Is the convergence in Hausdorff distance closed under union and intersection?

Question

Let $(X,d)$ be a metric space.
Let $\{A_n\}$ and $\{B_n\}$ be two sequences of subsets.
Suppose that $\{A_n\}$ and $\{B_n\}$ are both converge under Hausdorff distance.

Is it true that
$\lim(A_n\cup B_n)$=$(\lim_nA_n)\cup(\lim_nB_n)$?
$\lim(A_n\cap B_n)$=$(\lim_nA_n)\cap(\lim_nB_n)$?

Answer 1

Your second question is false. Let $A_i = \{a_i\}$ where $a_i \to c$ and let $B_i =\{b_i\}$ where $b_i \to c$. Then $$\lim (A_n \cap B_n) = \lim \emptyset = \emptyset \neq \{c\} = \lim(A_n) \cap \lim(B_n)$$

Answer 2

Limits are preserved under unions (but not under intersections as Kyle Gannon proved). Indeed, let $A=\lim A_n$ and $B=\lim B_n$. Write $N_\epsilon(A) $ for the $\epsilon$-neighborhood of $A$.

For any $\epsilon>0$, there exists $N$ such that
$$ A\subset N_\epsilon(A_n), \quad B\subset N_\epsilon(B_n), \quad A_n\subset N_\epsilon(A), \quad B_n\subset N_\epsilon(B) $$ for all $n\ge N$. Hence, $$ A\cup B\subset N_\epsilon(A_n\cup B_n),\qquad A_n\cup B_n\subset N_\epsilon(A\cup B) $$ for all $n\ge N$, which proves $\lim(A_n\cup B_n)=A\cup B$.