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Quick question - finishing a convergence proof

  1. Oct 8, 2009 #1
    1. The problem statement, all variables and given/known data
    Prove [tex]c_0[/tex] is closed in [tex]l^{\infty}[/tex].


    2. Relevant equations
    A set is closed iff it contains all its accumulation points.


    3. The attempt at a solution
    Let [tex]\left\{x_n\right\}[/tex] be an accumulation point of [tex]c_0[/tex]. Then for all [tex]\epsilon > 0[/tex], there exists [tex]\left\{y_n\right\}[/tex] in [tex]c_0[/tex] such that [tex]d(x_n, y_n) = sup|x_n - y_n| < \epsilon[/tex]. Then [tex]|x_n - y_n| < \epsilon[/tex] for all n. Now since [tex]\left\{y_n\right\}[/tex] converges to 0, there exists an N such that [tex]|y_n| < \epsilon[/tex] for all n bigger than N. But then by reverse triangle inequality,
    [tex] |x_n|\leq |x_n - y_n| + |y_n| < 2\epsilon. [/tex]

    Now I want to conclude that [tex]\left\{x_n\right\}[/tex] converges to 0, but I'm not sure how to say it. It's clear that since [tex]\epsilon[/tex] was arbitrary, [tex]\left\{x_n\right\}[/tex] should converge, but somehow I don't feel this is satisfactory enough for my 3rd year analysis course. Suggestions?

    Edit: why does my LaTex look so ugly?
     
  2. jcsd
  3. Oct 8, 2009 #2

    Office_Shredder

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    You can also choose n big enough and a sequence yk so that d(xk, yk) < epsilon/2, and |yn| < epsilon/2. Then |xn| < epsilon. But epsilon is arbitrary, so this is the actual definition of converging to zero.
     
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