Let $f(x): (0,1) \to \mathbb{R} $ be Uniformly continuous. Prove that $f(x)$ is bounded. Let $f(x): (0,1) \to \mathbb{R}  $ be Uniformly continuous. Prove that $f(x)$ is bounded. 
My way so far: 
Suppose $f(x)$ is not bounded. WLOG is not bounded from above. Hence for every $M>0$ there is $x\in(0,1)$ that $f(x)>M$.
In particular for every $x,y \in(0,1)$ such that $|x-y|<\delta$ 
$(\delta>0)$ exist $M_0>0$ such that $f(x),f(y)>M_0$.
Now I want to show that $f(x)$ is not Uniformly continuous because of the assumption, but I'm stuck at the stage:
$|f(x)-f(y)|\ge||f(x)|-|f(y)||\ge ||M_0|-|f(y)||\ge $ ??
 A: As $\;f\;$ is u.c. on $\;(0,1)\;$ , for every $\;\epsilon >0\;$ there exists $\;\delta_\epsilon>0\;$ such that 
$$\;|x-y|<\delta_\epsilon\implies |f(x)-f(y)|<\epsilon\;,\;\;x,y\in (0,1)$$
Suppose $\;f\;$ isn't bounded. Then there exists $\;\{x_n\}_{n\in\Bbb N}\subset (0,1)\;$ s.t.  $\;|f(x_n)|>n\;,\;\;n\in\Bbb N$ . But the sequence $\;\{x_n\}_{n\in\Bbb N}\;$ is bounded, then by Bolzano-Weierstrass's theorem there exists a subsequence $\;\{x_{n_k}\}_{k\in\Bbb N}\;$ s.t. $\;x_{n_k}\xrightarrow[k\to\infty]{}x_0\;$ . And we can even take $\;x_0=0,1\notin(0,1)\;$ , it really doesn't matter. Try now to take it from here  to show a contradiction...
A: Let $\epsilon=1$, by definition it exists $\alpha>0$ such as
$$
\forall\left(x,y\right) \in \left[0,1\right[^2, \ \left|y-x\right|\leq \alpha \Rightarrow \left|f\left(x\right)-f\left(y\right)\right|\leq 1 
$$
For $x \in \left[1-\alpha,1\right[$ you have $\left|\left(1-\alpha\right)-1\right|=\alpha \leq \alpha$ so it comes $\left|f\left(x\right)-f\left(1-\alpha\right)\right|\leq 1$ hence
$$
\left|f\left(x\right)\right|\leq 1+f\left(1-\alpha\right)
$$
The function is continuous hence bounded on $\left[0,1-\alpha\right]$ by $m$.
Then $f$ is bounded by $\text{max}\left(m,1+\left|f\left(1-\alpha\right)\right|\right)$.
A: Following what you started and supposing that $f$ is not bounded, you can find a sequence $(x_n)$ of reals in $(0,1)$ such that $f(x_{n+1}) >2\max(n,f(x_n))$. Therefore for all $n \in \mathbb N$ you have $f(x_{n+1}) -f(x_n) >n$
As $[0,1]$ is compact, you can find a subsequence $(x_{\varphi(n)})$ that converges to $a \in [0,1]$. Hence $\lim\limits_{n \to \infty} (x_{\varphi(n+1)} -x_{\varphi(n)}) = 0$ while $\lim\limits_{n \to \infty} (f(x_{\varphi(n+1)} -x_{\varphi(n)}) = \infty$, in contradiction with the fact that $f$ is uniform continuous.
A: How do you know that such a $\delta>0$ exists? What you have is that:

For every $M>0$ there exists a $x_M\in(0,1)$ such that $f(x_M)>M$

So, you can find a sequence $x_n\in(0,1)$ such that:
$$f(x_n)>\max\{n,f(x_{n-1})+1\},\ \forall\ n\in\mathbb{N}$$
So, in particular $$|f(x_n)-f(x_{n-1})|>1\tag{1}$$
Now, since $(x_n)\subseteq(0,1)$, it is bounded and, hence, it has a converging sub-sequence - Bolzano-Weierstrass - let $(x_{k_n})$. So, it is also true that, for every $n\in\mathbb{N}$:
$$f(x_{k_n})\geq x_{k_n}$$
Since $(x_{k_n})\subseteq(0,1)$ and it is convergent, it is also a Cauchy sequence.
Let now $\epsilon=1>0$. From the uniform continuity of $f$, there exists a $\delta=\delta(\epsilon)>0$ such that:
$$|x-y|<\delta\Rightarrow|f(x)-f(y)|<1$$
For that $\delta>0$, there exist a $n_0\in\mathbb{N}$ such that for every $n>m\geq n_0$ we have that:
$$|x_{k_n}-x_{k_m}|<\delta$$
So, we also have, from uniform continuity that:
$$|f(x_{k_n})-f(x_{k_m})|<1$$
which is a contradiction to $(1)$.
So, $f$ must be bounded.
