I know how to prove that any open subset of $\mathbb R$ is a countable union of disjoint open intervals. See this question.
I would like to prove this union is unique. Is there a straightforward way to prove this?
[Math] Any open subset of $\mathbb R$ is an uniquely countable union of disjoint open intervals
analysisgeneral-topologyreal-analysis
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Best Answer
It essentially follows from the fact that open intervals are connected spaces.
Let $S$ be open open subset of $\mathbb R$. Assume that $\{U_i\}_{i\in I}$ and $\{V_j\}_{j\in J}$ are two such representations of $S$ as the union of open intervals.
Fix $i\in I$, and note:
$$U_i = \bigcup_{j\in J} (U_i\cap V_j)$$
Then the $U_i\cap V_j$ are disjoint open subsets of $U_i$. But $U_i$ is connected. So only one of $U_i\cap V_j$ can be non-empty, and thus $U_i\cap V_j=U_i$ for some $V_j$ - that is $U_i\subseteq V_j$.
Then prove similarly that for each $j\in J$ there must be exactly one $i$ so that $V_j\subseteq U_i$ and $V_j\cap U_{i'}=\emptyset$ if $i'\neq i$.
Use these results to prove that the two partitions are the same.
We are essentially reducing it to the case where $S$ is an open interval - connectedness then shows that $S$ can't be represented as the non-trivial union of disjoint open sets.