TRANSITIVITY, LOWNESS, AND RANKS IN NSOP$_{1}$ THEORIES

Fact 2.11 [Reference Kim, Kim and Scow14, Theorem 4.3].

Let $I_{s}$ denote the $L_{s,\omega }$ -structure

$$ \begin{align*} I_{s} = (\omega^{<\omega}, \unlhd, <_{lex}, \wedge, (P_{\alpha})_{\alpha < \omega}) \end{align*} $$

with all symbols being given their intended interpretations and each $P_{\alpha }$ naming the elements of the tree at level $\alpha $ . Then $I_{s}$ -indexed indiscernibles have the modeling property.

Proof Let $\kappa $ be sufficiently large with respect to $|A|$ . By induction on $\gamma < \kappa $ , we will choose $(a_{\eta ,\gamma })_{\eta \in \mathcal {T}_{\alpha }}$ such that, taking $I_{\gamma }$ to be the sequence $\langle (a_{\eta ,i})_{\eta \in \mathcal {T}_{\alpha }} : i < \gamma \rangle $ , we have that $I_{\gamma }$ starts with $(a_{\eta })_{\eta \in \mathcal {T}_{\alpha }}$ , is mutually s-indiscernible over A, and satisfies

for all $i < \gamma $ . The sequence $I_{1}$ is already specified and trivially satisfies the requirements.

Assume we are given $(a_{\eta ,i})_{\eta \in \mathcal {T}_{\alpha }}$ for all $i < \gamma $ and set $I_{\gamma } = \langle (a_{\eta ,i})_{\eta \in \mathcal {T}_{\alpha }} : i < \gamma \rangle $ . Apply extension to get some $(b_{\eta })_{\eta \in \mathcal {T}_{\alpha }} \equiv _{A} (a_{\eta })_{\eta \in \mathcal {T}_{\alpha }}$ such that

By the modeling property, we can take $(b^{\prime }_{\eta })_{\eta \in \mathcal {T}_{\alpha }}$ to be locally based on $(b_{\eta })_{\eta \in \mathcal {T}_{\alpha }}$ and s-indiscernible over $AI_{\gamma }$ , then we we still have $(b^{\prime }_{\eta })_{\eta \in \mathcal {T}_{\alpha }} \equiv _{A} (a_{\eta })_{\eta \in \mathcal {T}_{\alpha }}$ , as $(a_{\eta })_{\eta \in \mathcal {T}_{\alpha }}$ was assumed to be s-indiscernible over A, and local basedness and strong finite character of non-forking imply

Now by induction on $i < \gamma $ , we will choose $(a^{\prime }_{\eta ,i})_{\eta \in \mathcal {T}_{\alpha }}$ satisfying the following conditions:

1. (1) $(a^{\prime }_{\eta ,i})_{\eta \in \mathcal {T}_{\alpha }}$ is s-indiscernible over

   $$ \begin{align*}A \cup \{a^{\prime}_{\eta,j} : \eta \in \mathcal{T}_{\alpha}, j < i\} \cup \{a_{\eta,k} : \eta \in \mathcal{T}_{\alpha}, k> i\} \cup \{b^{\prime}_{\eta} : \eta \in \mathcal{T}_{\alpha}\}. \end{align*} $$

2. (2) $(b^{\prime }_{\eta })_{\eta \in \mathcal {T}_{\alpha }}$ is s-indiscernible over

   $$ \begin{align*}A \cup \{a^{\prime}_{\eta,j} : \eta \in \mathcal{T}_{\alpha}, j \leq i\} \cup \{a_{\eta,k} : \eta \in \mathcal{T}_{\alpha}, k> i\}. \end{align*} $$

3. (3) $(a^{\prime }_{\eta ,j})_{\eta \in \mathcal {T}_{\alpha }, j \leq i} (a_{\eta ,k})_{\eta \in \mathcal {T}_{\alpha }, k> i} \equiv _{A} (a_{\eta ,j})_{\eta \in \mathcal {T}_{\alpha }, j \leq i} (a_{\eta ,k})_{\eta \in \mathcal {T}_{\alpha }, k > i}$ .

4. (4) The following independence holds:

   

Fix $i < \gamma $ and suppose we have chosen $(a^{\prime }_{\eta ,j})_{\eta \in \mathcal {T}_{\alpha }}$ for all $j < i$ . Pick $(a^{\prime }_{\eta ,i})_{\eta \in \mathcal {T}_{\alpha }}$ s-indiscernible over $A \cup \{a^{\prime }_{\eta ,j} : \eta \in \mathcal {T}_{\alpha }, j < i\} \cup \{a_{\eta ,k} : \eta \in \mathcal {T}_{\alpha }, k> i\} \cup \{b^{\prime }_{\eta } : \eta \in \mathcal {T}_{\alpha }\}$ and locally based on $(a_{\eta ,i})_{\eta \in \mathcal {T}_{\alpha }}$ . Then (1) is satisfied and (2) is easy to check using local basedness and the inductive assumption. We assumed $I_{\gamma }$ was mutually s-indiscernible over A and hence by (3) of the inductive hypothesis, we know that $(a_{\eta ,i})_{\eta \in \mathcal {T}_{\alpha }}$ is s-indiscernible over

$$ \begin{align*}A \cup \{a^{\prime}_{\eta,j} : \eta \in \mathcal{T}_{\alpha}, j < i\} \cup \{a_{\eta,k} : \eta \in \mathcal{T}_{\alpha}, k> i\}, \end{align*} $$

and therefore $(a^{\prime }_{\eta ,i})_{\eta \in \mathcal {T}_{\alpha }}$ has the same type over this set, which establishes (3). Finally, (4) follows by local basedness, (3), and the invariance of non-forking independence. More explicitly, suppose there is a finite tuple b from $(b^{\prime }_{\eta })_{\eta \in \mathcal {T}_{\alpha }}$ , a finite tuple a from $\{a^{\prime }_{\eta ,j} : j < i\} \cup \{a_{\eta ,k} : k> i\}$ , and a finite tuple $\overline {\eta }$ from $\mathcal {T}_{\alpha }$ such that

$$ \begin{align*}\models \varphi(b;a^{\prime}_{\overline{\eta},i},a), \end{align*} $$

where $\varphi (x;y,z) \in L(A)$ is a formula such that $\varphi (x;a^{\prime }_{\overline {\eta },i},a)$ forks over A. