Please, I need some clarification on this. If $f$ is continuous and bounded on $\Bbb{R}$, does it mean that $f$ is uniformly continuous on $\Bbb{R}$?
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2No. ${}{}{}{}{}$ – May 08 '18 at 22:22
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@ T. Bongers: Why? – Omojola Micheal May 08 '18 at 22:22
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@ T. Bongers: Any counterexample? – Omojola Micheal May 08 '18 at 22:23
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1Because there are tons of counterexamples; can you share what research you've done about this question, and what you've tried to answer it? For example: googling the phrase "continuous bounded function uniformly continuous" gave an answer from this very site. – May 08 '18 at 22:23
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Have you tried playing around at all @Mike – operatorerror May 08 '18 at 22:24
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@ qbert: Yes, I'm playing around right now! The question requires me to prove or provide a counterexample. – Omojola Micheal May 08 '18 at 22:25
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What about $h(x)=\frac{\sin x}{x},;x\neq 1.$ The function $\bar{h}$ defined on $\Bbb{R}$ by $\bar{h}(x)=\begin{cases}\frac{\sin x}{x}, & x \neq 0\1 & x=0\end{cases}$ – Omojola Micheal May 08 '18 at 22:37
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Your post has been closed so I answer this way hoping you can read it. Take the function $f(x)= \sin(e^x)$ as a good counterexample. – Ataulfo May 08 '18 at 23:23
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@Piquito: Thanks a lot! – Omojola Micheal May 09 '18 at 09:28
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You are welcome. (The example is very good really, I like it. In my answer I told you is important keep in mind that EACH OPEN BOULE of radius $\delta$ is contained in an open ball of radius $\epsilon$. With this idea you can try to get another good example). – Ataulfo May 09 '18 at 23:48