Uniform continuity on (0,1) implies boundedness I need to prove that if $f: (0,1) \rightarrow \mathbb{R}$ is Uniformly continuous then it is  bounded. 
Thank you.
 A: Here's one approach.  You can use uniform continuity (with $\varepsilon=1$, say) to show that $f$ is bounded on $(0,\delta)$ and $(1-\delta,1)$.  You probably already know the theorem that implies that $f$ is bounded on $[\delta,1-\delta]$.  
If you don't know about boundedness of continuous functions on $[a,b]$, then what you can do here is cover $(0,1)$ with a finite number of tiny intervals where you know (using uniform continuity) that $f$ can't vary by more than $1$.
Uniformly continuous functions can also be extended to the closure, so an approach that would actually do more would be to show that $\lim_{x\to1}f(x)$ and $\lim_{x\to 0}f(x)$ exist, so that you may consider $f$ to be a restriction of a continuous function on $[0,1]$.
A: Recall that if $f$ is uniformly continuous, then given $\epsilon>0$ we can find $\delta>0$ such that $|x-y|<\delta$ implies $|f(x)-f(y)|<\epsilon$.  
To show it is bounded, it doesn't really matter what $\epsilon$ is, so let it be some fixed constant.  Let $N=\lfloor\frac{1}{\delta}\rfloor$ and take
$$x_1=\delta,\  x_2=2\delta,\dots,\  x_n=n\delta,\dots, x_N=N\delta.$$  Notice every $y\in (0,1)$ satisfies $|y-x_i|<\delta$ for some $i$. Then $|f|$ will be bounded by $$\text{max}_{1\leq i\leq N} \{|f(x_i)|+\epsilon \}$$ 
which follows by applying the definition of Uniformly Continuous.
Hope that helps,
A: Suppose $A$ is a bounded set such that $f(A)$ is unbounded. We can choose a sequence $(f(x_n))$ in $f(A)$ such that $|f(x_n)|>n$. Now $X=(x_n)$ is a bounded sequence and hence admits a convergent subsequence $Y=(x_{n_k})$. But $Y$ is a Cauchy sequence whose image under $f$ (a uniformly continuous function) is not Cauchy. 
