# equality of unions and intersections

Could you show me how to prove this?

Let $A_i \subset N$, $i \in I$, where $I$ is an arbitrary nonempty set. Prove that there exists at most countable set $J \subset I$ such that:

$\bigcup_{i \in I}A_i = \bigcup_{j \in J} A_j$ and $\bigcap_{i \in I}A_i = \bigcap_{j \in J} A_j$.

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HINT: I’m assuming that $N$ is $\Bbb N$, the set of natural numbers.
For each $n\in\bigcup_{i\in I}A_i$ choose an index $i(n)\in I$ such that $n\in A_{i(n)}$, and let $$J=\left\{i(n):n\in\bigcup_{i\in I}A_i\right\}\;;$$ can you finish the argument from here?
For the second result, apply the first result to $\{\Bbb N\setminus A_i:i\in I\}$ and use the De Morgan laws.
 I'm really sorry. Could you explain it a bit further? I'm afraid I'm rather slow-thinking today. – Bilbo Jan 7 at 16:18 @Anna: Is it a problem seeing why $J$ is countable, or why $\bigcup_{i\in J}A_i=\bigcup_{i\in I}A_i$? Or both? – Brian M. Scott Jan 7 at 16:22 I see why J is countable. I just don't see why the unions are equal. If you showed me that, I guess I could do the intersections. – Bilbo Jan 7 at 17:02 @Anna: Suppose that $n\in\bigcup_{i\in I}A_i$; then $n\in A_{n(i)}\subseteq\bigcup_{i\in J}A_i$, so $\bigcup_{i\in I}A_i\subseteq\bigcup_{i\in J}A_i$. But $J\subseteq I$, so $\bigcup_{i\in J}A_i\subseteq\bigcup_{i\in I}A_i$, and the two unions are therefore equal. – Brian M. Scott Jan 7 at 17:09