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Let $E$ be a bounded measurable set of $\mathbb{R}^n$. If every countinuous function $f:E\to\mathbb{R}$ is also uniformly continuous, then $E$ is compact.

Since I do not know topological property of $f(E)$, it seems that I must use some measure property to show $E$ is closed. However, measure and topology seems not quite relevant.

Thank you!

Leonardo
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  • Is measurability of $E$ just a disguise to make the question look a little bit different? What if $E$ is a non-measurable set? – Leonardo Sep 05 '18 at 08:16
  • @Leonardo note that while measurability is superfluous such a set will be, a posteriori, necessarily measurable since it's compact – Alessandro Codenotti Sep 05 '18 at 09:46
  • You are right! In other words, if $E$ is non-measurable, then it is definitely not compact by regularity of Lebesgue measure. The question is then meaningless at all. – Leonardo Sep 05 '18 at 14:57

1 Answers1

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Suppose $x_n \in E, x_n \to x$ and $x \notin E$. Let $f(y)=\frac 1 {\|y-x\|}$. Then $f$ is continuous, hence uniformly continuous by hypothesis. Since $\|x_n-x_m\| \to 0$ it follows from uniform continuity that $\frac 1 {\|x_n-x\|}-\frac 1 {\|x_m-x\|} \to 0$. Since Cauchy sequences in $\mathbb R^{n}$ are bounded it follows that $\frac 1 {\|x_n-x\|}$ is bounded. But this sequence tends to $\infty$. This contradiction shows that $E$ is closed. Since it is bounded, it is compact.

Kavi Rama Murthy
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