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Show that the following function is uniformly continuous on $(-1,1)$

$$f(x) = \begin{cases} {x \sin \frac{1} {x}}, & \text{ } x\in(-1,0)\cup(0,1) \\ 0, & \text{ }x = 0. \end{cases} $$

We cannot use the theorem that a continuous function on a compact set K is continuous on K, because we don't have a compact set. I was told the following hint: "if a function is uniformly continuous on a set then it is also uniformly continuous on any subset of this set". I don't know exactly what to do with this information, can you help me ? :)

I know the definition of uniform continuity, I (should) know what open, closed, compact sets are.

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1 Answer 1

up vote 5 down vote accepted

Hint: Show that you can extend the definition to $[-1,1]$ and that it is continuous on the closed interval. Then use the theorem about uniform continuity.

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Very fast way to have the conclusion (+1). Good to know that $f(x)=\sin(1/x)$ fails to be U.C. in $(-1,1)$. –  Babak S. Oct 20 '12 at 19:14
    
So you suggest that I define f(1)= sin(1) and f(-1)= - sin(-1), then show that the function is continious. Continious functions on a compact set are uniformly continious, and then use the hint, namely a (-1,1) is a subset of [-1,1] ? –  MSKfdaswplwq Oct 20 '12 at 19:33
2  
Yes, there is no problem with that. The "magic" happens at $x=0$ anyway. –  Hagen von Eitzen Oct 20 '12 at 19:39
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@Hempo: Yes, exactly. –  Asaf Karagila Oct 20 '12 at 20:15
    
@Hempo: Is there a particular reason for which you unaccepted the answer? –  Asaf Karagila Nov 3 '12 at 22:26

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