In this particular post : Any open subset of $\Bbb R$ is a at most countable union of disjoint open intervals. [Collecting Proofs] , it is shown that any open subset of $\Bbb R$ is a countable union of disjoint open intervals. I am getting confused. I know that connected sets in $\Bbb R$ are intervals, say $(a,b) \subset \Bbb R$ and that the definition of connected says that a connected set cannot be decomposed into two disjoint open sets. So how can every open subset be a disjoint countable union of open sets?
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At most countable may as well mean one -- this is the case for the open intervals. Then there is no contradiction with the conectedness.
ajr
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Right. Good point. – IntegrateThis Sep 16 '18 at 20:14
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There is no contradiction: all open subsets of $\mathbb{R}$ are at most countable unions of open intervals (if we want a disjoint union, we should also include open segments in this). All connected open subset of $\mathbb{R}$ are open intervals, open segments (like $(a,\infty)$ and $(-\infty,b)$) and $\mathbb{R}$ and $\emptyset$ itself. The connected ones are the building blocks for all open sets (this is what locally connected means).
Henno Brandsma
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