# Which statement is stronger?

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 ?

• 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 $B.$ – DanielWainfleet Nov 13 '17 at 14:48
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.