# A way to prove that Hausdorff distance is complete

I have to solve an exercise the result of which will lead to the completeness of the Hausdorff distance. Basically it si the fulfilling of the details of this answer https://math.stackexchange.com/q/2197008

My definitions:

In all the following $(X,d)$ will be a metric space and $B(x,r) = \{ y \in X : d(x,y)<r \}$ for any $x \in X$ and $r>0$.

.Let $C \subset X$ and $r>0$, then $C_r := \bigcup_{y \in C}B(y,r)$.

.Let $\mathcal{X}$ be the collection of the nonempty, closed and bounded subsets of $X$. Then for every $C,D \in \mathcal{X}$ the Hausdorff distance is defined as $$h(C,D) := \inf \{ r | C \subset D_r , D \subset C_r \}$$

The proposition I want to prove is:

If $(X,d)$ is complete, then $(\mathcal{X},h)$ is complete.

The book suggest to consider a sequence $\{A_n \} \subset \mathcal{X}$ s.t. $h(A_n, A_{n+1})<2^{-n}$ and to prove that it converges to the closure of the limit points of the sequences $\{ x_n \}$ s.t. $x_n \in A_n$ for all $n \in \mathbb{N}$.

My attempt

Let $\{A_n \} \subset \mathcal{X}$ be a Cauchy sequence. I can extract a subsequence (which I will call again $\{A_n \}$) s.t. $h(A_n, A_{n+1})<2^{-n}$. If I prove the convergence of this subsequence I have the thesis. Following the suggestion let $A \subset X$ be defined as $$A:= \{ x \in X | \exists \{x_n\} s.t. x_n \in A_n \forall n \in \mathbb{N} \, , \, x_n \to x \}$$ and I want to prove $\overline{A} \in \mathcal{X}$ and $\{A_n \}$ converges to $\overline{A}$.

1. $A \ne \emptyset$ : every sequence $\{x_n\}$ s.t. $x_n \in A_n$ for all $n \in \mathbb{N}$ is Cauchy and then it has a limit. Note every $A_n$ is nonempty then those sequences actually exist.
2. $A$ is bdd: first of all observe that (by Cauchy property) $\exists N \in \mathbb{N}$ s.t. $h(A_N, A_n) < 1$ for all $n \ge N$. Since $A_N$ is bdd then $\exists R>0$ and $y \in X$ s.t. $A_N \subset B(y, R)$. Then $A_n \subset B(y, R+1)$ for all $n \ge N$. Suppose $A$ is not bdd, then $\exists x \in A$ s.t. $d(y,x) > 3(R+1)$. By definition of $A$, $x$ is the limit point of some sequence $\{x_n\}$ with $x_n \in A_n$ for all $n \in \mathbb{N}$. Then $\exists M \in \mathbb{N}$ s.t. $d(x,x_n)<R+1$ for all $n \ge M$. Take any $n \ge \max\{N,M\}$, then $d(y,x_n) \ge 2(R+1)$ but then $A_n$ is not included in $B(y,R+1)$. Contradiction.
3. By 1. and 2. we have $\overline{A}$ is nonempty, bdd and closed hence an element of $\mathcal{X}$.

Now I want to prove $\{A_n \}$ converges to $\overline{A}$. Let $\epsilon>0$ be fixed, I have to prove $\exists N_{\epsilon} \in \mathbb{N}$ s.t. $h(A_n, \overline{A})< \epsilon$ for all $n \ge N_{\epsilon}$ and it is enough to prove that $\exists N_{\epsilon} \in \mathbb{N}$ s.t.

a. $\forall \, x \in A_n \, \, \exists z \in \overline{A}$ s.t. $d(x, z)< \epsilon$

b. $\forall \, y \in \overline{A} \, \, \exists w \in A_n$ s.t. $d(y, w)< \epsilon$

for all $n \ge N_{\epsilon}$.

a. Take $N_1$ s.t. $2^{-N_1}< \epsilon/2$, then $h(A_n, A_{n+1}) < 2^{-n}$ for all $n \ge N_1$. Take $m \ge N_1$ and $x \in A_m$. Then there exists a sequence $\{ x_n \}$ s.t. $x_n \in A_n$ for all $n \in \mathbb{N}$ s.t. $x_m=x$ with limit point $z \in A$ and s.t. $d(x_n, x_{n+1})<2^{-n}$. If $n>m \ge N_1$ then $d(x_m,x_n) < \sum_{j=m}^{n-1}d(x_j,x_{j+1}) =2^{-m}\sum_{j=0}^{n-m-1}2^{-j} <2^{-m+1} < \epsilon$ and then $d(x_m,z) < \epsilon$.

b. Take $N_2$ s.t. $h(A_n, A_{n+1}) <2^{-N_2}<\epsilon/4$ for all $n \ge N_2$. Let $y \in \overline{A}$, then take $u \in B(y, \epsilon/2) \cap A \ne \emptyset$. Then take any sequence $\{x_n\}$ s.t. $x_n \in A_n$ for all $n \in \mathbb{N}$ and $x_n \to u$ and s.t. $d(x_n,x_{n+1})<2^{-n}$ for all $n \ge N_2$. If $m>n \ge N_2$ then $d(x_m,x_n) < \sum_{j=n}^{m-1}d(x_j,x_{j+1}) =2^{-n}\sum_{j=0}^{m-n-1}2^{-j} <2^{-n+1} < \epsilon/2$ and then $d(x_m,u) < \epsilon/2$. Then if $w=x_m$, $d(y,w) \le d(y,u)+d(u,w) <\epsilon$.

Then it is enough to take $N_{\epsilon} = \max \{N_1, N_2 \}$.

My doubts

Have I used in some crucial way the subsequence with distance $2^{-n}$? Maybe If not I'm not sure that every $\{x_n \}$ s.t. $x_n \in A_n$ is Cauchy?

Your proof that $A\neq\emptyset$ is wrong: not every (arbitrary) sequence like that is automatically a Cauchy sequence. But start with $x_0\in A_0$ and then find $x_1\in A_1$ with $d(x_0,x_1)<2^0$, then $x_2\in A_2$ with $d(x_1,x_2)<2^{-1}$, $x_3\in A_3$ with $d(x_2,x_3)<2^{-2}$, ..., $x_{n+1}\in A_{n+1}$ with $d(x_n,x_{n+1})<2^{-n}$, ... that will be a Cauchy sequence.
Neither does your proof in b: $N_2$ depends ultimately on $y$ (via $u$ and the sequence converging to $u$); it should be independent of the point in the closure of $A$.
Your doubts: you have used the $2^{-n}$ when going from $R$ to $R+1$, because $\sum_{n\ne N}2^{-n}\le1$.
• Thank you very much for your answer. For the part of $A \ne \emptyset$ you are right, I realized it by myself yet. In the proof af a. I wrote "there exists a sequence" , not "every sequence". In the proof of b, you are right: I have edited the question. Is it ok now? – Bremen000 Sep 11 '18 at 12:52
• Yes, you are right. I didn't change point 1. because, as you have said in your answer, I can extract a Cauchy sequence from the $\{A_n \}$ s.t. $x_n \in A_n$ for all $n \in \mathbb{N}$. It is something I already knew because it was the statement of another point of this exercise on the book, and I used exactly your argument to prove it. So my concerns were only about points a. and b. . I think that, with your correction of point 1. , everything should work! Does it? – Bremen000 Sep 12 '18 at 18:49