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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.

  • @JackD'Aurizio The definition of $I_n$ ought to be $[\frac{1}{2n+1},\frac{1}{2n}]$. – Alex Ortiz Jun 17 '17 at 18:13
  • @JackD'Aurizio Why should the function $f$ you consider be uniformly continuous? I am under the impression there is only one definition of uniform continuity; I've never heard of a definition tailored for disconnected domains. – Alex Ortiz Jun 17 '17 at 18:20

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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.

Alex Ortiz
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  • Thanks for your answer. I've not done up to results related to compactness. Is it possible to solve it without using the open-cover definition of compactness, using only simpler results and definitions? –  Jun 18 '17 at 05:00
  • @Dragon that's a great question. I'm sure there is a way to do it with the sequential definition of compactness, or perhaps even with something more elementary, but I'm not sure how. – Alex Ortiz Jun 18 '17 at 16:55