# Thread: Metric space and boundedness

1. ## 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?

2. Originally Posted by manjohn12
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 $\displaystyle - K < f(x) < K \implies |f(x)| < K$.