# Cauchy Sequence in metric space problem

• Sep 14th 2007, 08:00 AM
Cauchy Sequence in metric space problem
For each positive integer k, define $\displaystyle f_{k}(x)=x^{k}$ for $\displaystyle 0 \leq x \leq 1$. Is the sequence $\displaystyle \{ f_{k} : [0,1] \rightarrow R \}$ a Cauchy sequence in the metric space $\displaystyle C([0,1],R)$?

My proof:

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

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 $\displaystyle |x^{k} - 0| \rightarrow |0-0| = 0 < E$,
so $\displaystyle f_{k} \rightarrow 0$ when x=0.

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

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

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

Q.E.D.

Is this convincing? Thanks.
• Sep 14th 2007, 10:28 AM
Plato
I did not answer before because I did not quite know the exact question.
Of course $\displaystyle \left( {f_n } \right) \to f(x) = \left\{ {\begin{array}{cr} 0 & {0 \le x < 1} \\ 1 & {x = 1} \\ \end{array}} \right.$ .
But that is point-wise convergence and not uniform.
Moreover, $\displaystyle f$ is not continuous at $\displaystyle x = 1$.

Being point-wise convergent implies it is point-wise Cauchy.
If it were uniformly Cauchy it would be uniformly convergent.
• Sep 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.
• Sep 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.
• Sep 15th 2007, 06:36 PM
Convergence of a sequence in [math[C[0,1][/tex] with the sup-norm, is equivalent to uniform convergence. That is, $\displaystyle f_n\rightarrow f$ in $\displaystyle C[0,1]$ implies $\displaystyle f_n\rightarrow f$ 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 $\displaystyle f_n\rightarrow f$ in $\displaystyle C[0,1]$, check the limit function. The is not continuous, so not in [tex]C[0,1][/math.
This means the sequence cannot be Cauchy in $\displaystyle C[0,1]$, as then it would converge to a function belonging to that space, a contradiction.