Show that $T^n\left( \left[ \frac{k}{2^n},\frac{k+1}{2^n}\right]\right) = [0,1] , \forall n\in\mathbb N, 0<\frac{k}{2^n}<1$ 
Show that $T^n\left( \left[ \cfrac{k}{2^n},\cfrac{k+1}{2^n}\right]\right) = [0,1] , \forall n\in\mathbb N, 0<\cfrac{k}{2^n}<1$

We define the "tent" map $T:[0,1]\longrightarrow[0,1],$
$$T(x)= \begin{cases}
2x , \quad0\le x \le 1/2 \\
2(1-x), \quad 1/2\le x \leq 1
\end{cases}$$
Note that $T^n(x) = T\circ T\circ \dots\circ T(x),$ $n$-times.

I've proved that the set $A=\{ k/2^n: 0\leq k/2^n \le 1,n\in\mathbb N\}$ is dense in $[0,1]$ 
I'm not sure how to continue, i was wondering if induction works here.
 A: Indeed, induction is a good way: for $n=1$ this can checked directly.
Now, suppose that $T^n\left( \left[ \cfrac{k}{2^n},\cfrac{k+1}{2^n}\right]\right) = [0,1] , \forall n\in\mathbb N, 0<\cfrac{k}{2^n}<1$ and let us show it for $n+1$.
Let $0\leqslant k\leqslant 2^{n+1}-1$.

*

*If $k+1\leqslant 2^n$, then on $\left[ \cfrac{k}{2^{n+1}},\cfrac{k+1}{2^{n+1}}\right]$, $T(x)=2x$ hence  $$T\left(\left[ \cfrac{k}{2^{n+1}},\cfrac{k+1}{2^{n+1}}\right]\right)=\left[\cfrac{2k}{2^{n+1}},\cfrac{2\left(k+1\right)}{2^{n+1}}\right]=\left[\cfrac{k}{2^{n}},\cfrac{ k+1 }{2^{n}}\right]$$
and we can apply the induction assumption.

*If $2^n+1\leqslant k\leqslant 2^{n+1}-1$, then on $\left[ \cfrac{k}{2^{n+1}},\cfrac{k+1}{2^{n+1}}\right]$, $T(x)=2-2x$ hence
$$T\left(\left[ \cfrac{k}{2^{n+1}},\cfrac{k+1}{2^{n+1}}\right]\right)=\left[2- \cfrac{2(k+1)}{2^{n+1}},2-\cfrac{2k}{2^{n+1}}\right]=\left[\cfrac{2^{n+1}-k-1}{2^{n}},\cfrac{ 2^{n+1}-k }{2^{n}}\right]=\left[\cfrac{k'}{2^{n}},\cfrac{ k'+1 }{2^{n}}\right],$$
where $k'=2^{n+1}-k-1$, which satisfies $0\leq k'\leqslant 2^n-1$. Apply the induction assumption to conclude.

