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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.

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2 Answers 2

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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.

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  • $\begingroup$ Thank you so much! I got it. $\endgroup$
    – Akira
    Commented Mar 8, 2019 at 8:30
  • $\begingroup$ Hi @William, I have a closely related and similar question here. I have tried to replicate your method but have failed. If you don't mind, please help me tackle it! $\endgroup$
    – Akira
    Commented Mar 10, 2019 at 13:08
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I added the proof that $C$ is closed here.


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 $\lambda \in C$.

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