# Please explain the authors' reasoning in a proof about stationary set

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:

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

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

to

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

$$f_n$$ 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.