Throughout this question, I shall let $A^{\mathcal{U}}$ denote the ultrapower of a structure $A$ by an ultrafilter $\mathcal{U}$. Suppose that $T$ is an Aronszajn tree and $\mathcal{U}$ is an ultrafilter on a countable set $I$. Let $T\_{\alpha}$ denote the $\alpha$-th level of a tree $T$. Let $T^{(\mathcal{U})}$ denote the tree where the $\alpha$-th level of $T^{(\mathcal{U})}$ is the ultrapower $(T\_{\alpha})^{\mathcal{U}}$ and where if $(x\_{i})\_{i\in I}/\mathcal{U}\in(T\_{\alpha})^{\mathcal{U}},(y\_{i})\_{i\in I}/\mathcal{U}\in (T\_{\beta})^{\mathcal{U}}$, then $(x\_{i})\_{i\in I}/\mathcal{U}<(y\_{i})\_{i\in I}/\mathcal{U}$ iff $\{i\in I|x\_{i}<y\_{i}\}\in\mathcal{U}$. Then $T^{(\mathcal{U})}$ is clearly a tree. In other words, we trim the ultrapower $T^{\mathcal{U}}$ so that it becomes a tree $T^{(\mathcal{U})}$.


Does the tree $T^{(\mathcal{U})}$ have a branch of length $\omega\_{1}$? What about when we assume that $T$ is a Suslin tree? What happens when we consider ultraproducts of countably many Aronszajn trees instead of ultrapowers?



Does an ultrapower of an Aronszajn tree have an $\omega_{1}$-branch?
Joel David Hamkins asked for the saturation version of Martin Goldstern's argument; here it is. Like Martin, I assume that every node of $T$ has successors at all higher levels. I'll build, by induction on ordinals $\xi<\omega\_1$, a branch $(x\_\xi)$ in $T^{(U)}$, where $x\_\xi$ is at level $\xi$. Of course, I start with the root of $T$ as $x\_0$, and at successor steps I let $x\_{\xi+1}$ be any immediate successor of $x\_\xi$. The existence of an immediate successor follows from my initial assumption about $T$ because $T^U$ is an elementary extension of $T$, and $T^{(U)}$ agrees with $T^U$ at standard levels like $\xi$ and $\xi+1$. Now consider a limit ordinal $\delta$, and suppose we have $x\_\xi$ for all $\xi<\delta$, forming a chain in $T^{(U)}$. In the ultrapower with respect to $U$ of the universe (or a part of the universe adequate for what I'm about to do), apply saturation to the type, in one variable $v$, that says "$v$ is an element of the ${}^\*\delta$ level of ${}^\*T$ and is above each $x\_\xi$ for $\xi<\delta$." That's a countable set of formulas, using parameters ${}^\*\delta$ (the element of the ultrapower corresponding to $\delta$ in the standard universe), ${}^\*T$, and the earlier members $x\_\xi$ of the branch. This type is finitely satisfiable; a finite subset requires $v$ to be above only finitely many of the $x\_\xi$'s, and the highest of thhose $x\_\xi$'s has a successor at level ${}^\*\delta$ because of the initial assumption about $T$ (and elementarity). So, by saturation, the whole type is satisfiable, and an element satisfying it serves as $x\_\delta$.


