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 $f$ ?



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


