I'm trying to prove the following result which seems correct to me, but I'm not sure how to proceed. The proposition is:
Let $M$ be a structure, $A\subseteq M$, $(a_1,\dots,a_n)$ be a sequence from $M$ such that for all $i\le n$ the type $tp(a_i\,\vert\,A\cup\{a_1,\dots,a_{i-1}\})$ is isolated by $\varphi_i$, then $tp(a_1,\dots,a_n\,\vert\,A)$ is isolated by $\bigwedge_{i=1}^n\varphi_i$.
What I have so far:
Let $\Phi_i$ denote $\bigwedge_{j=1}^i\varphi_i$. By induction, in the base case $tp(a_1\,\vert\,A)$ is isolated by $\Phi_1=\varphi_1$ by hypothesis.
Now suppose that $\Phi_i$ isolates $tp(a_1,\dots,a_i\,\vert\,A)$ and $\varphi_{i+1}$ isolates $tp(a_{i+1}\,\vert\,A\cup\{a_1,\dots,a_i\})$. Let $\psi\in tp(a_1,\dots,a_{i+1}\,\vert\,A)$ and $(c_1,\dots,c_{i+1})$ from $M$ such that $M\vDash\Phi_i(c_1,\dots,c_i)\land\varphi_{i+1}(a_1,\dots,a_i,c_{i+1})$. We want to show that $M\vDash\psi(c_1,\dots,c_i,c_{i+1})$. Now since $\psi(a_1,\dots,a_i,x)\in tp(a_{i+1}\,\vert\,A\cup\{a_1,\dots,a_i\})$, we have $M\vDash\psi(a_1,\dots,a_i,c_{i+1})$.
Now because $M\vDash\Phi_i(c_1,\dots,c_i)$ I know that $tp(c_1,\dots,c_i\,\vert\,A)=tp(a_1,\dots,a_i\,\vert\,A)$, but it seems that I need to show that the two tuples realize the same type over $A\cup\{c_{i+1}\}$ in order to conclude. I'd appreciate any hint to how to proceed from here, or a counterexample if the result is not correct.
Answer
There's an issue with the statement of the proposition you're trying to prove: The formula $\varphi_i$ is a formula over $A\cup\{a_1,\dots,a_{i-1}\}$, not over $A$, so it can't appear as a conjunct in a formula purporting to isolate $\text{tp}(a_1,\dots,a_n/A)$.
It is true that if $\text{tp}(a_i/A\cup\{a_1,\dots,a_{i-1}\})$ is isolated for all $i$, then $\text{tp}(a_1,\dots,a_n/A)$ is isolated, and an inductive argument of the type you wrote down will work. I would recommend trying first trying to prove this, and then extracting the formula that works from the proof.
No comments:
Post a Comment