# $f:\aleph_{\omega_1}\to\aleph_{\omega_1}$ strictly increasing and continuous with $\aleph_1$ fixed points. Can $f$ exist?

Continuous function: $$\forall \lambda$$ limit ordinal $$f(\lambda)=\underset{\gamma<\lambda}\bigcup{f(\gamma)}$$

If I prove that $$Fix(f):=\{\alpha\in\aleph_{\omega_1}|f(\alpha)=\alpha\}$$ is unlimited on $$\aleph_{\omega_1}$$ then $$|Fix(f)|\geq cf(\aleph_{\omega_1})=cf(\omega_1)=\aleph_1$$.

$$\forall \alpha\in \aleph_{\omega_1}$$ I define for countable recursion $$\begin{cases} a_0=\alpha \\ a_{n+1}=f(a_n) \end{cases}$$ .

$$\{a_n\}_{n\in\omega}$$ is a strictly increasing sequence, so $$\underset{n\in\omega}{\bigcup}a_n=\lambda \;$$ is a limit ordinal and

$$f(\lambda)=f(\underset{\gamma\in\lambda}{\bigcup}\gamma)=\underset{\gamma\in\lambda}{\bigcup}f(\gamma)=\underset{n\in\omega}{\bigcup}f(a_n)\leq\underset{n\in\omega}{\bigcup}a_{n+1}=\lambda$$ and $$f(\lambda)\geq\lambda$$ because $$f$$ goes

from a well-ordered set to itself.

Now, my problem is to prove that $$|Fix(f)|\leq\aleph_1$$ and show that kind of function exists.

• Why not just use the identity map? Jul 7, 2020 at 18:04
• @Chickenmancer: Because it has too many fixed points: $\aleph_{\omega_1}>\aleph_1$. Jul 7, 2020 at 18:08
• Ah, I misread that. Thanks! Jul 7, 2020 at 20:11

Define $$f:\omega_{\omega_1}\to\omega_{\omega_1}$$ as follows:
$$f(\xi)=\begin{cases} \omega_2+\xi,&\text{if }\xi\le\omega_1\\ \omega_{\alpha+2}+\eta,&\text{if }\xi=\omega_\alpha+\eta\in(\omega_\alpha,\omega_{\alpha+1}]\text{ for some }\alpha<\omega_1\\ \xi,&\text{if }\xi=\omega_\gamma\text{ for some }\alpha<\omega_1\text{ such that }\operatorname{cf}\gamma=\omega\;. \end{cases}$$