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[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

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    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.
     
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