1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    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

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

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

    [tex]\|z_i-z_j\|[/tex]

    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

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Compactness of sets in Banach spaces
  1. Banach space (Replies: 5)

  2. Compact Sets (Replies: 6)

Loading...