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Converse Statement of Uniform Continuity

  1. Nov 10, 2009 #1
    Recently, I proved that Given [tex]f:A \rightarrow \mathbb R[/tex] is uniformly continuous and [tex](x_{n}) \subseteq A[/tex] is a Cauchy Sequence, then [tex]f(x_{n})[/tex] is a Cauchy sequence, which really isn't too difficult a proof, however I'm having issues with the converse statement... More specifically, Suppose [tex]A \subseteq \mathbb R[/tex] and [tex]f: A \rightarrow \mathbb R[/tex] with the property that it preserves Cauchy sequences in [tex]A[/tex], i.e. [tex]x_{n} \in A[/tex] and [tex]{x_{n}}[/tex] is Cauchy, [tex]{f(x_{n})}[/tex] is Cauchy, then prove that [tex]f[/tex] is uniformly continuous on [tex]A[/tex]

    My idea was to show by way of contradiction that the boundedness of [tex]A[/tex] cannot be removed, which would then imply that [tex]x_{n}[/tex] is not Cauchy if it were removed, which implies that [tex]f(x_{n})[/tex] is not Cauchy. From this point, I'm a bit stuck... I was under that impression that I could use Cauchy Criterion, which states that a sequence converges iff it is a Cauchy sequence, and since [tex]x_{n}[/tex] isn't a Cauchy sequence, then it doesn't converge, which implies that it cannot be continuous on all points in [tex]\mathbb R[/tex], which implies that it is not continuous on [tex]\mathbb R[/tex], therefore cannot be uniform continuous on [tex]\mathbb R[/tex].

    Perhaps contrapositive would work better? Any suggestions?
     
  2. jcsd
  3. Nov 10, 2009 #2
    I think the best way to handle this is to look at the contrapositive, since that's a good way to generate sequences. From the perspective of the contrapositive, we can basically always pick two points x and y arbitrarily close while their images are a fixed distance apart. It shouldn't be too hard to show that this would imply that we can find a sequence that is Cauchy (eventually any two terms are arbitrarily close), but the image sequence is not Cauchy (images of terms are a fixed distance apart).
     
  4. Nov 10, 2009 #3

    LCKurtz

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    What about f(x) = x2?
     
  5. Nov 10, 2009 #4
    Good point, you need A to be compact for this to work.
     
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