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$$ A=\sum^n_{i=1}A_i,B=\sum^m_{j=1}B_j \Rightarrow A \cap B=\sum^n_{i=1}\sum^m_{j=1} A_i\cap B_j $$

Σ means it as a direct sum of a set

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The statement makes no sense, I’m afraid: $A_i\cap B_j$ isn’t a statement, so it can’t be implied by anything. –  Brian M. Scott Nov 17 '12 at 23:32
Are you looking to show that $A \cap B = \sum^n_{i=1}\sum^m_{j=1} A_i\cap B_j$? –  Ross Millikan Nov 17 '12 at 23:37
Perhaps you could do something about that 25 percent accept rate? –  Gerry Myerson Nov 18 '12 at 0:22
It looks as if you don't mean direct sum, but union. –  André Nicolas Nov 18 '12 at 1:00
A stitch in time saves nine. –  Will Jagy Nov 18 '12 at 2:11

1 Answer 1

The simplest method would be to chase elements, namely take $x$ in the LHS, unwind the definitions and derive that it is in the RHS, then the same process in the other direction:

If $x\in A\cap B$ then $x\in A$ and $x\in B$, therefore for some $i$, $x\in A_i$ and some $j$, $x\in B_j$. Therefore $x\in A_i\cap B_j$ and therefore $x$ is in the sum.

On the other hand, let $x$ be an element of the sum. Therefore there exists $(i,j)\in\{1,\ldots,n\}\times\{1,\ldots,m\}$ such that $x\in A_i\cap B_j$, and therefore $x\in A$ and $x\in B$ as wanted.

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@coffeemath: Err, yes. That was a typo which I have corrected elsewhere when I wrote the answer, but apparently not everywhere. Thanks! –  Asaf Karagila Nov 18 '12 at 0:10
Yes, I figured you had it right and maybe typo or something; I'll delete previous comment. –  coffeemath Nov 18 '12 at 0:20
In the version I now see there is still the confusing "In parcitular $x \in A_1$ and $x \in B_1$ [which might not hold]. But once you have chosen $i,j$ you can stop there and say $x \in A_i \cap B_j $ and so $x$ is in the sum. –  coffeemath Nov 18 '12 at 0:45
@coffeemath: Yes. It's true. I'm just tired and had difficult few days... I think that it's good now. –  Asaf Karagila Nov 18 '12 at 0:49

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