In Exercise 6.2.13 page 162 Abbott define Cantor Function.

He defines it as follows:

$f\_0(x)=x $ for all $ x \in [0,1]$,

$f\_1(x)=(3/2)x$, for $ x \in [0,1/3]$,

$f\_1(x)=1/2$, for $x \in (1/3,2/3)$,

$f\_1(x)=(3/2)x-1/2$, for $x \in [2/3,1]$.

Then defines 

$f\_n(x)=1/2f\_{n-1}(3x)$, for $ x \in [0,1/3]$,

$f\_n(x)=f\_{n-1}(x)$, for $x \in (1/3,2/3)$,

$f\_n(x)=1/2f\_{n-1}(3x-2)+1/2$, for $x \in [2/3,1]$.

I have proved that for every $x \in [0,1] \ (f\_n(x))$ is increasing and 
$|f\_1(x)-f\_n(x)|<1/2$ for every n and for every x in our domain.

Then in order to prove uniform convergence I have to prove 
$|f\_m(x)-f\_n(x)|<1/2^m$ for $m<n$ AND THIS IS THE POINT THAT I HAVE STUCK

Please any help. THANKS!!