I have a sequence of countable sets $(A_{\gamma} \colon \gamma < \omega_1)$ for which I know that if I take any $x\in\bigcup\limits_{\gamma<\omega_1} A_{\gamma}$ then the set $$\{\beta < \omega_{1} \colon x\in A_{\beta}\}$$ is non stationary. I'm asked to prove there is a club $C$ such that for any $\alpha, \beta\in C$ we have $A_{\alpha}$ has an empty intersection with $A_{\beta}$.

What I thought of is to take, say $A_0$, all $x\in A_0$ and for each find a club $C_x$ that contains these indices which denotes the sets $x$ does not belong to. Then I wanted to do intersect all $C_x$-s. This would give me a club of indices that denotes the sets $A_0$ has an empty intersection with.

But this seems to be a dead-end as I'd have to do this construction for all $i<\omega_1$ and intersect all resulting clubs. I don't know if an uncountable intetsection of clubs would render a club. Would it?

Or perhaps there is an easier way or my idea is incorrect?

  • 2
    $\begingroup$ In general, the intersection of $\omega_{1}$-many club subsets of $\omega_{1}$ is not a club. It may even be empty - take for example $C_{\alpha} := \omega_{1} \setminus \alpha$ for all $\alpha < \omega_{1}$. Clearly $\bigcap \{ C_{\alpha} \mid \alpha < \omega_{1} \} = \emptyset$ is not a club. $\endgroup$ Jun 11, 2016 at 19:34

1 Answer 1


HINT: First, show that for each $A_\alpha$, the set $I_\alpha=\{\beta: A_\alpha\cap A_\beta\not=\emptyset\}$ is nonstationary (note that stationary sets have cardinality $\omega_1$, each $A_\theta$ is countable, and the countable union of nonstationary sets is nonstationary).

Next, work in stages. Say that a state is a pair $(X, D)$ where

  • $X\subseteq\omega_1$ is countable,

  • $X\cup D\subseteq\omega_1$ is club,

  • the smallest element of $D$ is greater than every element of $X$, and

  • for $\alpha\in X$, $\beta\in X\cup D$, we have $A_\alpha\cap A_\beta=\emptyset$.

Write $(X, D)\le (Y, E)$ if $X\supseteq Y$ and $D\subseteq E$, and $(X, D), (Y, E)$ are states.

Then if you can show that

  • There is no $\le$-least state


  • If $(X_0, D_0)\ge (X_1, D_1)\ge (X_2, D_2)\ge...$ is an $\omega$-sequence of decreasing states, then $(cl(\bigcup_{i\in\omega} X_i), E)$ is also a state (where "$cl$" is the closure of a set of ordinals) for some $E\subseteq \bigcap_{j\in\omega} D_j$.

you'll be able to build your $C$ by taking an appropriate chain of states of length $\omega_1$ and looking at the union of the left components of the elements of the chain!


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