Let $f:[a,b]→[a,b]$ be a differentiable and strictly increasing function (so it is a bijection) where $b$ is the unique fixed point of $f$ and $f([a,b])⊂[a,b]$. Then we deduce that the iterations $$x_{k+1}=f(x_{k})$$ converges to $b$. Let $(w_{k})$ be a known sequence. Assume we know that $$w_{n+1}=f(w_{n})........(*)$$ for several values of $n$ (the number of those $n$ is very big: about $800,000,000,000,000$) (also those $n$ are not successive) and we know also that there exist infinitely many $k$ (many many of them are between the values of $n$) such that $$w_{k+1}≠f(w_{k})$$ Let $Ω$ be the set of positive integers $n$ such that (*) is verified.
My question is: Is it possible to conclude that the number of elements of $Ω$ infinite?
I have no idea to start.
Just from the fact that the function is a bijection and a contraction you can show that the iteration starting from any $x_0$ that is not the fixed point never cycles, so it has an infinite number of values converging to the fixed point. You just show that each iteration takes you closer to the fixed point but cannot achieve the fixed point because you have a bijection. The only value for which $f(x)=b$ is $x=b$.
You haven't told us how the $w_n$ are chosen, so there could be finitely or infinitely cases of $w_{n+1}=f(w_n)$. It could be that the sequence has $w_{n+1}=f(w_n)$ for all even $n$ and not for all odd $n$. It could be true for the first $800,000,000,000,000$ even terms and never again.