I understand that in an open interval the only functions that are continuous but not uniformly are functions whose limits are singularities. But when we have a function $f:H\rightarrow\mathbb{R}$ and $H$ is not an open interval, what is an example a function that satisfies continuity but not uniform continuity ($H$ and $f$, please)?
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asked Jan 5, 2014 at 16:34
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$\begingroup$ If $H$ is a closed interval, it can't be done. If $H$ is not an interval at all, well, there are a lot of different $H$ around. What did you have in mind? $\endgroup$– Gerry MyersonJan 5, 2014 at 16:38
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$\begingroup$ Is $H$ a union of open intervals? $\endgroup$– Karolis JuodelėJan 5, 2014 at 16:38
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$\begingroup$ $\exp\cdot$ grows too rapidly to be uniformly continuous $\endgroup$– obatakuJan 5, 2014 at 16:54
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2 Answers
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I think you have in mind that the open interval is also bounded when you say "I understand that in an open interval the only functions that are continuous but not uniformly are functions whose limits are singularities".
Let $\mathbb{R}=(-\infty,\infty)$ (you can view $\mathbb{R}$ as an interval which hopefully explains my previous comment) then $f(x)=x^{2}$ is not uniformly continuous. (You can negate the definition of uniform continuity and show that the statement you obtain holds true. It will take some doing, but the basic idea is that as long as you can pick your numbers from the real line, you can pick two numbers really close together such that their squares are far apart. For example $10^{10}+\frac{1}{10^{10}}$ and $10^{10}$ are really close together but their squares are $2$ apart.)
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$H$ be the half open interval $(0,1]$ and $f(x)=\sin\left(\frac{1}{x}\right)$.
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$\begingroup$ That's kind of the same as 1/x though, just bounded. Not really what I was looking for... $\endgroup$ Jan 5, 2014 at 18:19
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2$\begingroup$ @ElliotG Not really. The "bad" behaviour of this function is different from that of $x^{-1}$, maybe more interesting, I'd say. Can you prove it is not uniformly continuous? $\endgroup$– Pedro ♦Jan 5, 2014 at 22:53