# Regressive functions on countable ordinals

Question: Are the following statements 1-7 correct ?

Let $$\Lambda$$ be the set of countable limit ordinals (0 added), $$\psi:\omega_1\to\Lambda$$ maps $$\alpha\mapsto\omega\cdot\alpha$$. Then:

1. $$\psi$$ is normal (i.e. strictly increasing and continuous) bijection and $$\psi(x)\geq x$$ for all $$x$$.
2. $$S=\{\omega^{\omega^\alpha}\cdot\beta\}_{\alpha,\beta<\omega_1}$$ is exactly the set of fixed points of $$\psi$$, i.e. $$\psi(x)=x\iff x\in S$$ (not used further)
3. $$\{\varepsilon_\alpha\}_{\alpha<\omega_1}\subset S$$, where $$\varepsilon_\alpha$$ are fixed points of $$\beta\mapsto\omega^\beta$$.

If 1,3 are correct let $$\varphi=\psi^{-1}:\Lambda\to\omega_1$$. Obviously, $$S$$ is the set of all fixed points of $$\varphi$$ and $$\varphi(x)\leq x$$. Let $$S_\alpha=(S\cap[0;\varepsilon_\alpha])\setminus 0$$ be the set of all nonzero fixed points of $$\varphi$$ less than $$\varepsilon_\alpha+1$$, $$~B_\alpha=([0;\varepsilon_\alpha]\setminus S_\alpha)\cap\Lambda$$ be the set of all points of $$\Lambda_\alpha=\Lambda\cap[0;\varepsilon_\alpha]$$ which are not fixed. So, $$\Lambda_\alpha=S_\alpha\cup B_\alpha$$. Denote $$\sigma_\alpha,\beta_\alpha$$ the order types of $$S_\alpha,B_\alpha$$ respectively.

1. $$\beta_\alpha\geq\sigma_\alpha$$, ($$\beta_\alpha=\sigma_\alpha$$ ?)

Let $$f_\alpha:S_\alpha\to B_\alpha$$ be monomorphism on initial segment (isomorphism?) of well-ordered sets.

1. $$f_\alpha(x) for all $$x\neq 0$$

Finally, we define the following family of functions $$g_\alpha:\Lambda\to\omega_1$$: $$g_\alpha(x)= \begin{cases} \varphi(x),~ \text{ if } ~x\in B_\alpha \\ \varphi(f_\alpha(x)),~ \text{ if } ~x\in S_\alpha \\ \varepsilon_\alpha,~\text{ otherwise, i.e.} ~x>\varepsilon_\alpha \end{cases}$$

1. $$g_\alpha(x) for all $$x\neq 0,~\alpha$$ ($$g_\alpha$$ is regressive)
2. $$\omega_1$$-sequence $$g_\alpha(\varepsilon_\alpha)$$ is strictly increasing
• I always thought that $\omega\cdot\omega^{\omega+1}=\omega^{1+\omega+1}=\omega^{\omega+1}$, but I also thought that $\omega+1$ does not have the form $\omega^\alpha$. – Asaf Karagila Apr 25 '19 at 9:47
• @AsafKaragila: Sorry. Corrected. But this statement (2) not used further and can be completely excluded. – user2935704 Apr 25 '19 at 17:00