Hausdorff distance and inclusion

Question

Let $A_n \subset \mathbb{R}^2$ be subsets that converge to $A$ for the Hausdorff distance (https://en.wikipedia.org/wiki/Hausdorff_distance) and let $B_n \supset A_n$ be subsets which converge to $B$ for the Hausdorff distance. Do we have necessarily $A\subset B$ ? (If extra conditions are needed, we can assume that $A_n \subset A_{n+1}$ for all $n$ and the same for $B_n$)

Answer 1

I'm using the following here: Suppose $X_n\to X$ in Hausdorff metric. Then $$X=\{\,x\mid \forall U\ni x\colon\exists N\colon \forall n>N\colon U\cap X_n\ne\emptyset\,\}.$$

Consider $a\in A$. Let $U\ni a$ be an open neighbourhood. Then $U$ intersects almost all $A_n$, hence $U$ intersects almost all $B_n$. As $U$ was an arbitrary open neighbourhood of $a$, we conclude $a\in B$.

Answer 2

Hint: It's not very hard to show that if $A_n$ is convergent then $$\lim A_n = \cap_{n\ge 0} (\overline{ \cup_{m\ge n} A_m} )$$

(the expression on the right is really the $\limsup A_n$ for the lattice of closed subsets)

Answer 3

They could converge to the same set. Trivial example: Let $A_n$ be $B_{n+1}$ for each $n$.

However, it is true that (given the extra conditions!) $A$ must be a subset of $B$, just not necessarily a proper subset: $A$ must be the intersection of all $A_n$, so each $B_n$ contains $A$, hence the intersection of all $B_n$ - i.e $B$ - contains $A$. I believe this still holds without the extra conditions, but I cannot see a proof offhand.