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

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

I'm trying to prove that if $f$ if is uniformly continuous on the inteval $(a,b)$ then it is bounded on the interval.

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

Basically if $f$ isn't bounded, then there must be a point $c$ in $(a,b)$such that at $c$ $lim f$ = 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 $f$ is uniformly continuous.

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

2. 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 $\left\{x_n\right\}$ such that $|f(x_n)|\geq n$.
You can extract from $\left\{x_n\right\}$ a Cauchy subsequence since it's a bounded one. $|f|$ is also uniformly continuous hence $\left\{|f(x_n)|\right\}$ is also a Cauchy sequence. We get a contradiction.

3. I'm afraid I haven't learned anything about sequences yet...is there another way?

4. We will show the result in the case $(a,b)=(0,1)$. We apply the definition of uniform continuity for $\varepsilon =1$. We get a $\delta$ such that if $x,y\in (0,1)$ and $|x-y|\leq \delta$ then $|f(x)-f(y)|\leq 1$. We choose an integer $n_0$ such that $\dfrac 1{n_0}\leq \delta$. We put $x_j := \dfrac j{n_0}$ for $j\in\left\{1,\ldots,n_0-1\right\}$. For $x\in (0,1)$, exists $j \in\left\{1,\ldots,n_0-1\right\}$ such that $|x-x_j|\leq\dfrac 1{n_0}\leq \delta$. Hence we get $\displaystyle|f(x)| \leq |f(x)-f(x_j)|+|f(x_j)|\leq 1+\max_{1\leq j\leq n_0-1}|f(x_j)|$, a bound which doesn't depend on $x$.

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

6. Originally Posted by moses
I'm not sure I understand why $|x-x_j|\leq\dfrac 1{n_0}\leq \delta$ necessarily...
Given $\delta > 0$, we can find a positive integer such that $\frac{1}{n_0}<\delta$.
Look at the collection $x_j=\frac{j}{n_0},~j=0,1,\cdots,n$.
They partition the interval $[0,1]$ into subintervals of length $\frac{1}{n_0}$.
So if $x\in (0,1)$ it is in one of the subintervals.

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uniform continuous on open interval

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