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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?

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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.

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For $x\in S$ let $V(x)$ be the set of all open intervals $J$ such that $x\in J\subset S.$ Let $U(x)=\cup V(x).$ Each $U(x)$ is an open interval containing $x.$ And $\cup_{x\in S}U(x)=S.$

Let $\{U_i:i\in I\}$ be a pair-wise disjoint family of open intervals with $\cup_{i\in I}U_i=S$ and such that $U_i \ne U_{i'}$ for distinct $i,i'\in I.$

For $x\in S$ there exists a unique $j(x)\in I$ such that $x\in U_{j(x)}.$ We have $U_{j(x)}\in V(x)$ so $U_{j(x)}\subset \cup V(x)=U(x).$

Suppose $U_{j(x)}\ne U(x)$. Because $U_{j(x)}\subset U(x)$ we may let $y=\sup U_{j(x)}$ if $\sup U_{j(x)}\in U(x)\setminus U_{j(x)},$ or, otherwise, let $y=\inf U_{j(x)}.$ In either case we have $y\in U(x)\setminus U_{j(x)}.$ But $y\in S$ because $y\in U(x)\subset S,$ so there exists $r>0$ and some $i\in I\setminus \{j(x)\}$ such that $(-r+y,r+y)\subset U_{i}.$ From the definition of $y$ it should be clear that this implies $U_i\cap U_{j(x)}\ne \phi,$ which is absurd because $i\ne j(x).$

Therefore, by contradiction, we have $U_{j(x)}=U(x).$

Therefore $\{U_i:i\in I\}=\{U_{j(x)}:x\in S\}=\{V(x):x\in S\}.$

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