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 $\overline{A}$.
- $A \ne \emptyset$ : 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.
- $A$ 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.
- 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?
Best Answer
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.
Your proof of a does not hold water: as above, there is no guarantee that an arbitrary sequence will satisfy the condition that you specify.
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 $\sum_{n\ne N}2^{-n}\le1$.