# "Good limit" of an uncountable sequence of elements of an ultrafilter

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 $$f$$ ?

• Did you try Fodor's Lemma? Apr 26, 2021 at 11:35

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.