given any open cover $\mathcal{U}$ of $\mathbb{R}$ fix for each $n \in \mathbb{Z}$ a finite $\mathcal{V}n \subseteq \mathcal{U}$ which covers $[n,n+1]$. Then $\bigcup{n \in \mathbb{Z}} \mathcal{V}_n$ is a countable subcover of $\mathcal{U}$. – user642796 Feb 05 '14 at 16:39
  • Yes of course... sorry, you are absolutly right.... – topsi Feb 05 '14 at 16:46

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