Cauchy Sequence in metric space problem

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September 14th 2007, 08:00 AM
tttcomrader

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.
September 14th 2007, 10:28 AM
Plato

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.
September 14th 2007, 11:14 PM
Rebesques

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.
September 15th 2007, 04:52 PM
ThePerfectHacker

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.
September 15th 2007, 06:36 PM
tttcomrader

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?
September 17th 2007, 10:40 AM
Rebesques

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.