Boundedness of Continuous Function

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gajohnson
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Homework Statement



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

Homework Equations



NA

The Attempt at a Solution



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

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

Thanks!
 
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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?
 
gajohnson said:

Homework Statement



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

Homework Equations



NA

The Attempt at a Solution



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

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

Thanks!

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.
 
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.)
 
Dick said:
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.

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