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 $C$.
Then the authors provide an example
Because the reasoning given by the authors is too concise for me, I would like to make it more detailed. Please help me check if my understanding is correct or not.
- $f$ is well-defined
If $C \cap \alpha = \emptyset$, then $f(\alpha)=\sup (C \cap \alpha) = \sup \emptyset =0$.
If $C \cap \alpha \neq \emptyset$, then $f(\alpha)=\sup (C \cap \alpha) = \bigcup (C \cap \alpha)$.
- $\alpha >0 \implies f(\alpha) < \alpha$
- $C \cap \alpha = \emptyset$
Then $f(\alpha)=0 <\alpha$.
- $C \cap \alpha \neq \emptyset$
$C \supseteq (C \cap \alpha)$ is countable and $C$ is closed $\implies \sup (C \cap \alpha) \in C$. It follows from $f(\alpha)=\sup (C \cap \alpha)$ that $C \ni f(\alpha) \le \alpha$. If $f(\alpha) = \alpha$, then $\alpha \in C$. Moreover, $\alpha \in A$. Thus $\alpha \in C \cap A$, which contradicts to the fact that $C \cap A = \emptyset$. Hence $f(\alpha) < \alpha$.
- $f$ does not have the same value for uncountably many $\alpha$'s
For each $\gamma < \omega_1$, let $\delta = \min \{\lambda \in C \mid \lambda>\gamma\}$. Then $C \ni \delta > \gamma$. For all $\alpha \in A$, we have two possible cases.
$\alpha > \delta$. Then $\alpha \ni \delta > \gamma$ and $C \ni \delta > \gamma$. Thus $f(\alpha)=\sup (C \cap \alpha) \ge \delta > \gamma$ and thus $f(\alpha) > \gamma$. This means that $f$ does not have value $\gamma$ for all $\alpha > \delta$.
$\alpha \le \delta$. Then $|\{\alpha \le \delta \mid f(\alpha)=\gamma\} | \le |\delta| < \aleph_1$. Hence $|\{\alpha \le \delta \mid f(\alpha)=\gamma\} | \le \aleph_0$. This means that $f$ has value $\gamma$ for countably many $\alpha$'s.
Both cases imply that for any $\gamma < \omega_1,$ $f$ has value $\gamma$ for at most countably many $\alpha$'s. In other words, $f$ does not have the same value for uncountably many $\alpha$'s.
Yes, you understand correctly, but for (1) there is no reason to put the emptyset separately($\bigcup\emptyset=\emptyset$)
And for (3) there is no reason to have cases:
If $|f^{-1}(\gamma)|=\aleph_1$ then it is unbounded(in $\omega_1$), in particular $B=\{\alpha\in f^{-1}(\gamma)\mid \alpha>\delta\}\ne\emptyset$, so if $x\in B$ then $\gamma=f(x)>\gamma$.