# Problem with proving equation. (Sets.)

I would like to ask you for helping me out with this problem. I had to prove this equation.

$$\left(\bigcap_i A_i\cap\bigcup_{i\text{ odd}}A_i\right)\triangle\bigcap_{i\text{ odd}}A_i=\left(\bigcap_i A_i\triangle\bigcap_{i\text{ odd}}A_i\right)\cap\bigcup_i A_i$$

I did following:

\begin{align*} \left(\bigcap_i A_i\cap\bigcup_{i\text{ odd}}A_i\right)&=\bigcap_i A_i\\ \left(X\cap\bigcup_i A_i\right)&=X \end{align*}

$X\subseteq\bigcup A_i$

And after this I modified the main equation. But I didn't prove that, that 2 equations are true in common. Is it a valid? Or I have to prove that? If yes, how to prove that?

Sorry for my english. Thanks

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 What does $\Delta$ mean? Also, is $i$ ranging over integers? – Braindead Nov 21 '12 at 20:02 I'd guess $\Delta$ is the symmetric difference, but definitely would be nice to see clarification on that – Thomas Andrews Nov 21 '12 at 20:03 Yea sorry for that, Δ is symmetric difference, and i is over integers. – Dávid Kaya Nov 21 '12 at 20:07 @Noturno - The first equality you wrote (just after "I did the following:") is false. ($A_1 \cap A_2 \cup A_1 \ne A_1 \cap A_2$) – Braindead Nov 21 '12 at 20:08 @Braindead - It's (($A_1$ $\cap$ $A_2$) $\cap$ $A_1$)= $A_1$ $\cap$ $A_2$ – Dávid Kaya Nov 21 '12 at 20:12

As $\ \bigcap_i A_i\subset \bigcup_{i\ {\rm odd}} A_i$ we have $$LHS=\bigcap_i A_i\ \triangle \ \bigcap_{i\ {\rm odd}} A_i\ .$$ On the other hand, since every $x$ in the considered universe belongs to $\bigcup_i A_i$ the $RHS$ is equal to its large parenthesis, i.e., equal to $LHS$.
Note that $\ \bigcap_i A_i\subset \bigcap_{i\ {\rm odd}} A_i\$. Therefore both sides of the stated equality describe the set of all $x$ which belong to all $A_i$ with $i$ odd, but not to all $A_i$ with $i$ even.
Suppose $A,B$ and $C$ are arbitrary sets such that $A\subseteq C\subseteq B$. Then $(A\cap B)\triangle C = A\triangle C$ holds because $A\subseteq B$ and $A\triangle C\subseteq B$ holds because $A,C\subseteq B$. But this implies $(A\triangle C)\cap B = A\triangle C$. Taking $A = \bigcap_i A_i, B=\bigcup_{i\text{ odd}}A_i$ and $C=\bigcap_{i\text{ odd}}A_i$, the proof is complete.