Can you help me to prove this proposition?
We saw this after having proved the Sharkovsky theorem.
$f : [0, 1] \mapsto [0, 1]$ continuous has a period four orbit: $\left\{x_1 < x_2 <x_3 < x_4\right\}$ s.t. $f(x_i) = x_{i+1}$ for $i < 4$ and $f(x_4) = x_1$. Then f has orbits of all periods.
To get the claim is enough to show a period three periodic orbit (thaks to the Sharkovsky theorem) but I'm having difficulties in doing so. My idea was to find two intervals $I_1$ and $I_2$ such that $$I_1 \mapsto I_2 \mapsto I_2 \mapsto I_1$$ so that we know it exists a period three periodic orbit.
The symbol $I \mapsto J$ means that the set $I$ $f$-covers $J$ i.e. $J \subset f(I)$.
Thank you
Consider $g(x)=f^{\circ 3}(x)-x$. Then $g(x_1)=x_4-x_1>0$ and $g(x_2)=x_1-x_2<0$. As $g$ is continuous, we find $x_0$ with $f^{\circ3}(x_0)=x_0$. This $x_0$ might but be a fixpoint -