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Jan 18, 2015 at 18:25 vote accept CommunityBot
Jan 18, 2015 at 4:53 review Suggested edits
Jan 18, 2015 at 5:11
Jan 18, 2015 at 4:51 answer added Tad timeline score: 3
Jan 18, 2015 at 1:54 comment added user198185 Ah I see, using the Power set. Now I see how to use the Pidgeonhle Principle. Thank you!
Jan 18, 2015 at 0:22 comment added Tad You have to be a little more careful than this. While there are $2^{n+1}$ choices for $I$ and only $2^n$ choices for the union, that only tells you there must be lots of pairs $I$ and $J$ which give the same union -- but it doesn't guarantee that $I$ and $J$ will be disjoint.
Jan 17, 2015 at 23:07 comment added user198185 So would the number of choices for I, J be $n!$? Unsure of the number of options for the union, would it be $n!$ as well?
Jan 17, 2015 at 22:46 history edited user198185 CC BY-SA 3.0
I noticed the question has been re-tagged as "combinatorics", but it should be noted I have never taken a course in combinatorics (this question is from an algebra course) and so I have no familiarity with many of its concepts.
Jan 17, 2015 at 22:00 comment added Fiktor Count amount of choices for $I$ and amount of options for $\bigcup_{i\in I} A_i$.
Jan 17, 2015 at 22:00 comment added user207868 Oh, I see: it is time to go to bed. A hint for the OP: use the pigeonhole principle.
Jan 17, 2015 at 21:58 comment added Tad @LeonAragones: your set $A$ does not have $n=2$ elements.
Jan 17, 2015 at 21:56 comment added user207868 I'm not sure if it is true. Let's set $n = 2$, $A = \{1, 2, 3\}$ and $A_k = \{k\}$ for $k \in \{1,2,3\}$. Can you find disjoint and nonempty $I$ and $J$?
Jan 17, 2015 at 21:50 history edited Asaf Karagila
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Jan 17, 2015 at 21:28 history asked user198185 CC BY-SA 3.0