Let $f$ be real uniformly continuous function on the bounded set $E$ in $\mathbb{R}^1$. Prove that $f$ is bounded on $E$. Show that the conclusion is false if boundedness of $E$ is omitted from the hypothesis.

Proof: Since $f$ is uniformly continuous on $E$ then for $\varepsilon=1$ $\exists \delta>0$ such that for any $x,y\in E$ and $|x-y|<\delta$ implies that $|f(x)-f(y)|<1$. Hence by triangle inequality $|f(x)|<1+|f(y)|.$

As $E$ is bounded then $\exists r>0$ such that $E\subset (x-r,x+r)$. It's easy to check that that $(x-r,x+r)$ we can write as disjoint union of intetvals each lenght is $\frac{\delta}{2}$. Namely $$(x-r,x+r)=(x-r,x-r+\frac{\delta}{2})\cup \bigcup_{k=1}^{[\frac{4r}{\delta}]-1}[x-r+\frac{k\delta}{2},x-r+\frac{(k+1)\delta}{2}]\cup [x-r+\frac{\delta}{2}\left[\frac{4r}{\delta}\right],x+r).$$ Using that $E\cap(\cup_{i=1}^{n} A_i)=\cup_{i=1}^{n} (E\cap A_i)$ and define $(a,b)_E=(a,b)\cap E$ we get: $$E=(x-r,x-r+\frac{\delta}{2})_E\cup \bigcup_{k=1}^{[\frac{4r}{\delta}]-1}[x-r+\frac{k\delta}{2},x-r+\frac{(k+1)\delta}{2}]_E\cup [x-r+\frac{\delta}{2}\left[\frac{4r}{\delta}\right],x+r)_E.$$ We see that $|f(x)|< 1+|f(x-r)+\frac{\delta}{2}|$ on the first interval, $|f(x)|<1+|f(x-r+\frac{\delta(k+1)}{2})|$ for $k=\overline{1,[\frac{4r}{\delta}]-1}$ and $|f(x)|<1+f(x+r)$ on the last half-segment. If $M=\max\limits_{k=1,..,[\frac{4r}{\delta}]}\{|f(x-r+\frac{\delta}{2})|,|f(x+r)|,|f(x-r+\frac{\delta(k+1)}{2})|\}$. Then $|f(x)|<1+M$ for any $x\in E$.

But if we omit boundedness the conclusion is false because $f(x)=x$ is uniformly continuous at $E=[0,+\infty)$ but it's unbounded on $E$.

Is my proof true? Sorry if this topic repeated.

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    $\begingroup$ Looks good. You may choose $r=4N\delta$ with $N$ large enough. That would reduce notation a bit. $\endgroup$ – Quang Hoang Sep 22 '15 at 16:15
  • $\begingroup$ Yeah right. But is the proof of first part is right? $\endgroup$ – ZFR Sep 22 '15 at 16:18
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    $\begingroup$ Did I forget to say it looks good? Oh I did :-) $\endgroup$ – Quang Hoang Sep 22 '15 at 16:20
  • $\begingroup$ Thanks a lot for checking, dear Quang Hoang! :) $\endgroup$ – ZFR Sep 22 '15 at 16:22

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