In Kirillov,Gvishiani, Theorems and Problems in Functional Analysis(Springer,1982) Exercise 71 (b) states

$(A_1\cap A_2)\Delta(B_1\cap B_2)\subset(A_1\Delta B_1)\cap(A_2\Delta B_2)$.

I tried with different methods, but I could prove only $(A_1\cap A_2)\Delta(B_1\cap B_2)\subset(A_1\Delta B_1)\cup(A_2\Delta B_2)$ (union instead of intersection in the right-hand-side). It can be a typo, but they remark that this inclusion means that the operation of intersection is continuous with respect to the "distance" $d(A,B):=A\Delta B$. I would appreciate for any proof or disproof. I don't understand this type of continuity.

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    $\begingroup$ You are right. If $A_1=A_2=B_1$ then the right-hand side is empty, but the left-hand side not necessarily. $\endgroup$ – njguliyev Sep 4 '13 at 13:33
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    $\begingroup$ By the way, the second Russian edition has the correct form of this formula (with $\cup$). $\endgroup$ – njguliyev Sep 4 '13 at 13:38

You are correct about your suspicion that there is a typo in your edition of the text, and more importantly, you are right with your conclusion that, in fact:

$$(A_1\cap A_2)\Delta(B_1\cap B_2)\subset(A_1\Delta B_1)\cup(A_2\Delta B_2)$$

DeMorgan's comes into play when expanding the definition of the symmetric difference of sets, and introduces "disjunctions" in set membership (and hence, introduces the need for set union), for example $x \notin B_1 \cap B_2 \iff (x\notin B_1 \lor x\notin B_2)$, and likewise $x \notin A_1 \cap A_2 \iff (x \notin A_1 \lor x\notin A_2)$. In the end, this, plus "unpacking" the other clauses implies $$(x \in A_1 \Delta B_1) \lor (x \in B_1\Delta B_2) \iff x \in (A_1\Delta B_1)\cup (A_2\Delta B_2)$$

  • $\begingroup$ You are so excellent at these! :-) +1 $\endgroup$ – Amzoti Sep 5 '13 at 1:31

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