Let $(M,d)$ be a compact metric space and $d_H(A,B)$ represent the Hausdorff distance between $A,B\subset M$. Let $A_n$ be a sequence of non-empty increasing compact subsets of $M$ (i.e. $A_n\subseteq A_m$ if $m>n$). Assume that $A=\text{cl}(\cup_nA_n)$ is non-empty, compact and bounded. Is it true that:
$$d_H(A_n,A)\xrightarrow{n\rightarrow \infty}0$$
I thought this was true but then I came across this document (which makes me think it isn't) and this question (which makes me think it is) and now I am unsure. If it is untrue what would be a good counterexample?
Best Answer
True.
Let $A_n$ be an increasing sequence of nonempt compact sets. Assume that $A := \text{cl}\;\bigcup_n A_n$ is compact.
For all $n$, we have $A_n \subseteq A$, so $$ \forall x \in A_n\quad \inf\{d(x,y) : y \in A\} = 0 \tag1$$ Now let $\epsilon > 0$. For each $n$, let $$ B_n = \bigcup_{x \in A_n}\{y : d(y,x) < \epsilon\} $$ Now $B_n$ is an open set, the sequence $B_n$ is increasing. By the definition of "closure", $\bigcup_n B_n \supseteq A$. Since $A$ is compact, there is a finite subcover; since $(B_n)$ is increasing, there exists $n_\epsilon$ so that $B_{n_\epsilon} \supseteq A$. Then for all $n \ge n_\epsilon$ we also have $B_n \supseteq A$. Thus: $$ \forall n \ge n_\epsilon\quad \forall y \in A\quad\inf\{d(x,y): x \in A_n\} \le \epsilon . \tag2$$ Combining this with $(1)$: for all $n \ge n_\epsilon$ we have $d_H(A_n,A) \le \epsilon$.
That is true for all $\epsilon > 0$, so $\lim_n d_H(A_n,A) = 0$.