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If every continuous funtion $f:X\to \mathbb{R}$ is uniformly continuous, where $X \subset \mathbb{R}$. Then $X$ is closed. But not necessarily compact

It's like the reciprocal of continuous $f$ and $X$ compact, then $f$ is uniformly continuous. But I do not know where to start

help

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  • $\begingroup$ $X$ is closed in what topology? $\endgroup$ – SMJK Oct 29 '16 at 21:27
  • $\begingroup$ With the usual. $X \in R$ $\endgroup$ – Jhon Kevin Astoquillca Aguilar Oct 29 '16 at 21:31
  • $\begingroup$ Have you tried the contrapositive? $\endgroup$ – Eric Towers Oct 29 '16 at 21:33
  • $\begingroup$ You should add to the question that $X \subset \mathbb{R}$. $\endgroup$ – SMJK Oct 29 '16 at 21:33
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If $X$ is not closed then there exists a limit point $x_0 \in \mathbb{R}$ not contained in $X$. Then, the function

$$f(x) := \frac{1}{x-x_0}$$

is continuous on $X$ but not uniformly continuous. Note that since this argument does not rely on $X$ being bounded or not, we cannot make any deduction about compactness of $X$.

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If $X$ were not closed, so that $a\in \mathbb{R} \backslash X$ is a limit point for $X$, the function $f(x)=\frac{1}{x-a}$ defined on $X$ is continuous but not uniformly continuous, since it is unbounded on any neighborhood of $a$.

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