Boundedness of Continuous Function

1. Mar 14, 2013

gajohnson

1. The problem statement, all variables and given/known data

Let $f$ be a real, uniformly continuous function on the bounded set $E$ in $R^1$. Prove that $f$ is bounded on $E$. Show that the conclusion is false if boundedness of $E$ is omitted from the hypothesis.

2. Relevant equations

NA

3. The attempt at a solution

Ok, so the second part is easy. We simply let $E=R^1$ and $f(x)=x$.

For the first part, I feel like there is a proof-by-contradiction to be had, but I can't quite find it. Any help in the right direction (including telling me that I'm barking up the wrong tree), would be helpful.

Here's what I have so far (pretty much just the definitions, currently searching for where to create the contradiction):

Assume $f$ is unbounded on $E$. Then, for all $M>0$, there exists some ${x}\in{E}$ s.t. $\left|f(x)\right|>M$.
Now, because $E$ is bounded, there exists ${x}\in{R^1}$ and $r>0$ s.t. $d(x,s)<r$ for all ${s}\in{E}$

Thanks!

2. Mar 14, 2013

jbunniii

If $E$ is bounded, then it is contained in some compact set $K$. Compact sets allow you to reduce any open cover to a finite subcover. What would be a good open cover to use here?

3. Mar 14, 2013

Dick

You haven't used uniform continuity! E contained in the interval [s-r,s+r]. State the definition of uniform continuity and split the interval into a lot of parts.

4. Mar 15, 2013

HallsofIvy

Staff Emeritus
And you do need to use uniform continuity. The function f(x)= 1/x is continuous on (0, 1) but not bounded.

(I point this out because my first thought was that continuity was sufficient. I was trying to remember "a function continuous on a closed and bounded set is bounded on that set" and momentarily forgot that E is not necessarily closed.)

5. Mar 15, 2013

gajohnson

Ah, got it! Of course. It's pretty simple from there. Thanks everyone!