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Thread: Metric space and boundedness

  1. #1
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    Metric space and boundedness

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

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

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

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