I'm working with equivalent definitions of properties of formulas/theories, in particular I found these two definitions of $TP_2$ and I would like to prove the equivalence. A professor told me that he thinks that these definitions are not equivalent but, for my purpose, I only need one implication. Let's call $TP_2$ the first definition and $tp_2$ the second one.
Definition of $TP_2$:
A formula has the $TP_2$ if there are $(a^i_j)_{i,j < \omega}$ such that:
- for all $\sigma \in \omega^\omega$ $$ \{\phi(x,a^n_{\sigma(n)}) \ : \ n < \omega\} \text{ is consistent} $$
- for all $n < \omega$ and $i<j<\omega$ $$ \{\phi(x,a^n_i),\phi(x,a^n_j)\} \text{ is inconsistent}$$
A theory $T$ has $TP_2$ if some formula does.
Definition of $tp_2$:
$T$ has $tp_2$ if there are tuples $a,b$ and $(b_n^m)_{m,n<\omega}$ and $k < \omega$ such that:
- for all $m< \omega$ and $n_1<n_2<...<n_k < \omega$ there is no $a_*$ such that $$ a_*b_{n_i}^m \equiv ab \text{ for all } 1 \leq i \leq k$$
- for all $\sigma \in \omega^\omega$ there is $a_*$ such that $$ a_*b_{\sigma(m)}^m \equiv ab \text{ for all } m < \omega$$
At first look it seems like an easy compactness argument, but I'm really struggling with this.
I would like to know:
(i) Is $TP_2$ equivalent to $tp_2$ for a theory?
(ii) Does $tp_2$ imply $TP_2$ for a theory? How can I show it?
EDIT : I know that "my" definition of $TP_2$ (the first one) is an instance of the definition of $k$-$TP_2$ ($k < \omega$) for $k = 2$, obtained by replacing condition 2. with the following :
for all $n<\omega$ $$\{\phi(x,a^n_i) \ : \ i < \omega\} \text{ is $k$-inconsistent}$$
And I found out that a theory has $TP_2$ iff it has $k$-$TP_2$ for some $k \geq 2$. I think that this could be useful but I'm struggling anyway.
For $tp_2 \Rightarrow TP_2$ I've tried this way :
Consider $p(x,y) = tp(a,b)$. Now, for each $m < \omega$ and $n_1 < ... < n_k < \omega$ we have that $$ p(x,b_{n_1}^m) \cup ... \cup p(x,b_{n_k}^m) \text{ is inconsistent}$$
Using compactness we can find a formula $\phi(x,y)$ in this type such that $$\{\phi(x,b^m_{n_i}) \ : \ 1 \leq i \leq k\} \text{ is inconsistent}$$
But this doesn't imply (at least to me) that $\{\phi(x,b^m_n) \ | \ n < \omega\}$ is $k$-inconsistent and also the same formula should satisfy that for each $m < \omega$ in order to witness $TP_2$ (iff $k$-$TP_2$). How can i proceed?
Here's an example showing that $\text{tp}_2$ does not imply $\text{TP}_2$.
Let $L = \{\leq_m\mid m\in \omega\}$, and let $T$ be the theory of independent linear orders. This is the model companion of the theory which says that each $\leq_m$ is a linear order. In the reduct to each finite sublanguage of size $n$, $T$ is the theory of the Fraïssé limit of the class of finite structures equipped with $n$ linear orders.
$T$ is $\text{NIP}$, hence $\text{NTP}_2$, i.e. no formula has $\text{TP}_2$.
Proof: $T$ has quantifier elimination, since any reduct of $T$ to a finite sublanguage has quantifier elimination (being the theory of a Fraïssé limit). So every $L$-formula is equivalent modulo $T$ to a boolean combination of atomic formulas. Every atomic formula is $\text{NIP}$ (it's an instance of equality or one of the orders $\leq_m$), and $\text{NIP}$ formulas are closed under Boolean combinations. Thus every $L$-formula is $\text{NIP}$ modulo $T$.
$T$ has $\text{tp}_2$.
Proof: Pick $a$, $b$, and $c$ such that $b<_m a <_m c$ for all $m\in \omega$. Now we pick an array $(b^m_n,c^m_n)_{m,n\in \omega}$ carefully. For each $m\in \omega$, we define the row $(b^m_n,c^m_n)_{n\in \omega}$ so that $b^m_n <_k b$ and $c<_k c^m_n$ for all $n$ and all $k\neq m$, but $b<_m b^m_n <_m c^m_n <_m c$ for all $n$, and the $<_m$-intervals $(b^m_n,c^m_n)$ are pairwise disjoint.
Letting $p(x,b,c)$ be $\text{tp}(a/bc)$, we have $p(x,b^m_n,c^m_n)\cup p(x,b^m_{n'}c^m_{n'})$ is inconsistent for all $m\in \omega$ and $n\neq n'$, since this type says $b^m_n<_m x <_m c^m_n$ and $b^m_{n'}<_m x <_m c^m_{n'}$, but these intervals are disjoint.
But for all $\sigma\in \omega^\omega$, the type $\bigcup_{m\in \omega}p(x,b^m_{\sigma(m)},c^m_{\sigma(m)})$ is consistent. It suffices to see that what it says about each order is consistent. Fixing an order $<_k$, the type says $b^m_{\sigma(m)} <_k x <_k c^m_{\sigma(m)}$ for all $m\in \omega$. But for all $m\neq k$, we have $b^m_{\sigma(m)}<_k b <_k b^k_{\sigma(k)} <_k c^k_{\sigma(k)}<_k c <_k b^m_{\sigma(m)}$. So picking some $a_*$ in the $<_k$-interval $b^k_{\sigma(k)} <_k c^k_{\sigma(k)}$ for all $k$ works.
On the other hand, $\text{TP}_2$ implies $\text{tp}_2$.
Proof: Suppose $\varphi(x,y)$ has $\text{TP}_2$. By standard tricks, we may assume that the array $(b^i_j)_{i,j\in \omega}$ witnessing this has mutually indiscernible rows, and that the sequence of rows is indiscernible (see, for example, Lemma 7.4 here, where Casanovas calls this kind of array very indiscernible). Now we can stretch the indiscernible sequence of rows to a sequence $(b^\alpha_i)_{\alpha<\kappa,i\in \omega}$ of length $\kappa$. where $\kappa>|S_{xy}(T)|$, the number of complete types in variables $xy$ over the empty set.
By compactness, there is some $a_*$ such that $\varphi(a_*,b^\alpha_0)$ is true for all $\alpha < \kappa$. For each such $\alpha$, let $p_\alpha = \text{tp}(a_*b^\alpha_0)$. By pigeonhole, some type $p$ must appear infinitely many times. So we can refine our array back down to $(b^m_n)_{m,n\in \omega}$ and assume that $a_*b^m_0$ realizes $p$ for all $m\in \omega$. But now by mutual indiscernibility of the rows of the array, the existence of $a_*$ realizing $\bigcup_{m\in \omega} p(x,b^m_0)$ implies that for any $\sigma\in \omega^\omega$, the type $\bigcup_{m\in \omega} p(x,b^m_{\sigma(m)})$ is consistent. And on the other hand, for any fixed $m$ and any $n\neq n'$, the type $p(x,b^m_n)\cup p(x,b^m_{n'})$ is inconsistent, since this type contains both $\varphi(x,b^m_n)$ and $\varphi(x,b^m_{n'})$. So we have a witness to $\text{tp}_2$ (taking $a = a_0$, $b = b^0_0$, and $k = 2$).