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Math Help - Metric space and boundedness

  1. #1
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    Metric space and boundedness

    Prove: Let  X be a metric space, let  a \in X be a limit point of  X , and let  f: X \to \mathbb{R} be a function. Suppose that  \lim_{x\to a} f(x) = L . For all  K \in \mathbb{R} such that  |L| < K , there exists  \delta > 0 such that if  x \in B_{\delta}(a) \ \backslash \{a \} then  |f(x)| < K

    In other words  -K < f(x) < K for some deleted  \delta -ball about  a . How would you show this? What value of  \delta would you pick?
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    Quote Originally Posted by manjohn12 View Post
    Prove: Let  X be a metric space, let  a \in X be a limit point of  X , and let  f: X \to \mathbb{R} be a function. Suppose that  \lim_{x\to a} f(x) = L . For all  K \in \mathbb{R} such that  |L| < K , there exists  \delta > 0 such that if  x \in B_{\delta}(a) \ \backslash \{a \} then  |f(x)| < K
    Pick \epsilon > 0 so that |L| + \epsilon < K. This means that |L| < K - \epsilon \implies - K + \epsilon < L < K - \epsilon.

    Now this means there is \delta>0 so that 0 < d(x,a) < \delta \text{ and }x\in X\implies |f(x) - L| < \epsilon .

    But that means L - \epsilon < f(x)  < L + \epsilon \implies (-K + \epsilon) - \epsilon < f(x) < (K+\epsilon) - \epsilon so  - K < f(x) < K \implies |f(x)| < K.
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