I want to prove completeness of $C[a,b]$ with metric induced by max norm, but need help

Question

I added a picture, please check it.

Can someone please explain the last three line of the proof?

I thought by finding a cauchy sequence that converges to a continuous function we are done with proving completeness of $C[a,b]$ with metric induced by max norm. Why did they add the last three lines "highlighted" and how do I prove them? Please, help, thanks.

Answer 1

Let $(f_n)_{n\in\Bbb N}$ be a Cauchy sequence of elements of $C[a,b]$. There is a $N_1\in\Bbb N$ such that$$m,n\geqslant N_1\implies\|f_m-f_n\|_\max<1.$$And there is a $N_2\in\Bbb N$ such that$$m,n\geqslant N_2\implies\|f_m-f_n\|_\max<\frac12.$$You can assume, without loss of generality, that $N_2>N_1$. Now, there is a $N_3\in\Bbb N$ such that$$m,n\geqslant N_3\implies\|f_m-f_n\|_\max<\frac1{2^2}$$and you can assume, without loss of generality, that $N_3>N_2$. But then $(f_{N_k})_{k\in\Bbb N}$ is a subsequence of $(f_n)_{n\in\Bbb N}$ and$$(\forall k\in\Bbb N):\left\|f_{N_{k+1}}-f_{N_k}\right\|<\frac1{2^{k-1}}.$$So, by the first part of the proof, $\left(f_{N_k}\right)_{k\in\Bbb N}$ converges uniformly. In particular, the sequence $(f_n)_{n\in\Bbb N}$ has a convergent subsequence. But then $(f_n)_{n\in\Bbb N}$ itself is convergent, since, as the author remarks, in order to prove that a Cauchy sequence converges it is enough to prove that it has a convergent subsequence.

Answer 2

The proof begins with a particular case: it assumes that the norm $$\|f_{n+k}-f_{k}\|_{\infty}\leq a_{k}$$ is controlled by $a_{k}$ real number which is the addend of a converging sum $\sum_{k}a_{k}$. The last $3$ lines are telling you how to adapt the particular case to the general one.