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Math Help - prove a lemma for a metric space

  1. #1
    Junior Member hercules's Avatar
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    prove a lemma for a metric space

    I need help proving this

    Lemma: A sequence \left( \bold{ x^{(n)}} \right) in \mathbb{R}^k converges if and only if for each j = 1,2,...,k, the sequence \left( x_j^{(n)}\right) converges in \mathbb{R}.

    Thanks guys
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  2. #2
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    Quote Originally Posted by hercules View Post
    I need help proving this

    Lemma: A sequence \left( \bold{ x^{(n)}} \right) in \mathbb{R}^k converges if and only if for each j = 1,2,...,k, the sequence \left( x_j^{(n)}\right) converges in \mathbb{R}.

    Thanks guys
    Say |\bold{x}_k - \bold{y} | <\epsilon.
    Then it means,
    \sqrt{(x^{(1)}_k - y_1)^2+...+(x^{(n)} - y_k)^2} < \epsilon
    But that means,
    \max | x^{(i)}_k - y_i| < \epsilon.
    Thus, each component converges.

    Conversely if each component converges it means,
    \max |x^{(i)}_k - y_i| < \epsilon.
    That means,
    \sqrt{(x^{(1)}_k - y_1)^2+...+(x^{(n)} - y_k)^2} \leq \sqrt{ n \max (x^{(i)}_k - y_i)^2} = \sqrt{n}\max | x^{(i)} - y_i| < \sqrt{n} \epsilon.
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