# Thread: Metric space and boundedness

1. ## 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?

2. 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 $- K < f(x) < K \implies |f(x)| < K$.