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 $ ??
Best Answer
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...