# Union of intersection of families

I'm studying Halmos' Naive Set Theory. In Section 9, Families, he (essentially) mentions a following exercise (on page 35).

Exercise. If $$\{A_i\}$$ and $$\{B_j\}$$ are both nonempty families, then $$(\bigcap_iA_i)\bigcup(\bigcap_jB_j)=\bigcap_{i,j}(A_i\bigcup B_j)$$.

However, I think that this is wrong, and the equality should be replaced with inclusion, i.e., LHS should be a subset of (not generally equal to) the RHS. Correct?

• You may be missing the fact that the intersection on the righthand side is over all pairs of $i$ and $j$. Aug 3 '20 at 21:55
• Yes, I can now see it!
– Atom
Aug 3 '20 at 22:00

No, the equality is correct. Suppose that $$x\in \bigcap_{i,j}(A_i\cup B_j)$$. If $$x\in A_i$$ for all $$i$$, then $$x\in \bigcap A_i$$ and we are done. Otherwise, there exists $$i$$ with $$x\not\in A_i$$. As $$x\in A_i\cup B_j$$ for all $$j$$, it follows that $$x\in B_j$$ for all $$j$$, so $$x\in\bigcap_jB_j$$.