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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.

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

1 Answer 1

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Define $f:\omega_{\omega_1}\to\omega_{\omega_1}$ as follows:

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

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