My textbook Introduction to Set Theory 3rd by Hrbacek and Jech defines relevant 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 exercise

Let $\{A_\alpha \mid \alpha < \omega_1\}$ be a collection of subsets of $\omega_1$, each of size $n$, for some fixed number $n$. Let $C = \{\alpha < \omega_1 \mid \forall \xi < \alpha: \max (A_\xi) < \alpha\}$. Prove that $C$ is closed unbounded.

I am stuck at the second case where $f$ is not eventually constant. Please shed me some light! Thank you so much.

My attempt:

  1. $C$ is unbounded

Define a function $f: \omega_1 \to \omega_1$ by $$f(\alpha) = \begin{cases} 0 & \text{if } \alpha=0 \\ \sup (\bigcup_{\xi < \alpha}A_\xi) &\text{otherwise}\end{cases}$$

It follows that $f$ is a nondecreasing function. For any $\beta < \omega_1$, we prove that there exists $\alpha_0 \in C$ such that $\beta < \alpha_0$.

a. $f$ is eventually constant, i.e. there exists $\lambda < \omega_1$ such that if $\alpha \ge \lambda$ then $f(\alpha) = f(\lambda)$

Let $\lambda_0 = \min \{ \lambda <\ \omega_1 \mid \forall \alpha \ge \lambda:f(\alpha) = f(\lambda)\}$ and $\alpha_0 = \max \{f(\lambda_0)+1,\beta+1\}$. It follows that $\alpha_0 > \beta$ and $f(\alpha_0) < \alpha_0$. On the other hand, $\forall \xi <\alpha_0: \max (A_\xi) \le \sup (\bigcup_{\xi < \alpha_0}A_\xi)= f(\alpha_0) < \alpha_0$.

b. $f$ is not eventually constant, i.e. for all $\alpha < \omega_1$ there exists $\lambda < \omega_1$ such that $f(\lambda) > \alpha$


1 Answer 1


Assume $C$ is bounded in $\omega_1$, then $\sup C<\omega_1$, define the sequence $(x_\alpha\mid\alpha<\omega)$ with $x_0=\sup C+1$ and $x_{n+1}=\sup\{\max{A_\beta}+1\mid\beta<x_n\}$.

Because $\omega_1$ is regular we have $\sup x_\alpha<\omega_1$, and given $\xi<\sup x_\alpha$, there exists $\beta$ such that $\xi<x_\beta$ so $\max A_\xi<x_{\beta+1}\le\sup x_\alpha$, so $\sup C<\sup x_\alpha \in C$

  • $\begingroup$ It's great! I think there is a typo ... so $A_\xi<x_{\beta+1}\le\sup x_\alpha$ ..., it should be $\max A_\xi<x_{\beta+1}\le\sup x_\alpha$. $\endgroup$
    – Akira
    Mar 10, 2019 at 14:13
  • $\begingroup$ @LeAnhDung yes, thanks. Fixed $\endgroup$
    – ℋolo
    Mar 10, 2019 at 14:16

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