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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. $f_\alpha(x)<x$ 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. $g_\alpha(x)<x$ for all $x\neq 0,~\alpha$ ($g_\alpha$ is regressive)
  2. $\omega_1$-sequence $g_\alpha(\varepsilon_\alpha)$ is strictly increasing
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    $\begingroup$ 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 $\omega^\alpha$. $\endgroup$ – Asaf Karagila Apr 25 '19 at 9:47
  • $\begingroup$ @AsafKaragila: Sorry. Corrected. But this statement (2) not used further and can be completely excluded. $\endgroup$ – user2935704 Apr 25 '19 at 17:00

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