# Composition of everywhere discontinuous bijection with itself, limit, and continuity

Let $$\phi$$ be everywhere discontinuous bijection that maps $$\mathbb R$$ onto $$\mathbb R$$.

Define $$\phi^n=\underbrace{\phi\circ \phi\cdots \phi}_{n\text{ times}}$$ where $$\circ$$ is ordinary composition of functions.

Define $$\psi= \lim_{n \to + \infty} \phi^n$$

Can $$\psi$$ be everywhere continuous?

Some of my thoughts on this problem:

Since $$\phi$$ is a bijection so is $$\phi^n$$ for every $$n \in \mathbb N$$. Since limit should not destroy bijectivity I think that $$\psi$$ also is a bijection and if also everywhere continuous that means that it is everywhere strictly monotone so everywhere differentiable. And, I am not sure what to do next.

Sure. For instance, it's possible that $$\phi^n(x)\to 0$$ for all $$x$$. As a sketch of how to construct such a $$\phi$$, build its orbits one by one by a transfinite recursion of length $$\mathfrak{c}$$. In each step, you just pick a $$\mathbb{Z}$$-indexed sequence of new points converging to $$0$$ to be a new orbit. All you have to do is arrange that every point of $$\mathbb{R}$$ is eventually put in an orbit and you avoid continuity at any point, which is easy (for instance, you can easily arrange that the image of every interval is dense, using just the first $$\omega$$ steps of the construction).
Note that $$\lim\phi^n(x)=\lim\phi^n(\phi(x))$$ (assuming either limit exists), so $$\psi$$ must satisfy $$\psi(\phi(x))=\psi(x)$$. So $$\psi$$ can actually never be injective (assuming it is defined in the first place).
• How can $\phi^n \to 0$ if $\phi^n (\mathbb R)=\mathbb R$? – Grešnik Jun 18 at 22:37
• Ignoring the discontinuity condition, consider the example $\phi(x)=x/2$, for instance. – Eric Wofsey Jun 18 at 22:38