"continuity" of symmetric difference wrt intersection Question 71(b) of Kirillov and Gvishiani's problem book on functional analysis is to show:
$$(A_1\cap A_2)\triangle (B_1\cap B_2)\subset (A_1\triangle B_1)\cap (A_2\triangle B_2).$$
Isn't this a counter-example: Let $A_1=A_2=B_1=\{0\}$ and $B_2=\{1\}$. I am also hoping to better understand the way this result is described, as "continuity of $\triangle$ with respect to intersection." How does this connect to the literal definition of continuity?
 A: On RHS it should be $\cup$. Continuity will mean: if $A_i$ approximate well the $B_i$  ( the symmetric difference is small), then so will $A_1\cap A_2$ approximate well $B_1 \cap B_2$. So in fact this is : continuity of intersection.
$\bf{Added:}$ If you want to check whether one set is included in the other, look at the characteristic functions. Denote
$$a_i = \chi_{A_i}, b_i = \chi_{B_i}$$
Then we have
$$\chi_{A_i \Delta B_i} = a_i + b_i - 2 a_i b_i$$
and we need to check whether
$$a_1 a_2 + b_1 b_2 - 2 a_1 a_2 b_1 b_2 \le (a_1 + b_1 - 2 a_1 b_1)(a_2 + b_2 - 2 a_2 b_2)$$
for $a_i$, $b_i \in \{0,1\}$. We give different values and we see that for some choices of $a_i$, $b_j$ the inequality does Not hold. To get the explicit counterexample, choose $A_i$, $B_j$ to be the total set or $\emptyset$, according to the values of $a_i$, $b_j$.
However, one can see that
$$(A_1 \cap A_2) \Delta (B_1 \cap B_2) \subseteq (A_1 \Delta B_1) \cup (A_2 \Delta B_2)$$
$\bf{Added:}$ Let's also investigate the continuity of the operation "difference". We have
$$(A_1 \backslash A_2) \Delta (B_1 \backslash B_2) = (A_1 \cap A_2^c) \Delta (B_1 \cap B_2^c) \subseteq (A_1 \Delta B_1) \cup (A_2^c\Delta B_2^c) = \\ =(A_1 \Delta B_1)\cup (A_2 \Delta B_2)$$
So again in the statement of the Exercise it should be $\cup$, rather than $\backslash$.  Still, the book is interesting.
But what is the general idea: say we have some operation on sets
$$(A_1, \ldots, A_m) \to \phi(A_1, \ldots, A_m)$$
If $\phi(A_1, \ldots, A_m)$, and $\phi(B_1, \ldots B_m)$ differ "at some element $x$ ", then at least one of the $A_i$, $B_i$ have to differ at that element $x$. Therefore we have
$$\phi(A_1, \ldots, A_m)\, \Delta\, \phi(B_1, \ldots, B_m) \subseteq \bigcup_{i=1}^m (A_i \Delta B_i)$$
Unfortunately, we cannot replace on RHS $\bigcup$ with $\phi$.
