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Here's the two lemmas :

Lindelöf : Every open set of $\mathbb{R}$ can be written as a countable union of open intervals.

A famous mathematician maybe : Every open set of $\mathbb{R}$ can be written as a countable union of pairwise disjoint open intervals.

I thing that Lindelöf's lemma is weaker because we can chose the open intervals as pairwise disjoint.

Two interesting characteristics of the open sets of $\mathbb{R}$. In general which lemma do we use ?

Thanks in advance !

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  • $\begingroup$ For some purposes it suffices that $\Bbb R$ has a countable base (basis) $B$ of open intervals for the topology (e.g. $\{(a,b): a,b\in \Bbb Q\}$)...so every open set is the union of a (necessarily countable) subset of $B.$ $\endgroup$ – DanielWainfleet Nov 13 '17 at 14:48
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They are equivalent. Clearly the second implies the first. Given the first, look at all pairs of intervals. If they are disjoint we are done. If any pair is not disjoint, the pair can be replaced by a single interval that is the union of the two. Keep this up until the remaining intervals are pairwise disjoint. Given that they are equivalent, we might choose to use the first because the assumption is easier to verify.

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  • $\begingroup$ Ok interesting and the second statement is from which mathematician ? $\endgroup$ – Maman Sep 26 '16 at 17:10
  • $\begingroup$ @Maman: I have no idea $\endgroup$ – Ross Millikan Sep 26 '16 at 17:13
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There is another useful lemma: $\mathbb R$ is a hereditarily Lindelof space. That is, if $X \subset \mathbb R$ and $X\subset \cup F$ where $F$ is a family of open subsets of $\mathbb R,$ then there exists a countable $G\subset F$ such that $X\subset \cap G.$ In particular if $X=\cup F$ where $F$ is an open family then $X=\cup G$ where $G$ is a countable subset of $F.$

An example of its use: If $X$ is an uncountable subset of $R$ then there exists an uncountable $Y\subset X$ such that $U\cap Y$ is uncountable whenever $U$ is an open subset of $\mathbb R$ and $U\cap Y$ is not empty. Proof: Let $F$ be the family of open subsets that have only countable intersection with $X.$ Let $Y=X$ \ $\cup F.$ Now $\cap F=\cap G$ where $G$ is a countable subset of $F, $ so $( \cup F )\cap X =\cup_{g\in G}(g\cap X)$ is countable.... and the rest I leave to the reader.

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