As part of Theorem 2.6 Bollobas gives an example of a unique graph up to isomorphism for a countable vertex random graph with $P_k$ for each k (i.e. extension axiom) : $G_0$ = ($\mathbb{N}$,E) with E ={ij : i < j, $p_i$ | j}. Presumably this is the same as the Rado graph, which is isomorphic to the hereditary finite sets, with vertices interpreted as finite sets, with no sets duplicated. However $G_0$ appears to contain an infinite number of duplicates of each finite set. So for example vertex 1 appears as the only vertex connected to (i.e. as the only element in) vertex numbers 2, 4, 8, 16,...since vertex 1 is associated with the prime number 2 and vertex number 2^k is only ever divided by 2.
So is this correct - countable random graphs can contain an infinite number of duplicates of each set, and if so how does Theorem 2.6 get around this? (The proof of Theorem 2.6 appears to assume there are no duplicates when the back and forth isomorphism construction is made).
What you’re missing is that vertex $1$ isn’t the only vertex connected to vertex $2^k$: vertex $2^k$ is also connected to every vertex $n$ such that $p_{2^k}\mid n$, and the set of such $n$ depends on $k$. Thus, these vertices are not duplicates of one another in the sense that you have in mind.
And it’s not hard to verify that $G_0$ does have the extension property. Suppose that I have vertices $a_k$ for $k=1,\ldots,n$, and let $I\subseteq\{1,\ldots,n\}$. Let
$$a=p\prod_{k\in I}p_{a_k}\;,$$
where $p$ is a prime greater than $\prod_{k=1}^np_{a_k}$. Then for $k=1,\ldots,n$ there is an edge between $a$ and $a_k$ if and only if $k\in I$: the factors $p_{a_k}$ for $k\in I$ ensure that there is an edge between $a$ and $a_k$ if $k\in I$, and the factor of $p$ ensures that there is no edge between $a$ and $a_k$ if $k\in\{1,\ldots,n\}\setminus I$.