Prove that the set of all closure points of an increasing function is closed unbounded

Question

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 $\sup \{\alpha_n \mid n \in \omega\} \in C$.

Then there is an excercise

If $f:\omega_1 \to \omega_1$ is a strictly increasing function, then $\alpha < \omega_1$ is a closure point of $f$ if $f(\xi)< \alpha$ whenever $\xi < \alpha$. Let $C$ be the set of all closure points of $f$. Then $C$ is closed unbounded.

I have tried to prove that $C$ is unbounded, but to no avail.

For any $\beta <\omega_1$, we prove that there exists $\alpha \in C$ such that $\beta < \alpha$.

Assume the contrary that there does not exist $\alpha \in C$ such that $\beta < \alpha$. Then $\forall \alpha \in C:\alpha \le \beta$ and thus $\alpha_0 = \sup C= \bigcup C \le \beta$. It follows that $\forall \xi < \alpha_0:f(\xi) < \alpha_0$ and thus $\alpha_0 \in C$. Hence $\alpha_0 +1 \notin C$ and thus $f(\alpha_0) \ge \alpha_0 +1$.

After that, I am stuck for over a week. Please shed me some light! Thank you for your help.

Answer 1

Let $b < \omega_1$ be an upper bound of $C$ and $r = \sup \{ f^n(b) : n \in \Bbb N \}$.
If $x < r$, then there exists $n$ with $x < f^n(b)$.
$f(x) < f(f^n(b)) = f^{n+1}(b) < r$.
As $r \in C$, $b < r \le b$, a contradiction.

Answer 2

I added the proof that $C$ is closed here.

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Let $(\alpha_n)_{n \in \omega}$ be an increasing sequence in $C$ and $\alpha = \sup \{\alpha_n \mid n \in \omega\}$.

$\lambda < \alpha \implies\lambda < \alpha_n$ for some $n \in \omega$. Then $f(\lambda) < f(\alpha_n) < \alpha_{n+1} < \alpha$ since $\alpha_{n+1} \in C$. Hence $\lambda \in C$.