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.