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Homework Help: Compactness of sets in Banach spaces

  1. Oct 15, 2012 #1
    1. The problem statement, all variables and given/known data
    Working in a banach space [itex](X,\|\cdot\|)[/itex] we have a sequence of compact sets [itex]A_k\subset X[/itex].
    Assume that there exist [itex]r_k>0[/itex] such that [itex]\sum_{k\in\mathbb{N}}r_k<\infty[/itex] and for every [itex]k\in\mathbb{N}[/itex]: $$A_{k+1}\subset\{x+u|x\in A_k,u\in X,\|u\|\leq r_k\}.$$Prove that the closure of [itex]\bigcup_{k\in\mathbb{N}}A_k[/itex] 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 [itex]\{x_n\}_{n\in\mathbb{N}}[/itex] be a sequence from the closure of [itex]\bigcup_{k\in\mathbb{N}}A_k[/itex]. Then for each n there exists [itex]y_n[/itex] from [itex]\bigcup_{k\in\mathbb{N}}A_k[/itex] such that [itex]\|x_n-y_n\|<\frac{1}{n}[/itex] and it sufficent to show that [itex]\{y_n\}_{n\in\mathbb{N}}[/itex] has a convergent subsequence.

    This is the point where I lose my grip and have no idea what to do further
  2. jcsd
  3. Oct 15, 2012 #2
    Split up the situation in two parts:

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

    What is the convergent subsequence in each of those cases?
  4. Oct 16, 2012 #3
    In Case 1 existence of convergent subsequence follows immediately from compactness of [itex]A_k[/itex].

    In Case 2 I do not yet have a complete solution.
    I decided to define [itex]\{B_n\}[/itex] disjoint such that [itex]\bigcup_k B_k=\bigcup_k A_k[/itex] and subsequence [itex]\{z_m\}_{m\in\mathbb{N}}\subset\{y_n\}_{n\in \mathbb{N}}[/itex] such that for any two elements [itex]z_i\in B_{k_i}[/itex] and [itex]z_j\in B_{k_j}[/itex], [itex]j>i[/itex] leads to [itex]k_j>k_i[/itex] and vice versa.
    Now trying to show that [itex]\{z_m\}[/itex] has a Cauchy and therefore convergent subsequence as [itex]\{B_n\}[/itex]s shrink in size, but I feel this might not be the case for infinite-dimensional spaces unless compactness of [itex]\{A_n\}[/itex]s in an infinite-dimensional space means that the object itself is finite-dimensional (does it?)
  5. Oct 16, 2012 #4
    Ok, your proposal for the convergent sequence is a good one. Now, can you find an estimate for


    Try to find an estimate that uses the [itex]r_k[/itex]?? Use that

    [tex]A_{k+1}\subseteq \{x+u~\vert~x\in A_k,~u\in X,~\|u\|<r_k\}[/tex]
  6. Oct 16, 2012 #5
    I define [itex]\{B_k\}[/itex] in the following way: $$B_1:=A_1, B_k:=A_k\setminus B_{k-1}.$$ so that [itex]B_{k+1}[/itex] is in [itex]r_k[/itex] surrounding of [itex]B_k[/itex].

    Estimating [itex]\|z_i-z_{i+1}\|[/itex] from above I get [itex]r_i +[/itex] diam[itex](B_i)[/itex] which need not go to 0 as [itex]i\to\infty[/itex] and that's the problem.
  7. Oct 16, 2012 #6
    Well, what you did now is written [itex]z_{i+1}=x+u[/itex] with [itex]x\in A_k[/itex]. Can you write other [itex]z_j[/itex] also as [itex]x+u[/itex] with [itex]x\in A_k[/itex] and [itex]\|u\|\leq \sum_k r_k[/itex]?? That way you have transformed the sequence [itex](z_j)[/itex] into a sequence in the compact set [itex]A_k[/itex] and some other terms u.
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