Metric space and boundedness

**manjohn12** · December 6th 2008, 03:55 PM

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?

**ThePerfectHacker** · December 6th 2008, 05:41 PM

Originally Posted by manjohn12

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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