Why $C[0,1]$ with $d_\infty$ is a complete space?

Question

So my textbook says that it is a complete space, However in The same textbook i have found an example of a series that makes me doubt this.

If we use $f_n(t)=t^n$, after taking into account the max distance, we obtain that to the limit of n it should go to a function that is not continuos, that is $f(t)=1$, when $t=1$, while $f(t)=0$, when $t \in [0,1)$.

Then we conclude that since the succession must be continuos, it does not converge.

However doesn't this mean that the space is incomplete?

Answer 1

The sequence $f_n$ in not a Cauchy sequence for the "max distance" $$d_\infty(f,g) := \sup_{t \in [0,1]} |f(t) - g(t)|.$$

Indeed, take any $n \geq 1$. Then you can find $t_n$ close to $1$ such that $f_n(t_n) \geq \frac34$, and some larger $m \geq n$ such that $f_m(t_n) \leq \frac14$. In particular, $$ d_\infty(f_n, f_m) \geq |f_n(t_n) - f_m(t_n)| \geq \tfrac12. $$

Therefore, the sequence $\{f_n\}_{n \geq 1}$ is not a counterexample to the completeness of $C([0,1])$.

Answer 2

Your sequence is not a Cauchy sequence in $C[0,1]$. Completeness means that every Cauchy sequence converges.

Answer 3

Hint: let $f:[0,1]\rightarrow R$ defined by $f(x)=0, x\in [0,1), f(1)=0$. Consider $d_{\infty}(t^n,f)=sup_{t\in [0,1)}|t^n|=1$. Thus $t^n$ does not converge towards $f$ for $d_{\infty}$.

In fact $d_{\infty}(t^n,t^m)=sup_{t\in [0,1]}|t^n-t^m|$, this implies that $lim_md_{\infty}|t^n-t^m|=sup_{t\in [0,1]}|t^n|=1$. Thus for every $n>0$, there exists $m>n$ such that $d_{\infty}(t^n,t^m)>1/2$, we conclude that the sequence is not a Cauchy sequence.