Proofs:NOP = NIP + NsOP

Fix some complete theory $$T$$ with monster model $$\mathbb{U}$$.

A formula $$\phi(x;y)$$ has the order property if there exists $$a_1, b_1, a_2, b_2, \ldots$$ in $$\mathbb{U}$$ such that $$i < j \iff \models \phi(a_i;b_j)$$ for every $$i, j$$. $$T$$ is said to be NOP (or stable) if no formula has the order property.

A formula $$\phi(x;y)$$ has the strict order property if there exist $$b_1, b_2, \ldots$$ such that $$\phi(\mathbb{U};b_1) \subsetneq \phi(\mathbb{U};b_2) \subsetneq \phi(\mathbb{U};b_3) \subsetneq \cdots$$ Taking $$a_i$$ to be in $$\phi(\mathbb{U};b_{i+1}) \setminus \phi(\mathbb{U};b_i)$$, one sees that any formula having the strict order property has the order property. $$T$$ is said to be NsOP if no formula has the strict order property. Clearly NOP implies NsOP.

'''Remark. ''' An equivalent condition to NsOP is that $$T$$ has no interpretable partial order with an infinite chain.

Proof. If $$(P,\le)$$ is some interpretable poset, then $$P$$ is the quotient of some definable pre-order $$(P',\le)$$. The existence of an infinite chain implies that for each $$n$$, we can find $$b_1, \ldots, b_n$$ such that $$b_1 \le \cdots \le b_n$$ and $$b_n \not \le b_{n-1} \not \le \cdots \not \le b_1$$. By compactness we can find an infinite sequence $$b_1 \le b_2 \le \cdots$$ with $$b_i \not \ge b_{i+1}$$ for each $$i$$. Let $$\phi(x;y)$$ assert that $$x, y \in P'$$ and $$x \le y$$. Then $$\phi(\mathbb{U};b_1) \subsetneq \phi(\mathbb{U};b_2) \subsetneq \cdots$$, so $$T$$ has the strict order property.

Conversely, suppose that $$T$$ has the strict order property. Then there is a formula $$\phi(x;y)$$ and $$b_1, b_2, \ldots$$ with $$\phi(\mathbb{U};b_i) \subsetneq \phi(\mathbb{U};b_{i+1})$$. Let $$P'$$ be the sort of $$y$$ and let $$\le$$ be the pre-order on $$P'$$ given by $$b_1 \le b_2 \iff \phi(\mathbb{U};b_1) \le \phi(\mathbb{U};b_2)$$. If $$(P,\le)$$ is the quotient partial order, then $$P$$ has an infinite chain. QED

A formula $$\phi(x;y)$$ has the independence property if there exist $$a_i$$ for $$i \in \mathbb{N}$$ and $$b_S$$ for $$S \in \mathbb{N}$$ such that $$i \in S \iff \models \phi(a_i;b_S)$$ for every $$i$$, $$S$$. If $$\phi(x;y)$$ has the independence property, witnessed by $$a_i$$ and $$b_S$$, then $$\models \phi(a_i;b_{1,\ldots,j-1}) \iff i < j$$ so $$\phi$$ has the order property. $$T$$ is said to be NIP (or dependent) if no formula has the independence property. Clearly NOP implies NIP.

'''Theorem. ''' $$T$$ is NOP (stable) if and only if $$T$$ is both NIP and NsOP.

Proof. We have already noted that NOP implies NIP and NsOP. Conversely, suppose $$T$$ is NIP and NsOP.

'''Lemma. ''' Suppose $$T$$ is NsOP. Let $$b_1, b_2, \ldots$$ be a $$C$$-indiscernible sequence. Let $$\phi(x;y)$$ be a formula. Suppose there is $$a$$ such that $$\models \phi(a;b_2) \wedge \neg \phi(a;b_1)$$ Then there is some $$a' \equiv_C a$$ such that $$\models \phi(a';b_1) \wedge \neg \phi(a';b_2).$$

Proof. Suppose not. Then $$\operatorname{tp}(a/C)$$ is inconsistent with $$\phi(x;b_1) \wedge \neg \phi(x;b_2)$$. By compactness, some finite subtype of $$\operatorname{tp}(a/C)$$ is inconsistent with that formula. Therefore there is some $$c \in C$$ and formula $$\psi(x;z)$$ such that $$\psi(a;c)$$ holds but $$\psi(x;c) \wedge \phi(x;b_1) \wedge \neg \phi(x;b_2)$$ is inconsistent. Let $$\phi'(x;y,z)$$ be the formula $$\phi(x;y) \wedge \psi(x;z)$$. Then $$\phi'(x;b_1,c) \wedge \neg \phi'(x;b_2,c)$$ is inconsistent. On the other hand, $$\phi'(x;b_2,c) \wedge \neg \phi'(x;b_1,c)$$ is consistent, being satisfied by $$a$$.

