Let $(X,\rho)$ be a compact metric space and $\mathcal{M}$ be the set of all compact subsets of $X$. Let $A\in \mathcal{M}$ and $(K_n)$ be a sequence in $\mathcal{M}$ such that $K_n\subset A$ and $K_n\to A$ with respect to Hausdorff metric. Is it true that $A = \bigcup\limits_{n\geq 1}K_n$?
My thoughts:
$B:= \bigcup_{n\geq 1} K_n$. From the definition of the Hausdorff metric, $$d_H(C,D)= \max \{ \sup\limits_{c\in C} d(c,D), \ \sup\limits_{d\in D} d(d,C) \}$$
and since $K_n\subset A$, $B$ , for all $n\in \mathbb{N}$, we have that
$$d_H(A, K_n) = \sup\limits_{a\in A} d(a, K_n) = \sup\limits_{a\in A} \inf\limits_{k\in K_n} d(a, k)$$
and
$$d_H(B, K_n) = \sup\limits_{b\in B} d(b, K_n) = \sup\limits_{b\in B} \inf\limits_{k\in K_n} d(b, k)$$
By that and since $B\subset A$ we have that
$$d_H(B, K_n)\leq d_H(A, K_n)\underset{n\to \infty}{\longrightarrow} 0$$
Follows that $A=B$
Is this correct?
Best Answer
It is not true. Take $A=[0,1]$ and $K_n=[1/(2n), 1-1/(2n)]$, then $K_n$ converges to $A$ in the Hausdorff metric, but we have $A = [0,1] \neq (0,1) = \bigcup_{n\geq 1} K_n$.
In short, it fails as $B$ needs not to be compact.