1. Limited time only! Sign up for a free 30min personal 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 closed unit ball

  1. Apr 23, 2010 #1
    1. The problem statement, all variables and given/known data
    Let l∞ be the space of bounded sequences of real numbers, endowed with the norm
    ∥x∥∞ = supn∈N |xn | , where x = (xn )n∈N .
    Prove that the closed unit ball of l∞ , B(0, 1) = {x ∈ l∞ ; ∥x∥∞ ≤ 1} , is not compact.

    2. Relevant equations



    3. The attempt at a solution
    I'm thinking about using the notion of sequential compactness, since every sequence Xn has an upper limit here, but I'm not sure if that would help much. Could anyone please give me a hint? Any input is appreciated!
     
  2. jcsd
  3. Apr 23, 2010 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Sure, use the sequential form. Find a sequence in B(0,1) with no convergent subsequence.
     
  4. Apr 23, 2010 #3
    thanks for the reply, but I'm still not sure about which non-convergent sequence to choose, would something like 1/n work? I just don't know how to use the complete norm here.
     
  5. Apr 23, 2010 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Remember what the space is here. It's the set of bounded sequences. So a point in your space is itself a sequence. Thus a sequence in your space must in fact be a sequence of sequences. Give me some examples.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Compactness of closed unit ball
Loading...