Compactness of sets in Banach spaces

[TaPaKaH]

Homework Statement

Working in a banach space (X,\|\cdot\|) we have a sequence of compact sets A_k\subset X.
Assume that there exist r_k>0 such that \sum_{k\in\mathbb{N}}r_k<\infty and for every k\in\mathbb{N}: $$A_{k+1}\subset\{x+u|x\in A_k,u\in X,\|u\|\leq r_k\}.$$Prove that the closure of \bigcup_{k\in\mathbb{N}}A_k is compact.

Homework Equations

The Attempt at a Solution

While talking to the teaching assistant it all seemed very doable, but now that I am back home, I am still struggling.
Here is what I was suggested to do:

Since we are in a normed space, then compactness is equivalent to sequential compactness, i.e. existence of a convergent subsequence for every sequence.
Let \{x_n\}_{n\in\mathbb{N}} be a sequence from the closure of \bigcup_{k\in\mathbb{N}}A_k. Then for each n there exists y_n from \bigcup_{k\in\mathbb{N}}A_k such that \|x_n-y_n\|<\frac{1}{n} and it sufficent to show that \{y_n\}_{n\in\mathbb{N}} has a convergent subsequence.

This is the point where I lose my grip and have no idea what to do further

 

[micromass]

Split up the situation in two parts:

• Case I: Infinitely many terms of (y_n)_n are contained in a set A_k.
• Case II: There are only finitely many terms of (y_n)_n in each A_k.

What is the convergent subsequence in each of those cases?

 

[TaPaKaH]

In Case 1 existence of convergent subsequence follows immediately from compactness of A_k.

In Case 2 I do not yet have a complete solution.
I decided to define \{B_n\} disjoint such that \bigcup_k B_k=\bigcup_k A_k and subsequence \{z_m\}_{m\in\mathbb{N}}\subset\{y_n\}_{n\in \mathbb{N}} such that for any two elements z_i\in B_{k_i} and z_j\in B_{k_j}, j>i leads to k_j>k_i and vice versa.
Now trying to show that \{z_m\} has a Cauchy and therefore convergent subsequence as \{B_n\}s shrink in size, but I feel this might not be the case for infinite-dimensional spaces unless compactness of \{A_n\}s in an infinite-dimensional space means that the object itself is finite-dimensional (does it?)

 

[micromass]

Ok, your proposal for the convergent sequence is a good one. Now, can you find an estimate for

\|z_i-z_j\|

Try to find an estimate that uses the r_k?? Use that

A_{k+1}\subseteq \{x+u~\vert~x\in A_k,~u\in X,~\|u\|<r_k\}

 

[TaPaKaH]

I define \{B_k\} in the following way: $$B_1:=A_1, B_k:=A_k\setminus B_{k-1}.$$ so that B_{k+1} is in r_k surrounding of B_k.

Estimating \|z_i-z_{i+1}\| from above I get r_i + diam(B_i) which need not go to 0 as i\to\infty and that's the problem.

 

[micromass]

I define \{B_k\} in the following way: $$B_1:=A_1, B_k:=A_k\setminus B_{k-1}.$$ so that B_{k+1} is in r_k surrounding of B_k.

Estimating \|z_i-z_{i+1}\| from above I get r_i + diam(B_i) which need not go to 0 as i\to\infty and that's the problem.

Well, what you did now is written z_{i+1}=x+u with x\in A_k. Can you write other z_j also as x+u with x\in A_k and \|u\|\leq \sum_k r_k?? That way you have transformed the sequence (z_j) into a sequence in the compact set A_k and some other terms u.