Is this induced topology the product topology? Consider the pairs $(A_i,a_i)$ where $A_i$ is a set and $a_i: U(A_i)\rightarrow A_i$ is a relation where $U(A_i)$ is denotes the set of ultrafilters on $A_i$.
Now consider the topological spaces $(A_i,\tau_i)$ where $\tau_i$ is the topology generated as follow:

$U\in \tau_i \iff$ for every $x\in U$ whenever $(\mathcal{F},x)\in a_i
$ we have $U\in \mathcal{F}$ (here $\mathcal{F}$ is an ultrafilter
  on $A_i$). $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  \ \ \ \ \ \ \ \ \ \ \ \
 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (*)$

Now consider the pair $(\Pi_{i\in I}A_i, a_p)$ where $a_p: U(\Pi_{i\in I}A_i)\rightarrow \Pi_{i\in I}A_i$ is a relation between ultrafilters on the product and the product itself (as before). 
Suppose that the relation $a_p$ is described as follows:
$$\big(\mathcal{F},(x_i)_{i\in I}\big)\in a_p \iff ({\pi_i}_*(\mathcal{F}),x_i) \in a_i \text{ for every } i\in I$$
Here ${\pi_i}_*(\mathcal{F}) = \{X\subseteq A_i: \pi_i^{-1}(X)\in\mathcal{F}\} $ as usual where $\mathcal{F}$ is an ultrafilter on the product.
Question: is the topology generated by $a_p$ on $\Pi_{i\in I}A_i$ (by using the characterization of open sets in $(*)$ in the orange block) the product topology?
Background: the reason I suspect this is that if $\tau$ is the product topology on $\Pi_{i\in I}A_i$, because ultrafilter convergence in the topological sense satisfies the same relation as $a_p$.
 A: No.  For instance, let $a_i$ be the empty relation for all $i$.  Then $\tau_i$ is the discrete topology, and $a_p$ is also the empty relation (assuming $I$ is nonempty) so it induces the discrete topology as well.  But the product topology on $\prod A_i$ is not the discrete topology as long as each $A_i$ is nonempty and infinitely many of the $A_i$ have more than one point.

For an example where principal ultrafilters converge to the corresponding points, let $A$ be an infinite set and choose two distinct elements $s,t\in A$ and a nonprincipal ultrafilter $\mathcal{G}$ on $A$.  Define a relation $a:U(A)\to A$ by saying $(\mathcal{F},x)\in a$ iff $\mathcal{F}$ is the principal ultrafilter at $x$, or $\mathcal{F}\neq\mathcal{G}$ is nonprincipal and $x=s$, or $\mathcal{F}=\mathcal{G}$ and $x=t$.  Note that if $U$ is any open set containing $s$ in the induced topology, then $U$ is cofinite (since if $U$ is coinfinite, you can choose a nonprincipal ultrafilter $\mathcal{F}\neq\mathcal{G}$ which contains $A\setminus U$ and $(\mathcal{F},s)\in a$).  In particular, it follows that with respect to the induced topology, the ultrafilter $\mathcal{G}$ converges to both $s$ and $t$, so the induced topology is not Hausdorff.
Now consider the product relation $a_p$ on $A\times A$.  I claim that the set $\Delta=\{(x,x):x\in A\}$ is closed in the topology induced by $a_p$.  Indeed, suppose $\mathcal{F}$ is an ultrafilter on $A\times A$ such that $\Delta\in\mathcal{F}$.  For any $X\subseteq A$, $(X\times A)\cap \Delta=(A\times X)\cap\Delta$, and it follows that ${\pi_0}_*(\mathcal{F})={\pi_1}_*(\mathcal{F})$.  Thus if $\mathcal{F}$ converges to a point $(x,y)$ according to $a_p$, then the ultrafilter ${\pi_0}_*(\mathcal{F})={\pi_1}_*(\mathcal{F})$ converges to both $x$ and $y$ according to $a$.  Since no ultrafilter converges to two different points according to $a$, this means $x=y$, so $(x,y)\in\Delta$.  Thus $\Delta$ is closed in the topology induced by $a_p$.  However, $\Delta$ is not closed in the product topology since the topology on $A$ is not Hausdorff.
