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August 31st, 2015, 05:52 PM   #11
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Yes, zylo indeed it can. This statement is enough:

Consider $\displaystyle U\subset\mathbb R^m$ and $\displaystyle f:U\rightarrow\mathbb R^n$ uniformly continuous. Then there exist $\displaystyle \overline{f}:\overline{U}\rightarrow\mathbb R^n$ continuous such that $\displaystyle \overline{f}(x)=f(x)$ for all $\displaystyle x\in U$.

With this result in hand you have that if $\displaystyle U$ is bounded then $\displaystyle \overline{U}$ is compact and therefore $\displaystyle \overline{f}$ is bounded which gives us that $\displaystyle f$ is bounded.

As soon as my wife let go of my laptop I will post the proof. It's a little bit boring writing it all in the cell phone.
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Last edited by parasio; August 31st, 2015 at 05:58 PM.
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September 1st, 2015, 05:53 PM   #12
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As promissed here is my solution..


Afirmation: Consider $\displaystyle U\subset\mathbb R^m$ and consider $\displaystyle f:U\rightarrow\mathbb R^n$ a uniformly continuous function. Then for each $\displaystyle a\in U'$, $\displaystyle \displaystyle\lim_{x\rightarrow a}f(x)$ exists.

Proof: Take $\displaystyle a\in U'$ then there exist a sequence $\displaystyle (a_n) $ in $\displaystyle U\backslash\{a\}$ such that $\displaystyle \displaystyle\lim_{n\rightarrow\infty}a_n=a$ which gives us that $\displaystyle (a_n)$ is a Cauchy sequence. Since $\displaystyle f$ is uniformly continuous we have that the sequence $\displaystyle (f(a_n))$ is a Cauchy sequence. Since $\displaystyle \mathbb R^n$ is complete we have that there exist $\displaystyle L=\displaystyle\lim_{n\rightarrow\infty}f(a_n)$.

Take another sequence $\displaystyle (b_n)$ in $\displaystyle U\backslash\{a\}$ with $\displaystyle \displaystyle\lim_{n\rightarrow\infty}b_n=a$. Consider the sequence $\displaystyle (c_n)$ where $\displaystyle c_{2n-1}=a_n$ and $\displaystyle c_{2n}=b_n$. We have that $\displaystyle \displaystyle\lim_{n\rightarrow\infty}c_n=a$ and therefore it's a Cauchy sequence which gives us that there exist $\displaystyle M=\displaystyle\lim_{n\rightarrow\infty}f(c_n)$. Since $\displaystyle (f(a_n))$ is a subsequence of $\displaystyle (f(c_n))$ we have that $\displaystyle M=L$. Since $\displaystyle \displaystyle\lim_{n\rightarrow\infty}b_n=a$ we have that $\displaystyle (b_n)$ is a Cauchy sequence which gives us that there exist $\displaystyle N=\displaystyle\lim_{n\rightarrow\infty}f(b_n)$. Since $\displaystyle (f(b_n))$ is a subsequence of $\displaystyle (f(c_n))$ we have that $\displaystyle N=M=L$. Hence for any sequence $\displaystyle (a_n)$ in $\displaystyle U\backslash\{a\}$ converging to $\displaystyle a$ we have that $\displaystyle \displaystyle\lim_{n\rightarrow\infty}f(a_n)=L$ and therefore there exist $\displaystyle \displaystyle\lim_{x\rightarrow a}f(x)$ and $\displaystyle \displaystyle\lim_{x\rightarrow a}f(x)=L$.

Consider now the function $\displaystyle \overline{f}:\overline{U}\rightarrow\mathbb R^n$ (remember $\displaystyle \overline{U}=U'\cup U$) defined as $\displaystyle \overline{f}(a)=f(a)$ if $\displaystyle a\in U$ and $\displaystyle \overline{f}(a)=\displaystyle\lim_{x\rightarrow a}f(x)$ if $\displaystyle a\in U'\backslash U$. $\displaystyle \overline{f}$ is obviously continuous.
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