# Hausdorff distance union limit

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?

• Example 6 on page 113 in the linked document is a valid counterexample in $X=C([0,1])$ isn't it? And the linked question is refuted in the only answer. So whence the doubt? Commented Mar 3, 2021 at 13:27
• I'm a little confused by whether the example in the document satisfies all the given compactness and boundedness conditions... I have never studied compactness of function sets and am unsure what it would mean for this context. The linked question is refuted in the answer but it is then proven with the additional constraint that the limiting set is bounded; I have included this condition in my question Commented Mar 3, 2021 at 13:32
• That example is not compact, right? Commented Mar 3, 2021 at 13:46
• This document has an elementary exposition of properties of the Hausdorff metric on compact sets. It might be helpful. Commented Mar 3, 2021 at 13:50
• I see! Thank you both! Commented Mar 3, 2021 at 13:55

## 1 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 $$\lim_n d_H(A_n,A) = 0$$.

• Ah I see! Thank you! Yes, this makes sense to me! Thank you for the quick and clear answer! Commented Mar 3, 2021 at 13:52