The Problem : Let $f:B \subset \mathbb{R} \to \mathbb{R}$ be uniformly continuous on a bounded set $B$. To show that $f(B)$ is bounded.
I have done the simplified case when $B$ is an interval (using Weierstrass' theorem and local boundedness induced by continuity). But what if $B$ is not an interval? Can we always write B as finite union of intervals? Any help will be greatly appreciated. Thank you.
If $B\subset\mathbb R$ is bounded, then $\overline B$ is compact. Since $f$ is uniformly continuous, there is a unique continuous extension $\tilde f$ of $f$ defined on $\overline B$. Since $\overline B$ is compact, $\tilde f$ is uniformly continuous. Let $\delta$ be such that $|\tilde f(x)-\tilde f(y)| < 1$ if $|x-y|<\delta$. By compactness, there are finitely many $\delta$-neighborhoods that cover $\overline B$. Hence $\tilde f$ is bounded on $\overline B$. The restriction $\tilde f|_B = f$ is therefore also bounded.