Let $A_j$, $B_j$, $A$ and $B$ be sets in $\mathbb{R}^n$. Let $A_j \rightarrow A$ and $B_j \rightarrow B$ both in the Hausdorff distance (induced by the usual Euclidean metric in $\mathbb{R}^n$) and $A_j\cap B_j\neq \emptyset$ for all $j\in\mathbb{N}$. If $A_j \cap B_j \rightarrow E$ for some set $E\subset \mathbb{R}^n$, then do we have $E\subset A\cap B$?
It is known that $E\neq A\cap B$ in general, but the counterexample I could only find falls under the case $E\subset A \cap B$. So is it true in general?
No. It can help to notice that two sets which are relatively dense have Hausdorff distance zero. (Relatively dense means that for any point of one and any $\varepsilon > 0$ radius open ball of that point, there is a point of the other within that ball. "Dense" isn't the right idea because we do not require one set to be a subset of the other.)
Let $A = B = A \cap B$ be the rational points in the unit ball of $\mathrm{R}^n$. Let $A_j = B_j = E$ be the unit ball in $\mathrm{R}^n$. Clearly, the Hausdorff distance between $A_j = E$ and $A$ is zero (any $\varepsilon > 0$ neighborhood of one contains the other), so $A_j \rightarrow A$ and likewise $B_j \rightarrow B$. Then $E \smallsetminus (A \cap B)$ is the set of points in the unit ball with at least one irrational coordinate, so is definitely nonempty, so $E \not \subseteq A \cap B$.