By the definition of the logic topology, we see that this is an action by homeomorphisms. Finally, the $G$-orbit of $e/{\bar F}_\Delta$ equals $G/{\bar F}_\Delta$, which is dense in the logic topology by Remark \[remark: denseness very basic\].

We will also need the following general remark.

\[remark: added because of the Referee\] If $D$ is a type-definable subset of $G^*$ which is a union of sets of realizations of complete $\Delta$-types over $M$, then $D$ is $\Delta$-type-definable over $M$.

Let $\pi \colon G^* \to S_{G,\Delta}(M)$ be the map given by $\pi(a):= \operatorname{{tp}}_{\Delta}(a/M)$, and let $p(x)$ be a partial type defining $D$. Then $\pi[D]$ is the subset of $S_{G,\Delta}(M)$ consisting off all types consistent with $p(x)$, so $\pi[D]$ is closed, i.e. it is the set of all types in $S_{G,\Delta}(M)$ extending some partial $\Delta$-type $p'$ over $M$. Since $D$ is a union of sets of realizations of complete $\Delta$-types over $M$, we see that $D=\pi^{-1}[\pi[D]]$, so $D=p'(G^*)$ is $\Delta$-type-definable over $M$.

\[theorem: main theorem\] ${\bar F}_\Delta=E_\Delta'$, so $(G,G^*/{\bar F}_\Delta, e/{\bar F}_\Delta)$ is the universal $\Delta$-definable $G$-ambit.

First, we prove that ${\bar F}_\Delta \subseteq E_\Delta'$. For this, it is enough to show that ${F_0}_\Delta \subseteq E_\Delta'$, because then clearly $F_\Delta \subseteq E_\Delta'$ 