Certainly $A\subseteq\operatorname{dcl}(A)$. If $c\notin A$, its quantifier-free type over $A$ records only its adjacency or nonadjacency to each element of $A$. The random-graph extension axioms make every finite part of this type realizable away from any prescribed finite set; saturation therefore gives infinitely many distinct realizations. By part (a), maps fixing $A$ and moving $c$ among these realizations are elementary and extend to automorphisms. Thus $c$ has an infinite orbit over $A$ and does not belong to $\operatorname{acl}(A)$. Hence the stronger <algebraic and definable closure in the random graph> identity holds:
$$
\operatorname{acl}(A)=\operatorname{dcl}(A)=A.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 144 / 2 / d / Solution

{title2=$\operatorname{acl}(A)=\operatorname{dcl}(A)=A$}

For every small set $A$ in a monster random graph,
$$
\operatorname{acl}(A)=\operatorname{dcl}(A)=A.
$$
Any vertex outside $A$ has infinitely many conjugates over $A$ with the same adjacency pattern.


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