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Let $I$ be an uncountable set of indices and $U_\alpha\subset\mathbb{R}^3$ be an open subset for each $\alpha\in I$. We know that $U=\bigcup_{\alpha\in I} U_\alpha$ is an open subset.

My question is: does there exist a countable subset $J \subset I$ such that $U=\bigcup_{\alpha\in J} U_\alpha$?

In general it is not true without the openness condition.

Thanks!


As suggested by Chandru, this is a `named' property: Lindelöf space.

Every $\sigma$--compact set is Lindelöf: Let $K_0\subset\cdots K_n\to\subset U$ be an increasing sequence of compact subsets with $\bigcup_{n\ge0}K_n=U$. Each $K_n$ has a finite open cover. So $U$ has a countable open cover.

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en.wikipedia.org/wiki/Lindel%C3%B6f_space –  anonymous Feb 16 '11 at 5:48
    
YES! Thank you! –  Pengfei Feb 16 '11 at 5:57
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Now post it as an answer, wait a while, and eventually accept it. Then the question will not be marked as "unanswered". –  Arturo Magidin Feb 16 '11 at 6:11
    
Thanks! Henno's answer is also good enough to be marked. –  Pengfei Mar 18 '11 at 5:43

2 Answers 2

up vote 2 down vote accepted

If $X$ is a topological space, then if this property holds for every open subset of $X$, then this space is called hereditarily Lindelöf. All separable metric spaces have this property, and for metric spaces it's equivalent to separability.

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As suggested by Chandru, this is a `named' property: Lindelöf space.

Every $\sigma$--compact set is Lindelöf: Let $K_0\subset\cdots K_n\to\subset U$ be an increasing sequence of compact subsets with $\bigcup_{n\ge0}K_n=U$. Each $K_n$ has a finite open cover. So $U$ has a countable open cover.

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