# How to prove that the set $C$ is unbounded

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 $$f(\lambda) > \alpha$$

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 $$x_{n+1}=\sup\{\max{A_\beta}+1\mid\beta.
Because $$\omega_1$$ is regular we have $$\sup x_\alpha<\omega_1$$, and given $$\xi<\sup x_\alpha$$, there exists $$\beta$$ such that $$\xi so $$\max A_\xi, so $$\sup C<\sup x_\alpha \in C$$
• It's great! I think there is a typo ... so $A_\xi<x_{\beta+1}\le\sup x_\alpha$ ..., it should be $\max A_\xi<x_{\beta+1}\le\sup x_\alpha$. Mar 10, 2019 at 14:13