This means that $$\phi'(\mathbb{U};b_1,c) \subsetneq \phi'(\mathbb{U};b_2,c).$$ Since $$b_1, b_2, \ldots$$ is $$c$$-indiscernible, $$\phi'(\mathbb{U};b_i,c) \subsetneq \phi'(\mathbb{U};b_{i+1},c)$$ for each $$i$$. So $$T$$ has the strict order property, a contradiction. QED

'''Lemma. ''' Suppose $$T$$ is NsOP. Let $$\{b_i\}_{i \in \mathbb{Q}}$$ be an indiscernible sequence. Let $$\phi(x;y)$$ be a formula. Suppose that $$S$$ and $$S'$$ are subsets of $$\{1,\ldots,n\}$$ having the same cardinality. If there is an $$a$$ such that $$S = \{i \in \{1,\ldots,n\} : \models \phi(a;b_i)\}$$ then there is $$a'$$ such that $$S' = \{i \in \{1,\ldots,n\} : \models \phi(a';b_i)\}$$

Proof. It suffices to consider the case where the symmetric difference of $$S$$ and $$S'$$ is of the form $$\{j,j+1\}$$, since we can get between any two subsets of $$\{1,\ldots,n\}$$ of the same size via such steps. So $$j$$ is in one of $$S$$ and $$S'$$, and $$j+1$$ is in the other.

Replacing $$\phi$$ with $$\neg \phi$$, we may assume that $$S \setminus S' = \{j+1\}$$ and $$S' \setminus S = \{j\}$$. Let $$C$$ be $$\{b_1,\ldots,b_{j-1},b_{j+2},\ldots,b_n\}$$. Note that the sequence $$b_j, b_{j+1}, b_{j+2-1/2}, b_{j+2-1/3}, b_{j+2-1/4}, b_{j+2 - 1/5}, \ldots$$ is $$C$$-indiscernible. Also, $$\phi(a;b_{j+1})$$ holds and $$\phi(a;b_j)$$ does not, because $$j+1 \in S$$ and $$j \notin S$$. By the previous lemma, we can find $$a' \equiv_C a$$ such that $$\phi(a';b_j)$$ holds and $$\phi(a';b_{j+1})$$ does not hold. If $$i \in \{1,\ldots,n\}$$ is not one of $$j$$ or $$j+1$$, then $$b_i \in C$$, so $$\phi(a;b_i)$$ holds if and only if $$\phi(a';b_i)$$ holds, because $$a' \equiv_C a$$. Therefore, $$\{i \in \{1,\ldots,n\} : \models \phi(a';b_i)\} = S \setminus \{j+1\} \cup \{j\} = S'.$$ QED

Now suppose for the sake of contradiction that $$T$$ has the order property. Then there is $$a_1, b_1, \ldots$$ such that $$\models \phi(a_i;b_j)$$ if and only if $$i < j$$. Extracting an indiscernible sequence of length $$\mathbb{Q}$$ from the sequence $$a_1b_1, a_2b_2, \ldots$$, we obtain an indiscernible sequence $$\{(a_i,b_i)\}_{i \in \mathbb{Q}}$$ such that $$\models \phi(a_i;b_j)$$ holds if and only if $$i < j$$. The $$b_i$$ form an indiscernible sequence. For each $$n$$ and each $$0 \le k \le n$$, we can find an $$a$$ such that $$|\{i \in \{1, \ldots, n\} : \models \phi(a;b_i)\}| = k$$ namely $$a = a_{k + 0.5}$$. By the previous lemma, it follows that if $$S$$ is any subset of $$\{1,\ldots,n\}$$, then we can find an $$a$$ such that $$\{i \in \{1, \ldots, n\} : \models \phi(a;b_i)\} = S.$$ Now if $$S$$ is any subset of $$\mathbb{N}$$, then by compactness we can find an $$a_S$$ such that $$\{i \in \mathbb{N} : \models \phi(a_S;b_i)\} = S.$$ Letting $$\phi^\vee(y;x) = \phi(x;y)$$, the formula $$\phi^\vee$$ has the independence property, a contradiction. QED