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Math Help - co is closed in l∞

  1. #1
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    co is closed in l∞

    I want to show the Banach space co is closed in l∞ .

    So, I pick a convergent sequence xn in co that converges to x in l
    Now, x_n --> x: given e>0, there is an N_e s.t. for all n>N_e,
    ||x_n -x ||= Sup |x_n(k)-x(k)|<e (we're supping over k).

    Since x_n is a sequence in co , for each fixed n, x_n(k)-->0 as k--> infinity.
    So, given e>0, there is a K depending on n and e, such that for all k> K, we have |
    x_n(k)|<e.

    We want to show x is in
    co
    so we show there is a Ko such that for all k>Ko, |x(k)|<e

    I am having trouble getting this Ko.

    I know |x(k)| |x_n(k)| + |x_n(k)-x(k)| |x_n(k)| + Sup (over k) |x_n(k)-x(k)|

    we have |x_n(k)|<e as k>K, but Sup (over k) |x_n(k)-x(k)|<e for n>N_e.


    So I am not so sure how to get this Ko.


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  2. #2
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    Opalg's Avatar
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    I think you have essentially done this. It's just a matter of piecing the bits together.

    You know that x_n\to x in the sup norm. So given \varepsilon>0 there exists N such that n\geqslant N\,\Rightarrow\,\sup_k|x_n(k) - x(k)|<\varepsilon/2.

    Choose and fix a value of n greater than or equal to N. Then x_n\in c_0 and so \lim_{k\to\infty}x_n(k) = 0. Therefore there exists K such that k\geqslant K\,\Rightarrow\,|x_n(k)|<\varepsilon/2.

    Hence k\geqslant K\,\Rightarrow\,|x(k)|\leqslant|x_n(k)| + |x_n(k) - x(k)|<\varepsilon. Therefore x\in c_0.
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  3. #3
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    Thanks, it makes a lot of sense.
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