Club Sets and Stationary Sets

11 Nov 2025

This post was originally written in Korean, and has been machine translated into English. It may contain minor errors or unnatural expressions. Proofreading will be done in the near future.

Club Sets

Definition. $C \subseteq \omega_1$ is a club set if $C$ is closed and unbounded. That is,

• For any sequence of ordinals $(\alpha_n)_{n \in \omega} \subseteq C$, if $\alpha_n \to \alpha$, then $\alpha \in C$.
• For any $\gamma < \kappa$, there exists $\alpha \in C$ such that $\alpha > \gamma$.

The name is indeed quite straightforward.

Theorem. If $C_1, C_2 \subseteq \omega_1$ are two club sets, then $C = C_1 \cap C_2$ is also a club set.

Proof. It is easy to show that $C$ is closed. To prove that $C$ is unbounded, let us take an arbitrary $\gamma \in \omega_1$. Since $C_1$ is unbounded, there exists the smallest $\alpha_0 \in C_1$ such that $\gamma < \alpha_0$. Similarly, since $C_2$ is unbounded, there exists the smallest $\beta_0 \in C_2$ such that $\alpha_0 < \beta_0$. Again, since $C_1$ is unbounded, there exists $\alpha_1 \in C_1$ such that $\beta_0 < \alpha_1$. Repeating this process, we obtain sequences of ordinals $\lbrace \alpha_n \rbrace$ and $\lbrace \beta_n \rbrace$. These sequences converge to the same ordinal $\delta$, so $\delta \in C$ and $\delta > \gamma$. (Note: “the smallest” is used to avoid unnecessary reliance on the axiom of choice.) ■

By slightly modifying the above proof, we can derive the following theorem.

Theorem. If $\lbrace C_n \rbrace _{n < \omega}$ is a countable collection of club sets, then $C = \bigcap C_n$ is also a club set.

Proof. Exercise (Hint: Use a triangle argument). ■

Thus, given a countable collection of club sets, we can obtain a new club set by taking their intersection. On the other hand, if we are given an uncountable collection of club sets, the following result can be used.

Theorem. Let $\lbrace C_\xi \rbrace _{\xi < \omega_1}$ be a collection of club sets. The diagonal intersection of $\lbrace C_\xi \rbrace$ is defined as follows:

\[\Delta_{\xi < \omega_1} C_\xi = \{ \alpha \in \omega_1 : \forall \xi < \alpha \;\; \alpha \in C_\xi \}\]

$D$ is a club set.

Proof. First, let us show that $D$ is closed. Suppose $(\alpha_n)_{n \in \omega} \subseteq D$ and $\alpha_n \to \alpha$. For any $\xi < \alpha$, there exists $n \in \omega$ such that $\xi < \alpha_n$. By the definition of the diagonal intersection, $C_\xi$ contains $\lbrace \alpha_m : m \geq n \rbrace$. Hence, $\sup \lbrace \alpha_m : m \geq n \rbrace = \alpha \in C_\xi$, so $\alpha \in D$.

Next, let us show that $D$ is unbounded. For any $\gamma < \omega_1$, let $D_0 = \bigcap_{\xi < \gamma} C_\xi$. Since $D_0$ is a club set, it is unbounded. Thus, there exists the smallest $\alpha_0 \in D_0$ such that $\alpha_0 > \gamma$. Similarly, let $D_1 = \bigcap_{\xi < \alpha_0} C_\xi$. Since $D_1$ is a club set, there exists the smallest $\alpha_1 \in D_1$ such that $\alpha_1 > \alpha_0$. Repeating this process inductively, we obtain a sequence $(\alpha_n)$ whose limit is $\alpha$. For any $\xi < \alpha$, there exists $n$ such that $\xi < \alpha_n$, and for all $m \geq n$, $\alpha_m \in C_\xi$. Hence, $\alpha \in C_\xi$ and $\alpha \in D$. ■

Verify that $\Delta C_\xi \supseteq \bigcap C_\xi$.

The definitions and theorems discussed so far can be generalised as follows. Let $\kappa$ be an uncountable cardinal with $\operatorname{cf}\kappa > \omega$.

Definition. $C \subseteq \kappa$ is a club set if $C$ is closed and unbounded. That is,

• For any $\alpha < \kappa$, if $\sup (C \cap \alpha) = \alpha$ (i.e., $C$ is unbounded at $\alpha$), then $\alpha \in C$.
• For any $\gamma < \kappa$, there exists $\alpha \in C$ such that $\alpha > \gamma$.

Theorem. If $\lambda < \operatorname{cf}\kappa$ and $\lbrace C_\xi \rbrace _{\xi < \lambda}$ is a collection of club sets, then $C = \bigcap C_\xi$ is also a club set.

Theorem. If $\kappa$ is regular, then the diagonal intersection of a collection $\lbrace C_\xi \rbrace _{\xi < \kappa}$ of club sets is also a club set.

Proof. The proof is almost identical to the case $\kappa = \omega_1$. ■

Stationary Sets

By closure under finite intersections, the collection of club sets generates a filter. A set that does not belong to the dual ideal of this filter is called a stationary set. In other words:

Definition. Let $S \subseteq \omega_1$. If $C \cap S \neq \varnothing$ for every club set $C \subseteq \omega_1$, then $S$ is a stationary set.

If there exists a club set $C$ such that $C \cap S = \varnothing$, then $S^c \subseteq C$, so $C$ belongs to the dual ideal. Thus, the two definitions are equivalent.

