Union of a family $J$ of open sets in $\mathbb R^n$ equals union of a countable subfamily $K$ in $J$.

Question

Let $(U_j)_{j\in J}$ be a family of open subsets in $\mathbb R^n$. I'm asked to show that there exists a countable subset $K$ in $J$ such that $\bigcup_{j\in\ K}\left( U_j \right) = \bigcup_{j\in\ J}\left( U_j \right).$

Attempt: If we have the whole space in any of these $(U_j)_{j\in J}$ then we just pick it and it equals the union. We may assume that $J$ is not countable family. I was thinking to try to show that (and it must be the idea) we can only find countable many sets in $J$ that have an element which is not in any of the other open sets in $J$. I'm not sure how I would do it.

Answer 1

HINT: Let $\mathscr{B}$ be the family of all open balls in $\Bbb R^n$ with rational radii and centres whose coordinates are all rational; $\mathscr{B}$ is a countable base for the topology of $\Bbb R^n$. Let $V=\bigcup_{j\in J}U_j$. For each $x\in V$ choose a $j(x)\in J$ and a $B_x\in\mathscr{B}$ such that $x\in B_x\subseteq U_{j(x)}$. Let $\mathscr{B}_0=\{B_x:x\in V\}$, and for each $B\in\mathscr{B}_0$ pick one $x(B)\in V$ such that $B=B_{x(B)}$. Finally, let $K=\left\{j\big(x(B)\big):B\in\mathscr{B}_0\right\}$, and show that $K$ has the desired properties.

(The actual idea is simpler than the notation may suggest, but learning to wade through notational thickets is an important part of learning to read and write mathematics.)