Rado graph containing infinitely many isomorphic subgraphs

I realise that the question is almost three years old, but maybe that's not that long in Maths.

I am not sure who the following is originally due to, but as far as I am aware, it's the standard to show that the Rado graph contains every countable graph. The idea is to start with whatever graph you want and construct the Rado graph around it.

Let $G=G_0$ be any countable graph. For $i>0$ define a new Graph $G_i$ as follows:

• The vertices $V(G_{i})$ of $G_i$ are all of $V(G_{i-1})$ plus an extra vertex $v_A$ for every finite subset $A$ of $V(G_{i-1})$.




• All edges of $V(G_{i-1})$ are also edges of $V(G_{i})$




• Add an edge between $ain V(G_{i-1})$ and $v_{A}in V(G_{i})$ whenever $ain A$.

Let $R$ be the union of the graphs $G_i$ over all $i>0$. Then show that $R$ is the Rado graph. Clearly, $R$ contains $G=G_0$.

One of the nice things about this construction is that you can show without much difficulty that any automorphism of $G$ extends to an automorphism of $R$. So not only does the Rado graph contain every countable graph $G$, but it contains "special" copies of $G$ with the above extension property.