Stationary sets capture “significantly large” subsets of $\omega_1$. This can be better understood through the following (informal) analogy: if $\omega_1$ has measure 1, then club sets correspond to sets with $\mu(C) = 1$, and stationary sets correspond to sets with $\mu(S) > 0$.

The following facts are easily proven (note how 1 and 3 naturally align with the measure analogy):

Theorem. The following hold:

1. Every club set is a stationary set.
2. Every stationary set is unbounded.
3. If $\bigcup_{n < \omega} A_n$ is a stationary set, then there exists some $m \in \omega$ such that $A_m$ is a stationary set.

The name “stationary” for stationary sets arises from the following theorem.

Theorem. Let $S \subseteq \omega_1$. A function $f: S \to \omega_1$ is regressive if $f(\alpha) < \alpha$ for all $\alpha \in S$. The following are equivalent:

1. $S$ is a stationary set.
2. For any regressive function $f: S \to \omega_1$, $f$ is constant on some unbounded set $T \subseteq S$.
3. For any regressive function $f: S \to \omega_1$, $f$ is constant on some stationary set $T \subseteq S$.

Proof. 3 ⇒ 2 is trivial, so it suffices to prove 1 ⇒ 3 and 2 ⇒ 1.

1 ⇒ 3

We prove the contrapositive. Suppose there exists a regressive function $f: S \to \omega_1$ such that $f$ is not constant on any stationary subset of $S$. That is, for every $\xi \in \omega_1$, $A_\xi = f^{-1}(\lbrace \xi \rbrace)$ is not stationary. Hence, there exists a club set $C_\xi$ such that $A_\xi \cap C_\xi = \varnothing$.

Let $D = \Delta_{\xi < \omega_1} C_\xi$. Since $D$ is a club set, if $S$ is stationary, then $D \cap S \neq \varnothing$. Let $\alpha \in D \cap S$. Since $\alpha \in D$, for any $\xi < \alpha$, $\alpha \in C_\xi$. Thus, $\alpha \notin A_\xi$, so $f(\alpha) \neq \xi$. However, since $\alpha \in S$, $f(\alpha) < \alpha$. This is a contradiction, so $S$ is not stationary. □

2 ⇒ 1

We again prove the contrapositive. Suppose $S$ is not stationary. Then there exists a club set $C$ such that $S \cap C = \varnothing$. Define the following function:

\[f(\alpha) = \sup (C \cap \alpha)\]

For any $\alpha \in S$, $f(\alpha) \leq \alpha$. If $f(\alpha) = \alpha$, then $C$ contains a sequence of ordinals converging to $\alpha$. Since $C$ is closed, $\alpha \in C$. This contradicts $S \cap C = \varnothing$. Thus, $f(\alpha) < \alpha$, and $f$ is regressive.

If $f$ were constant on some unbounded set $T$, then for some $\gamma < \omega_1$, there would exist an unbounded sequence $(\alpha_n)$ such that $\sup (C \cap \alpha_n) = \gamma$. Since $C$ is unbounded, this implies $\sup \alpha_n = \gamma$. However, since $(\alpha_n)$ is unbounded, $\sup \alpha_n = \omega_1$. This is a contradiction. Thus, $f$ is a regressive function that is not constant on any unbounded set. ■

The above theorem can be generalised as follows:

Fodor’s Lemma. Let $\kappa$ be an uncountable regular cardinal. For any $S \subseteq \kappa$, the following are equivalent:

1. $S$ is a stationary set.
2. For any regressive function $f: S \to \omega_1$, $f$ is constant on some unbounded set $T \subseteq S$.
3. For any regressive function $f: S \to \omega_1$, $f$ is constant on some stationary set $T \subseteq S$.

Applications of Stationary Sets

As an application of stationary sets, consider the following combinatorial theorem:

Δ-System Lemma. Let $\lbrace A_\xi\rbrace _{\xi < \omega_1}$ be a collection of finite subsets of $\omega_1$. There exist an unbounded set $I \subseteq \omega_1$ and a finite set $A$ such that for any $i, j \in I$ with $i \neq j$, $A_i \cap A_j = A$.

Proof. By the pigeonhole principle, we may assume without loss of generality that $|A_\xi | = n$ for all $\xi < \omega_1$. Define:

\[C = \{ \alpha < \omega_1 : \forall \xi < \alpha \;\; \operatorname{max}C_\xi < \alpha \}\]

$C$ is a club set. For $0 \leq k \leq n$, define:

\[T_k = \{ \alpha \in C : |A_\alpha \cap \alpha | = k \}\]

Since $\bigcup_{0 \leq k \leq n} T_k = C$, there exists some $m$ such that $T_m$ is stationary. For $0 \leq l < m$, define:

\[f_l(\alpha) : T_m \to \omega_1; \quad \alpha \mapsto \text{$l$th element of $A_\alpha$}\]

Since $|A_\alpha \cap \alpha| > l$, $f_l(\alpha) < \alpha$. Thus, $f_l$ is regressive. Now, repeat the following steps:

• There exists a stationary set $S_0 \subseteq T_m$ such that $f_0$ is constant on $S_0$.
• There exists a stationary set $S_1 \subseteq S_0$ such that $f_1 |_{S_0}$ is constant on $S_1$.
• There exists a stationary set $S_2 \subseteq S_1$ such that $f_2 |_{S_1}$ is constant on $S_2$.
• …
• There exists a stationary set $S_{m-1} \subseteq S_{m-2}$ such that $f_{m-1} |_{S_{m-2}}$ is constant on $S_{m-1}$.

For each $0 \leq l < m$, $f_l$ is constant on $S_{m-1}$. It is now easy to verify that $S_{m-1} = I$ is the desired set. ■