1. ## Cauchy Sequences

Problems:
For each positive integer k define

a. $f_{k}(x)=x^k$
b. $f_{k})(x)=e^{x/k}$
c. $f_{k}(x)=cos(x/k)$

for $0 \leq x \leq 1$. Is the sequence ${ f_{k}:[0,1] \rightarrow R}$ a Cauchy Sequence in the metric space $C([0,1]),R$

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By definition, a sequence ${f_{k}}$ is Cauchy sequence in $C([0,1]),R$ if and only if there is an $\epsilon > 0$

such that

$|f_{k}(x)-f_{l}(x) | < \epsilon$ for all x in [a, b]

For a I said that $f_{k}(x)=x^k$ is only Cauchy for when $k < 1$ since the distance between the $f_{k}'s$ get smaller and smaller together. Thus this sequence is a Cauchy sequence.

For b, $f_{k})(x)=e^{x/k}$, I said that this is not Cauchy as it is distance is getting larger and larger since the sequence of function grows exponentially. So as k and l goes to infinity, $|f_{k}(x)-f_{l}(x) |$ does not go to zero.

For c, $f_{k}(x)=cos(x/k)$, I said that this is Cauchy with a similar argument with part (a).

I have no idea how to put this in a formal proof though, but the problem seems like it only wants a true or false type of answer, I think.

a. $f_{k}(x)=x^k$
If this was a uniform Cauchy sequence on $[0,1]$ then it would be a uniformly convergent sequence on $[0,1]$. If it was a uniformly convergent sequence on $[0,1]$ then the limit function would be continous. But it is not.