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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?

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

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  • $\begingroup$ Thanks, that was clear! $\endgroup$
    – Atom
    Aug 3 '20 at 21:57
  • $\begingroup$ Glad I could help :D $\endgroup$ Aug 3 '20 at 22:00

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