RG is not $\kappa$-stable for all $\kappa$

Question

I am trying to show for my home work that the theory of random graphs RG, is not $\kappa$-stable for every $\kappa$ i.e If $M\vDash RG$ and $A\subseteq |M|$ with $|A|\le\kappa$ then $S_1(A)\le\kappa$

I tried to build a $\kappa$-tree with $2^{\kappa}$ branches of consistent formulas,but i get stuck at the limit ordinal stage, and so i manage to do that only for $\kappa =\omega$

It must have something to do with the fact that there are $\kappa$ many finite subsets for $A\subseteq |M|$ with $|A|=\kappa$, and for each pair of such sets we can fined a member of $|M|$ that is in relation with all members of one of them, and is not in relation with any of the other set members. But i need help to formalise it.

thanks a head

Answer 1

So i think iv got this:

Let $M$ be a model of $RG$ of infinite cardinality $\kappa$ . We will note that given a partition of $M$ in to pairs of size less then $n>0$ sets $\left\{ A_{i},B_{i}\right\} _{i\in I}$ , if we define $\varphi_{A_{i},B_{i}}(x)$ to be the statement “$x$ is in relation with all members of $A_{i}$ and with non of the members of $B_{i}$ ” then a set of boolean combination that uses $\varphi_{A_{i},B_{i}}$ at most once, is finitely realizable in $M$ . (since $M\vDash RG$ ) Thus there are at least $2^{|I|}$ types in $S_{1}\left(M\right)$ and since it must be that $\left|I\right|\ge\kappa$ then $S_{1}\left(M\right)>\kappa$ and we are done. i.e $RG$ is not $\kappa$ -stable for any infinite $\kappa$ $ \square$

Answer 2

If you have talked about the order property (Assume you are working inside a monster model. $\varphi(x,y)$ has the OP iff there exists $(a_{i})_{i<\omega}$ and $(b_{i})_{i<\omega}$ s.t. $\models{\varphi(a_{i},b_{j})}$ iff $i<j$), and the fact that a theory is unstable iff there exists a model $M$ and a $L$ formula witnessing the order property inside of $M$, then you can finish this problem by showing that the formula $E(x,y)$ has the OP (.).

If not, you probably need to mine the ideas found in the proof of the above statement. The proof that OP implies unstable is usually pretty straight forward, though the other direction of the theorem is very involved. The problem is, authors usually prove a bunch of equivalent statements in one go, so mining exactly the ideas you need for this can be a little tricky.

Edit 1: As an alternate, if you have talked about the connections between ranks and stability, you should be able to show that $R^{E(x,y)}(z=z)$ is $\geq{\omega}$, which in turn implies that $S_{E(x,y)}(A)$ has cardinality $>\kappa$ for any cardinal $\kappa$. Thinking on it further this is probably the shortest way. These facts should be there in pretty much any book that is dedicated to stability theory. I would recommend glancing at Cassanova's text "Simple Theories and Hyper Imaginaries". The second chapter should have what you need in very concise form.