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.