Vtyjkyuk

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 $C_r := bigcup_y in CB(y,r)$.

.Let $mathcalX$ be the collection of the nonempty, closed and bounded subsets of $X$. Then for every $C,D in mathcalX $ the Hausdorff distance is defined as $$h(C,D) := inf r $$

The proposition I want to prove is:

If $(X,d)$ is complete, then $(mathcalX,h)$ is complete.

The book suggest to consider a sequence $ A_n subset mathcalX$ 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 $n in mathbbN$.

My attempt

Let $ A_n subset mathcalX$ 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
$$A:= exists x_n s.t. x_n in A_n forall n in mathbbN , , , x_n to x $$
and I want to prove $overlineA in mathcalX$ and $ A_n $ converges to $overlineA$.

1. $A ne emptyset$ : every sequence $ x_n$ s.t. $x_n in A_n$ for all $n in mathbbN$ 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 mathbbN $ 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 $n in mathbbN$. Then $ exists M in mathbbN$ s.t. $d(x,x_n)<R+1$ for all $n ge M$. Take any $n ge maxN,M$, 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 $overlineA$ is nonempty, bdd and closed hence an element of $mathcalX$.

Now I want to prove $ A_n $ converges to $overlineA$. Let $epsilon>0$ be fixed, I have to prove $exists N_epsilon in mathbbN$ s.t. $h(A_n, overlineA)< epsilon$ for all $n ge N_epsilon$ and it is enough to prove that $exists N_epsilon in mathbbN$ s.t.

a. $ forall , x in A_n , , exists z in overlineA$ s.t. $d(x, z)< epsilon$

b. $ forall , y in overlineA , , 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 $n in mathbbN$ 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 $d(x_m,x_n) < sum_j=m^n-1d(x_j,x_j+1) =2^-msum_j=0^n-m-12^-j <2^-m+1 < epsilon$ 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 $y in overlineA$, 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 $n in mathbbN$ 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 $d(x_m,x_n) < sum_j=n^m-1d(x_j,x_j+1) =2^-nsum_j=0^m-n-12^-j <2^-n+1 < epsilon/2$ 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 $Aneqemptyset$ is wrong: not every (arbitrary) sequence like that is automatically a Cauchy sequence. But start with $x_0in A_0$ and then find $x_1in A_1$ with $d(x_0,x_1)<2^0$, then $x_2in A_2$ with $d(x_1,x_2)<2^-1$, $x_3in A_3$ with $d(x_2,x_3)<2^-2$, ..., $x_n+1in A_n+1$ 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 $sum_nne N2^-nle1$.