i was thinking on the problem bellow but I couldn't fully solve the problem, here is the statement:
Let $f:[0,1] \rightarrow [0,1]$ be a continuous function such that $\forall x \in [0,1]$ there exists $ n_x \in \mathbb{N}: f^{n_x}(x)=0$, prove that there exists $n \in \mathbb{N}$ such that $f^n \cong 0$
my work on the problem:
first of all if $f(x_0)=x_0$ for some $x_0$ then $x_0=0$ as the condition suggests, now we can easily drive $f(x)<x$ for all $0<x \le 1$ and hence $f(0)=0$.
now let $E_k := \{x \in [0,1] | f^k(x)=0\}$, we have $E_1 \subseteq E_2 \subseteq \dots$ and $[0,1]= \bigcup E_i$(note that $E_i$ are closed) we define $K$ to be the union of "open" intervals $(a,b)$ such that for some $E_n$ we have $(a,b) \in E_n$ this clearly means that $K$ is open
if $K = [0,1]$ then we are done as these open covers form a finite subcover and we can choose the $n$ we want
now assume $K \subsetneq [0,1]$ and let: $F_n= E_n/K$ since $K$ is open and $E_n$ closed we have $F_n$ are closed and $\bigcup F_n = [0,1]/K$ since $F_n$ are closed they must also be no where dense, hence by Baire category theorem $[0,1]/K$ has empty interior and hence $K$ is dense in $[0,1]$
now I was trying to say that for some $n$ we have that $f$ is identically $0$ on $K$ which finishes the problem since $f$ was continuous.
I also reached that since $f(x) \le x$ and $f(x), f(f(x)), \dots \rightarrow 0$ we have that the sequence of functions $f^i(x)$ is uniformly convergent to $0$
Hints and solutions are appreciated.