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Homework Help: Boundedness of Continuous Function

  1. Mar 14, 2013 #1
    1. The problem statement, all variables and given/known data

    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.

    2. Relevant equations


    3. 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]

  2. jcsd
  3. Mar 14, 2013 #2


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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?
  4. Mar 14, 2013 #3


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    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.
  5. Mar 15, 2013 #4


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    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.)
  6. Mar 15, 2013 #5
    Ah, got it! Of course. It's pretty simple from there. Thanks everyone!
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