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How do I show that every family of open nonempty disjoint sets (in $\mathbb{R}$) is countable?

I think I should use the fact that every nonempty open set is the disjoint union of a countable collection of open intervals, but I'm not sure. The question here is pretty simple but I can't even start to think of a way to solve it.

H44S
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  • Open sets in which space? If you mean open sets in $\Bbb R^n$ with the usual topology, then show that any non-empty open sets contain a point all of whose coordinates are rational numbers. – Batominovski Oct 08 '19 at 12:37
  • Hint: Using axiom of choice for every element of a family ${U}$ of open nonempty disjoint subsets of $\mathbb{R}$, there exists an injective map $f:{U}\rightarrow\mathbb{Q}$ such that $f(U)\in U$. – Floris Claassens Oct 08 '19 at 12:47

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More generally:

Lemma. Assume that $X$ is a topological space and $D\subseteq X$ is a dense subset. Then any family of open, pairwise disjoint subsets of $X$ has cardinality at most $|D|$.

Proof. Let $\mathcal{U}$ be a family of open, pairwise disjoint subsets. Since $D$ is dense then $D\cap U\neq\emptyset$ for any $U\in\mathcal{U}$. Thus by the Axiom of Choice there is a function $f:\mathcal{U}\to D$ such that $f(U)\in U$. This function is injective because $\mathcal{U}$ contains pairwise disjoint subsets. Therefore $|\mathcal{U}|\leq |D|$. $\Box$

Now just apply the lemma to $X=\mathbb{R}$ and $D=\mathbb{Q}$.

freakish
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