Boundary, unions and intersections Let $(X,\tau)$ be a topological space. If $A,B\subseteq X$, then $\partial(A\cap B)\subseteq(\partial A\cap B)\cup(A\cap\partial B)\cup(\partial A\cap\partial B)$, where $\partial$ denotes the boundary operation. Is it possible to generalize this result? I. e., if $\{A_i\}_{i\in I}$ is a family of subsets of $X$, then
$$\partial(\bigcap_{i\in I}A_i)\subseteq\bigcup_{J\subseteq I, J\neq\emptyset}\cap_{j\in J}\partial A_j\bigcap\cap_{i\in I\setminus J}A_i?$$  

It seems the following. 
We show that the right hand side of the inclusion is equal to $$\bigcap\overline{A_i}\setminus\bigcap\operatorname{int} A_i.$$
Indeed, let $x\in \bigcap\overline{A_i}\setminus\bigcap\operatorname{int} A_i$. Then the set $J=\{i\in I:x\not\in\operatorname{int} A_i \}$ is non-empty. Then $$x\in \bigcap_{j\in J}\partial A_j\cap\bigcap_{i\in I\setminus J}\operatorname{int} A_i\subset 
\bigcap_{j\in J}\partial A_j\cap\bigcap_{i\in I\setminus J} A_i.$$ Conversely, let $x$ belongs to the right hand side of the inclusion.
Clearly, that $x\in \bigcap\overline{A_i}$. But  there exists an index $i$ such that $x\in\partial A_i$. Then $x\not\in\operatorname{int} A_i$ and therefore $x\not\in\bigcap\operatorname{int} A_i.$
Now the inclusion follows from 
$$\partial(\bigcap A_i)=\overline{\bigcap A_i}\setminus\operatorname{int}\left(\bigcap A_i\right)\subset \bigcap\overline{A_i}\setminus\bigcap\operatorname{int} A_i.$$
