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!

[a,b] is sequentially compact

  1. Oct 21, 2008 #1
    I'd like to show that [a,b] is sequentially compact. So I pick a sequence in [a,b] , say (xn).

    case 1:range(xn) is finite
    Then one term, say c is repeated infinitely often. Now we choose the subsequence that has infinitely many similar terms c. It converges to c.

    case 2: range of (xn) is infinite.
    I know I can prove this using the Bolzano Weierstrass theorem that says that every infinite bounded subset of R has a limit point. But in this case I'd have to prove the theorem. How?
    Or I may Try and subdivide [a,b] in two parts and say that one part must contain infinitely many points of the range, and pick a term x_n1 in that part. Next I can choose x_n2 since I have infinitely many terms to choose from. And then continue to subdivide.... But then I'm stuck. I am not sure what I need to do from there or how exactly I need to write this down (how the subdivision part works exactly).
    Thanks for your help.
     
  2. jcsd
  3. Oct 21, 2008 #2

    morphism

    User Avatar
    Science Advisor
    Homework Helper

    The subdivision part works like this. Say you divide [a,b] into two closed intervals I_1 and J_1 (for instance, I_1 = [a,c] and J_1 = [c,b]), where say I_1 contains infinitely many terms of the sequence. Then further split I_1 into I_2 and J_2 where I_2 contains infinitely many terms of the sequence. Proceed like this to get a decreasing sequence of closed intervals I_1 >= I_2 >= ... of decreasing 'length'.

    Show that the sequence {x_n_i} you formed is Cauchy.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: [a,b] is sequentially compact
Loading...