# Algebraic types are isolated

Let $\mathcal M$ be a $\mathcal L$-structure, and $A\subseteq M$. Let $p(x)\in S_1(A)$ be a complete 1-type of the theory $T_A := \mathrm{Th}(\mathcal M_A)$. We say that $p$ is algebraic if there exists an integer $k$ such that $p$ has at most $k$ realisations in every model of $T_A$.

I have to show that if $p$ is an algebraic 1-type then it is isolated. The only thing I could think of is the following: let $k$ be the upper bound on the number of realisations of $p$, and let $C = \{c_1,\ldots,c_{k+1}\}$ be new constants. We then have that $T_A\cup \bigcup_{i=1}^{k+1} p(c_i) \models \bigvee_{i\neq j} c_i=c_j$, and by compactness there exists a finite subset $\Phi$ of $p$ such that $T_A\cup \bigcup_{i=1}^{k+1} \Phi(c_i) \models \bigvee_{i\neq j} c_i=c_j$, i.e. (treating $\Phi$ as a conjunction a formulas of $p$) $$T_A \models \forall x_1,\ldots,x_{k+1} \left(\bigwedge_{i=1}^{k+1}\Phi(x_i) \Rightarrow \bigvee_{i\neq j} x_i=x_j \right)$$

This only tells me that in every model of $T_A$ at most $k$ points may satisfy $\Phi$, but I don't see why $\Phi$ would isolate $p$. On the other hand, I don't see how to extract more information from $p$ than this bound on the number of realisations. I'd really appreciate a hint!

Hint: if a $\Phi$ does not isolate $p$, then there is a formula $\Psi \in p$, such that $\Phi \land \Psi$ define a proper subset of the set defined by $\Phi$. Repeat this argument for $\Phi \land \Psi$. Can this process go to infinity?
• Since $\Phi$ defines a subset of size $\leq k$ in every model, this process terminates after at most $k$ steps. Letting $\Psi_1,\ldots,\Psi_l$ be the formulas obtained during the process, $\Phi\wedge \bigwedge_{i=1}^l \Psi_i$ isolates $p$. Thanks a lot! – zarathustra Jan 29 '14 at 13:08