If $f(x)$ is uniformly continuous at $(0,1)$ then is it bounded at $(0,1)$? 
Possible Duplicate:
Uniform continuity 

I have a question which is:

If ${f(x)}$ is uniformly continuous at ${(0,1)}$ then is it bounded at ${(0,1)}$?

This sound like it's correct to me but I can't see why exactly (Or maybe it's wrong :P).
Could someone help me figure out the truth here? :) Thanks!
 A: By uniform continuity there is an $n\in{\mathbb N}_{\geq 1}$ such that $0<x\leq y<x+{1\over n}<1$ implies $|f(y)-f(x)|<1$. Put $x_k:={k\over n+1}$ $\ (1\leq k\leq n)$. Then any $y \in \ ]0,1[\ $ is at distance ${}<{1\over n}$ from an $x_k$. Therefore we have
$$|f(y)| < \max_{1\leq k\leq n} |f(x_k)| + 1\ =:\ M$$
for all $y \in \ ]0,1[\ $.
A: $f$ would indeed be bounded.
$f$ is continuous on $(0,1)$; thus, if it were not bounded on $(0,1)$, there would exist $x_n\nearrow 1$ or $x_n\searrow 0$ with $f(x_n)\rightarrow\infty$ or $f(x_n)\rightarrow-\infty$. But, none of these options can happen due to the following facts: 1) uniformly continuous functions map Cauchy sequences to Cauchy sequences. 2) Cauchy sequences are bounded. 
Fact 1) is easily proven: Let $\{x_n\}$ be Cauchy and $\epsilon>0$. Let $\delta$ be such that $|f(x)-f(y)|\lt\epsilon$ whenever $|x-y|<\delta$.  Now choose $N$ so that $m,n> N$ implies $|x_n-x_m|<\delta$.  We then have for any $n, m>N$ that $|f(x_n)-f(x_m)|<\epsilon$.
Thus, $\{f(x_n)\}$ is Cauchy.
Fact 2) is easy to prove also, but I'll leave this to the interested reader.
