# Compactness of sets in Banach spaces

1. Oct 15, 2012

### TaPaKaH

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
3. 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

2. Oct 15, 2012

### micromass

Staff Emeritus
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?

3. Oct 16, 2012

### 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?)

4. Oct 16, 2012

### micromass

Staff Emeritus
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

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

5. Oct 16, 2012

### 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.

6. Oct 16, 2012

### micromass

Staff Emeritus
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.