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.