My textbook Introduction to Set Theory 3rd by Hrbacek and Jech defines some concepts as follows:

A set $C \subseteq \omega_1$ is closed unbounded if

$C$ is unbounded in $\omega_1$ , i.e., $\sup C=\omega_1$.

$C$ is closed, i.e., every increasing sequence $$\alpha_0 < \alpha_1 < \cdots < \alpha_n < \cdots \quad (n \in \omega)$$ of ordinals in $C$ has its supremum $\sup \{\alpha_n \mid n \in \omega\} \in C$.

A set $S \subseteq \omega_1$ is stationary if $S \cap C \neq \emptyset$ for every closed unbounded set $C$.

A function $f$ with domain $S \subseteq \omega_1$ is regressive if $f(\alpha)<\alpha$ for all $\alpha \neq 0$.

Then they go on to prove this theorem:

3.6 Theorem A set $S \subseteq \omega_1$ is stationary if and only if every regressive function $f:S \to \omega_1$ is constant on an unbounded set. In fact, $f$ has a constant value on a stationary set.

My question lie in the proof of the following example:

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In the proof, I am unable to understand how the authors go from

For each $n$, because $f_n(\alpha)<\alpha$ on $C$


There exists $\gamma_n$ such that the set $S_n = \{\alpha \in C \mid _n(\alpha) = \gamma_n\}$ is stationary.

Please helps me understand the authors' reasoning! Thank you for your help!


$f_n$ is regressive, and $C$ is a stationary set, so there exists some stationary subsets, $S_n$, of $C$ such that $f_n$ is constant on.


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