TITLE: Ultraproducts and Elementary Embeddings
QUESTION [6 upvotes]: Let $K= \{A_i: i\in \omega\}$ be a countable collection of $L-$structures. 
Suppose that for each $A_i, A_j$ in $K$, $\exists A_p \in K$ such that $i,j< p$ and $A_i \prec A_p $ and $A_j \prec A_p$. 
Suppose $D$ is a non-principal ultrafilter over $\omega$. Then, does it follow that for each $A_i \in K$, $A_i \prec \prod_D K$ (If not, could any strengthening of our hypothesis make it true)?
Notice that for every cofinite set $H$ of $\omega$, we have the following:
For any $A_i$ and $A_j$ ($i \in \omega$, $j \in H$), there exists a $p \in H$ such that $A_i, A_j \prec A_p$.

REPLY [2 votes]: I'd like to expand on tomasz's last paragraph, just to emphasize that you can get what you want without any further assumptions on the collection of structures, just by picking your ultrafilter more carefully.
For each $i\in \omega$, let $C_i = \{j\in \omega\mid A_i \prec A_j\}$. Let's call this the cone above $A_i$. Now let $\text{Cone} = \{X\subseteq \omega\mid \exists i\,C_i\subseteq X\}$. This is called the cone filter - it's the filter generated by the cones. To check that it's a proper filter, you just need to check that the set of cones has the finite intersection property, which is an easy consequence of your assumption that the set of structures has the joint embedding property.
Now let $A = \prod_{i\in\omega} A_i/U$, the using any ultrafilter $U\supseteq \text{Cone}$. For all $i$, we have a natural elementary embedding $A_i\prec A$, mapping an element $a\in A_i$ to the class $[a_i]_{i\in \omega}$, where $a_j$ is the image of $a$ in $A_j$ when $j\in C_i$ and arbitrary otherwise. This is well-defined, since $C_i\in U$.