1. Uniformly continuous #3

Prove that if f : (a,b)-->R is is uniformly continuous function in (a,b) so f is bounded function.

2. Assume it's not bounded then pick a (unbounded) sequence $(x_n) \subset (a,b)$ such that $f(x_n) for all $n$ then since this sequence is bounded we get a Cauchy subsequence $(x_{n_k})$ by Bolzano-Weierstrass. Since $f$ is uniformly cont. it sends Cauchy seq. into Cauchy seq. but what can you say about $f(x_{n_k})$ by construction?.

3. Originally Posted by Jose27
Assume it's not bounded then pick a (unbounded) sequence $(x_n) \subset (a,b)$ such that $x_n for all $n$ then since this sequence is bounded we get a Cauchy subsequence $(x_{n_k})$ by Bolzano-Weierstrass. Since $f$ is uniformly cont. it sends Cauchy seq. into Cauchy seq. but what can you say about $f(x_{n_k})$ by construction?.
Jose27, I think you want too 'clean up' that proof.
I know what you mean but you did not do it.

4. hmmm... I don't understand...
What is Cauchy subsequence ?
"Since f is uniformly cont. it sends Cauchy seq..." why is that?

5. Originally Posted by Jose27
Assume it's not bounded then pick a (unbounded) sequence $(x_n) \subset (a,b)$ such that $f(x_n) for all $n$. Since $\left(x_n\right)$ is bounded we get a Cauchy subsequence $(x_{n_k})$ by Bolzano-Weierstrass. Since $f$ is uniformly cont. it sends Cauchy seq. into Cauchy seq. but what can you say about $f(x_{n_k})$ by construction?.
I believe that now it is easier to understand it.

Originally Posted by Also sprach Zarathustra
hmmm... I don't understand...
What is Cauchy subsequence ?
"Since f is uniformly cont. it sends Cauchy seq..." why is that?
If $f$ is unif. cont and $\left(x_{n_k}\right)$ is Cauchy, then $f\left(x_{n_k}\right)$ is Cauchy too.

Let $\epsilon>0$. Since $f$ is unif cont, there exists $\delta>0$ such as $|x-y|<\delta\Rightarrow |f(x)-f(y)|<\epsilon.$ Since $\left(x_{n_k}\right)$ is Cauchy, there existe $K$ such as $|x_{n_i}-x_{n_j}|<\delta$ for all $i,j\geq K$.

6. Originally Posted by Plato
Jose27, I think you want too 'clean up' that proof.
I know what you mean but you did not do it.
So you know a Cauchy sequence is bounded, and Abu-Khalil proved that if $f$ is a unif. cont. function and $x_n$ is a Cauchy seq. then $f(x_n)$ is also a Cauchy seq. but by how we constructed them $f(x_n) \rightarrow \infty$ (or $-\infty$ in which case invert all inequalities in my first post) clearly contradicting the fact that $f$ is unif. cont. so we conclude that our initial hypothesis is false ie. $f$ is bounded.