# Proof that a uniformly continuous function on an open interval is bounded

• Apr 1st 2011, 01:33 AM
moses
Proof that a uniformly continuous function on an open interval is bounded
I'm trying to prove that if $\displaystyle f$ if is uniformly continuous on the inteval $\displaystyle (a,b)$ then it is bounded on the interval.

I thought to prove it by contradiction - assuming that there exists a uniformly continuous function $\displaystyle f$ on $\displaystyle (a,b)$ that is not bounded, and see that that leads to a contradiction.

Basically if $\displaystyle f$ isn't bounded, then there must be a point $\displaystyle c$ in $\displaystyle (a,b)$such that at $\displaystyle c$ $\displaystyle lim f$ = infinity. Therefore in any the neighborhood of the limit we can always find two values of the function which are very far away from each other, contradicting the assumption that $\displaystyle f$ is uniformly continuous.

Is this the right direction? And if it is, how do I got about formalizing it?

• Apr 1st 2011, 04:28 AM
girdav
It's the idea indeed. A point on which the limit is infinite can only be a or b.
Assume that f is not bounded. We can find a sequence $\displaystyle \left\{x_n\right\}$ such that $\displaystyle |f(x_n)|\geq n$.
You can extract from $\displaystyle \left\{x_n\right\}$ a Cauchy subsequence since it's a bounded one. $\displaystyle |f|$ is also uniformly continuous hence $\displaystyle \left\{|f(x_n)|\right\}$ is also a Cauchy sequence. We get a contradiction.
• Apr 1st 2011, 04:58 AM
moses
I'm afraid I haven't learned anything about sequences yet...is there another way?
• Apr 2nd 2011, 03:15 AM
girdav
We will show the result in the case $\displaystyle (a,b)=(0,1)$. We apply the definition of uniform continuity for $\displaystyle \varepsilon =1$. We get a $\displaystyle \delta$ such that if $\displaystyle x,y\in (0,1)$ and $\displaystyle |x-y|\leq \delta$ then $\displaystyle |f(x)-f(y)|\leq 1$. We choose an integer $\displaystyle n_0$ such that $\displaystyle \dfrac 1{n_0}\leq \delta$. We put $\displaystyle x_j := \dfrac j{n_0}$ for $\displaystyle j\in\left\{1,\ldots,n_0-1\right\}$. For $\displaystyle x\in (0,1)$, exists $\displaystyle j \in\left\{1,\ldots,n_0-1\right\}$ such that $\displaystyle |x-x_j|\leq\dfrac 1{n_0}\leq \delta$. Hence we get $\displaystyle \displaystyle|f(x)| \leq |f(x)-f(x_j)|+|f(x_j)|\leq 1+\max_{1\leq j\leq n_0-1}|f(x_j)|$, a bound which doesn't depend on $\displaystyle x$.
• Apr 3rd 2011, 04:32 AM
moses
Quote:

Originally Posted by girdav
For $\displaystyle x\in (0,1)$, exists $\displaystyle j \in\left\{1,\ldots,n_0-1\right\}$ such that $\displaystyle |x-x_j|\leq\dfrac 1{n_0}\leq \delta$. Hence we get $\displaystyle \displaystyle|f(x)| \leq |f(x)-f(x_j)|+|f(x_j)|\leq 1+\max_{1\leq j\leq n_0-1}|f(x_j)|$, a bound which doesn't depend on $\displaystyle x$.

I'm not sure I understand why $\displaystyle |x-x_j|\leq\dfrac 1{n_0}\leq \delta$ necessarily...
• Apr 3rd 2011, 05:30 AM
Plato
Quote:

Originally Posted by moses
I'm not sure I understand why $\displaystyle |x-x_j|\leq\dfrac 1{n_0}\leq \delta$ necessarily...

Given $\displaystyle \delta > 0$, we can find a positive integer such that $\displaystyle \frac{1}{n_0}<\delta$.
Look at the collection $\displaystyle x_j=\frac{j}{n_0},~j=0,1,\cdots,n$.
They partition the interval $\displaystyle [0,1]$ into subintervals of length $\displaystyle \frac{1}{n_0}$.
So if $\displaystyle x\in (0,1)$ it is in one of the subintervals.