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For an unbounded nonempty set of real numbers $D$, does there necessarily exist a continuous function $f$ that maps $D$ to the real numbers such that $f$ is not uniformly continuos?

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No, there doesn't. Let $D = \mathbf N$, and let $f \colon \mathbf N \to \mathbf R$. We will show that $f$ is uniformly continuous: Let $\epsilon > 0$, let $\delta = \frac 12$. If $x,x' \in \mathbf N$ are given such that $|x-x'| < \delta = \frac 12$, we have $x=x'$, hence $f(x) = f(x')$, so $|f(x)-f(x')| < \epsilon$.

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  • $\begingroup$ That's an answer that i was thinking about. but does that not only prove it for the case where $D$ is the set of real numbers? Because there are other unbounded sets of real numbers? $\endgroup$ – nEv3r Oct 5 '16 at 9:36
  • $\begingroup$ @nEv3r There are unbounded sets, where there is a non-uniformly continuous function. On $\mathbf N$ there isn't. So it proves there oppisite of "For all unbounded $D$, there is ...", which is: "There is some unbounded $D$, where there is not ....". $\endgroup$ – martini Oct 5 '16 at 9:41

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