# A simple series of set-theoretic operations (unions and intersections)

How can I evaluate the expression $$(A_1 \cup \ldots \cup A_n) \cap (B_1\cup \ldots \cup B_m) \cap (C_1 \cup \ldots \cup C_o) ?$$

I know that for $(A_1 \cup \ldots \cup A_n) \cap B_1$ we get $((A_1\cap B_1) \cup \ldots \cup (A_n\cap B_1))$, but don't knw how the generalize this to the above.

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In what sense do you mean "evaluate" - and in what context? –  Mark Bennet Dec 29 '12 at 9:46

## 1 Answer

1. Step 1: Replace $A=A_1\cup\ldots\cup A_n$, and similarly $B$ and $C$. We have $A\cap B\cap C$.
2. Step 2: Simplify. In this case this is simplified as it is.
3. Step 3: Put $A_1\cup\ldots\cup A_n$ back in, add parenthesis. We have $((A_1\cup\ldots\cup A_n)\cap B)\cap C$.
4. Step 4: Distribute. We have $(A_1\cap B\cup\ldots\cup A_n\cap B)\cap C$.
5. Step 5: Restore the $B_1\cup\ldots\cup B_m$. The indices are now long and tiresome.
6. Step 6: Distribute $A_i\cap(B_1\cup\ldots\cup B_m)$.
7. Step 7: Distribute the intersection with $C$. Arrive at the union small expressions of the form $(A_i\cap B_j\cap C)$.
8. Step 8: Place the union of $C_k$'s into the expression.
9. Step 9: Distribute.
10. Step 10: Have a beer and recall how much bookkeeping after indices is terrible!

We end up with the union of $A_i\cap B_j\cap C_k$ for $i\leq n, j\leq m, k\leq o$.

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Nice Asaf. Honestly I haven't seen such this generalization and wonder who can manage it as it should be. $\ddot\smile$ –  Babak S. Dec 29 '12 at 9:50
@AsafKaragila Would this also work, if we either had an arbitrary (not a finite) union of sets, or one or all of sets would consist of arbitrary intersections - or both ? –  arila Dec 29 '12 at 10:30
@arila: Yes, it would. Although it would be slightly harder to write it cleanly when you have an arbitrary intersection of arbitrary unions. –  Asaf Karagila Dec 29 '12 at 10:33
Sometimes these questions are best thought through in terms of elements. You could think about what it means for $x$ to be an element of the set given by the intersection - $x$ has to be a member of all of the component unions. –  Mark Bennet Dec 29 '12 at 10:34
@Mark: Yes, this is very true. I think that it's an excellent approach for the general case. I also think that everyone should at least one time keep track of indices, so they'll know how much it is preferred to be avoided! :-) –  Asaf Karagila Dec 29 '12 at 10:36