# Thread: Cauchy Sequence in metric space problem

1. ## 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.

2. I did not answer before because I did not quite know the exact question.
Of course $\left( {f_n } \right) \to f(x) = \left\{ {\begin{array}{cr}
0 & {0 \le x < 1} \\
1 & {x = 1} \\
\end{array}} \right.$
.
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.

3. 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.

4. 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.

5. 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?

6. 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.