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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.

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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.)

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$j(x)\in J$ are you just finding some $U_j$ which includes $x$? Also your family of those $B$ are all one-dimensional? And $x(B)$ is that just picking some x in $B$? – Raxel Jan 13 '13 at 20:47
@Raxel: (1) No, I’m finding a pair $B_x\in\mathscr{B}$ and $U_{j(x)}$ such that $x\in B_x\subseteq U_{j(x)}$; the sets $B_x$ are crucial, because there are only countably many of them. (2) When I wrote that I’d forgotten that you were working in $\Bbb R^n$, not $\Bbb R$; I’ve fixed it now. (3) No! It is very specifically choosing an $x$ for which $B$ is the chosen set $B_x$. – Brian M. Scott Jan 13 '13 at 20:55
@Raxel: I'm using the balls in $\mathscr{B}_0$ to pick out a countable subfamily of the $U_j$'s whose union is all of $V$. Each member of $V$ is contained in (at least) one member of $\mathscr{B}_0$, and each member of $\mathscr{B}_0$ is contained in a specific $U_j$. Those specific $U_j$'s therefore cover $V$, so their union is $V$. – Brian M. Scott Jan 13 '13 at 22:28
I understand what you are doing now to the line with $\mathscr{B}_0$. You are taking $B$ from $\mathscr{B}_0$ which is not countable and you must be trying to make it so but I don't understand how. Are you just making sure we reduce the family of $B_x$ by counting equal balls just once? – Raxel Jan 13 '13 at 22:51
Because if there where no equal balls we would have from the given that the rational radii and rational centers they would be countable? And after we have done that, counting each only once the family K is countable and from your last argument it's union must be $V$? – Raxel Jan 13 '13 at 23:32

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