A graph is homogeneous when every isomorphism between two finite induced subgraphs can be extended to an automorphism of the whole graph. I was reading Diestel's graph theory where he describes the following graph and says its clearly homogeneous. It's not immediately apparent to me why. If I have an isomorphism between two finite induced subgraphs, and map every other vertex outside the subgraphs to itself, does it become an automorphism of the entire graph? But wouldn't that make almost anything homogeneous? What is special about this one?
No, we cannot just say that we map every vertex outside the subgraphs to itself. For this to work, every one of those outside vertices would have to have the exact same adjacency relationship to both subgraphs. This is an unlikely coincidence for any graph, but for Rado-like graphs such as this one, the construction mechanism makes it impossible! (No matter which two finite subgraphs we start with, in some stage of the construction, we add a vertex which has neighbors in one, but not the other.)
Instead, we can extend a partial automorphism to a full one, one step at a time, by a back-and-forth argument. Let $S$ and $T$ be two finite subsets of $V(R^r)$, and let $\phi\colon S \to T$ be an isomorphism between the induced subgraphs $R^r[S]$ and $R^r[T]$. (For simplicity, I will stop distinguishing between vertex sets and their induced subgraphs.) We will also need to pick an ordering of the vertices of $R^r$.
The back-and-forth argument alternates between two steps. For one step, let $v$ be the first vertex of $R^r$ not yet in $S$; we identify the subset $S_v$ of all vertices in $S$ adjacent to $v$, and then apply $\phi$ to get an image $\phi(S_v) \subseteq T$. Note that because $v$ is adjacent to all of $S_v$, there cannot be an $(r-1)$-clique in $S_v$: together with $v$ it would form an $r$-clique, which we know cannot exist in $R^r$. So there is no $(r-1)$-clique in $\phi(S_v)$ either.
Because $T$ is finite, $T$ is contained in $\bigcup_{n=0}^N R^r_n$ for some finite $N$. Therefore, $R^r_{N+1}$ adds a vertex $w$ adjacent to every vertex in $\phi(S_v)$ and no other vertex in $T$. Now we can replace $S$ by $S \cup \{v\}$, $T$ by $T \cup \{w\}$, and set $\phi(v) = w$ to extend the isomorphism to subgraphs with one more vertex.
The step above guarantees that eventually, $S$ will grow to include each vertex of $R^r$. We alternate it with a reverse step, that lets $v$ be the first vertex of $R^r$ not yet in $T$, and adds $v$ to $T$ while adding a preimage of $v$ to $S$. It is done in essentially the same way. Alternating both steps, we eventually have both $S$ and $T$ grow to include every vertex of $R^r$, extending $\phi$ to an automorphism.
I understand Diestel's "clearly" as "if you understand the proof of the corresponding property for the Rado graph, it should be clear how to write a similar proof for $R^r$."