Motivation
Call a set $H$ of vertices of a graph $G$ homogeneous if it is either independent or complete. Every perfect graph of size $n$ has a homogeneous subset of size $\sqrt n$.
Noga Alon has shown that for every finite graph $G$ there is a finite graph $F$ of some size $n$ such that every induced subgraph of $F$ of size at least $\sqrt n$ contains an induced copy of $G$.
This implies the following: Let $G$ be any finite graph. Then there is a graph $F=(V,E)$ such that whenever $E'$ is another set of edges on the vertex set $V$ such that $(V,E')$ is perfect, then there is a set $H\subseteq V$ that is homogeneous with respect to $(V,E')$ such that the induced subgraph of $F$ on $H$ is isomorphic to $G$.
Let $C$ and $D$ be classes of finite graphs (closed under isomorphism). We say that $(C,D)$ is a Ramsey pair if the following holds: For every $G\in C$ there is $F\in C$ such that whenever $E'$ is another set of edges on the vertex set $V$ of $F$ such that $(V,E')\in D$, then there is a set $H\subseteq F$ that is homogeneous wrt $(V,E')$ such that the induced subgraph of $F$ on $H$ is isomorphic to $G$.
By the remark above, if $C$ is the class of all graphs and $D$ is the class of perfect graphs, then $(C,D)$ is a Ramsey pair. I know that the notion of a Ramsey triple,
where $G$ and $F$ come from different classes, is probably more natural, but the Ramsey
pairs did come up in some combinatorics related to set-theoretic forcing.
In some sense, $(C,D)$ being a Ramsey pair says that $C$ is substantially more complicated than $D$. In particular, if $C$ and $D$ are the same and nontrivial in the sense that they don't just consist of complete and empty graphs, then $(C,D)$ is not a Ramsey pair.
Here is the question
Can anyone come up with another nontrivial Ramsey pair?