# Do we need _strict_ monotonicity for the fixed point lemma for normal functions?

I was reading about the fixed-point lemma for normal functions and I was wondering if we could weaken strict monotonicity to non-decreasing. I see why it's necessary for the proof, but is there a continuous non-decreasing ordinal operation without a fixed point? I'm probably missing something obvious but I can't think of any.

You don't need that $f$ is strictly increasing to have at least one fixed point. Consider the sequence $\alpha_n = f^n(0)$. By induction it follows that this is non-decreasing. If $\alpha_n=\alpha_{n+1}$ for some $n$ then we are done as this means that $\alpha_n = f(\alpha_n)$. Else consider the sup of this sequence and follow by continuity that this is a fixed point.
Of course you cannot get arbitrarily large fixed points as witnessed by the constant $0$ function.
With $\ell(\alpha)=\min\{\,\lambda\mid \lambda\text{ is limit ordinal}, \lambda>\alpha\,\}$, let $$f(\alpha)=\begin{cases}\ell(\alpha)&\text{if }\alpha\text{ is limit ordinal}&\\\ell(\ell(\alpha))&\text{otherwise}\end{cases}$$ Then $f$ is continuous, non-decreasing, and has no fixed point.
• Why is this continuous? It looks discontinuous at $\omega\cdot\omega$, or am I missing something? Apr 1, 2018 at 9:45
• @Jonathan: It actually seems discontinuous at every limit ordinal, for example every finite $\alpha$ gives $f(\alpha)=\omega$, but $f(\omega)=\omega+\omega$. Apr 1, 2018 at 10:19
• @Asaf finite successors $\alpha$ give $\omega +\omega$, the next next limit. Apr 1, 2018 at 13:12