Let $U$ be an ultrafilter on $\mathcal{P}(\omega)$ and $\langle \sigma _\alpha \mid \alpha < \omega_1 \rangle$ be a sequence of elements of $U$.

I know that the limit sup of $\sigma _\alpha$'s ($= \{i \in \omega \mid (\forall \alpha \in \omega_1 )(\exists \beta \in \omega_1 \setminus \alpha )\ i \in \sigma_\beta \} $) is also an element of $U$.

My question is: Is there $\sigma \in U$ such that for every finite $F \subset \sigma$, $\{\alpha\in \omega_1 \mid F \subset \sigma _\alpha \}$ is uncountable ??

My Observation: Construct $\langle \chi _s \in U , T_s \in [\omega_1 ]^{\aleph_1} \mid s \in X \rangle$ for some infinite branching tree $X \subset \omega ^{<\omega}$ by induction. Let $\chi _{()} := \limsup _{\alpha \in \omega_1} \sigma _\alpha$ and $T_{(i)} := \{ \alpha \in \omega_1 \mid i \in \sigma _\alpha \} $ for each $i \in \sigma$. Suppose that uncountable $T_s \subset \omega_1$ has been constructed for some $s \in \omega ^{<\omega}$, then define $\chi _s := \limsup _{\alpha \in T_s} \sigma_\alpha$ and $T_{s\frown(i) } := \{ \alpha \in T_s \mid i \in \sigma_\alpha \} $ for each $i \in \chi _s$. The sequence has been constructed. Now, if $\bigcup _{n < \omega} \mathrm{ran}(f|n) \in U $ for some branch $f \in \omega^\omega$ of $X$, then $\bigcup _{n < \omega} \mathrm{ran}(f|n)$ is what we want; for each $N < \omega$, $\{ \alpha \in \omega_1 \mid f(0), \ldots , f(N-1) \in \sigma_\alpha \} $ contains an uncountable set $T_{f|N}$ . Is there such $f$ ?

  • $\begingroup$ Did you try Fodor's Lemma? $\endgroup$
    – Farmer S
    Apr 26, 2021 at 11:35

1 Answer 1


Let $U$ be an ultrafilter on $\omega$ (or any family of subsets of $\omega$) and let $\langle\sigma_\alpha:\alpha\lt\omega_1\rangle$ be a sequence of elements of $U$.

Call a set $F\subset\omega$ bad if $F$ is finite and $\{\alpha\in\omega_1:F\subset\sigma_\alpha\}$ is countable. Let $B$ be the collection of all bad sets.

For each $F\in B$ let $\beta_F=\sup\{\alpha\in\omega_1:F\subset\sigma_\alpha\}\lt\omega_1$.

Let $\beta=\sup\{\beta_F:F\in B\}$; then $\beta\lt\omega_1$ since $B$ is countable.

If $\beta\lt\xi\lt\omega_1$ then $\sigma=\sigma_\xi$ does the trick: $\sigma\in U$, and $\sigma$ contains no bad sets, so for every finite $F\subset\sigma$, $\{\alpha\in\omega_1:F\subset\sigma_\alpha\}$ is uncountable.

  • 1
    $\begingroup$ I see, you are right. I focused to a misplaced perspective. Thank you for your simple proof. $\endgroup$
    – Gawr Gura
    Apr 26, 2021 at 12:31

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