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Math Help - Cauchy Sequence in metric space problem

  1. #1
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    Cauchy Sequence in metric space problem

    For each positive integer k, define f_{k}(x)=x^{k} for 0 \leq x \leq 1. Is the sequence  \{ f_{k} : [0,1] \rightarrow R \} a Cauchy sequence in the metric space C([0,1],R)?

    My proof:

    Since f_{k} is continuous, it is in C([0,1],R).

    Now, for each E > 0, there exists N in the set of Natural numbers such that k >= N, we have:

    Case 1: x=0;
    then |x^{k} - 0| \rightarrow |0-0| = 0 < E,
    so f_{k} \rightarrow 0 when x=0.

    Case 2: 0 < x < 1;
    then |x^{k} - 0| \rightarrow |0-0| = 0 < E,
    so f_{k} \rightarrow 0 when 0 < x < 1.

    Case 3: x=1;
    then |x^{k} - 1| \rightarrow |1-1| = 0 < E,
    so f_{k} \rightarrow 1 when x = 1.

    Therefore f_{k} converges in C([0,1],R), thus proves it is a Cauchy sequence.

    Q.E.D.

    Is this convincing? Thanks.
    Last edited by tttcomrader; September 14th 2007 at 11:09 AM.
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  2. #2
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    I did not answer before because I did not quite know the exact question.
    Of course \left( {f_n } \right) \to f(x) = \left\{ {\begin{array}{cr}<br />
   0 & {0 \le x < 1}  \\<br />
   1 & {x = 1}  \\<br />
\end{array}} \right. .
    But that is point-wise convergence and not uniform.
    Moreover, f is not continuous at x = 1 .

    Being point-wise convergent implies it is point-wise Cauchy.
    If it were uniformly Cauchy it would be uniformly convergent.
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  3. #3
    Super Member Rebesques's Avatar
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    Plato got it. We know the pointwise limit to be f. Suppose it was Cauchy in C[0,1]; Since C([0,1]) is complete, the limit f would belong to C([0,1]) also. But this is not the case.
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  4. #4
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    Another way to see it is that a uniform limit of continous functions is continous. This is not continous (look at Plato) thus it cannot be uniform continous, i.e. uniform Cauchy.
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  5. #5
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    Then I'm quite lost, I don't really know how to prove this. Is the method I use to prove this, that is, prove it is convergence, possible?
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    Super Member Rebesques's Avatar
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    Convergence of a sequence in [math[C[0,1][/tex] with the sup-norm, is equivalent to uniform convergence. That is, f_n\rightarrow f in C[0,1] implies f_n\rightarrow f uniformly, and vice versa.

    What you did is to find the pointwise limit of the sequence. This does not imply uniform convergence. Now, to determine whether f_n\rightarrow f in C[0,1], check the limit function. The is not continuous, so not in [tex]C[0,1][/math.

    This means the sequence cannot be Cauchy in C[0,1], as then it would converge to a function belonging to that space, a contradiction.
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