Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $.

share|improve this question

1 Answer 1

up vote 2 down vote accepted

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.

share|improve this answer
    
I'm really sorry. Could you explain it a bit further? I'm afraid I'm rather slow-thinking today. –  Bilbo Jan 7 '13 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 '13 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 '13 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 '13 at 17:09

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.