I am trying to solve this exercise on ultrapower constructions and nonstandard objects.
The definition of the countable saturation property I am working with is: Suppose $ \lbrace B_n \rbrace_{n \in \mathbb{N}} \subseteq \, ^*A$ is a countable family of internal sets with the finite intersection property, then $\bigcap_{n \in \mathbb{N}} B_n \neq \emptyset $.
I want to show that countable saturation implies that every sequence $\langle B_n \mid n \in \mathbb{N} \rangle $ of internal sets can be extended to an internal sequence $\langle B_\nu \mid \nu \in \, ^* \mathbb{N} \rangle$ , that is, there exists an internal function $\sigma$ with domain $ ^* \mathbb{N}$ and such that $\sigma(n) = B_n$ forevery $ n \in \mathbb{N}$.
Let $C_n$ (for $n\in\mathbb N$) be the set of internal ${}^*$finite sequences $s$ of length at least $n$ and such that the first $n$ terms of $s$ coincide with the first $n$ terms of the given sequence $\langle B_n:n\in\mathbb N\rangle$. Apply countable saturation to get an $s\in\bigcap_{n\in\mathbb N}C_n$. This $s$ is almost what you need; its only defect is that its length is some nonstandard natural number $\xi$, whereas you want a sequence as long as all of ${}^*\mathbb N$. No problem; just extend $s$ with zeros at all places beyond $\xi$.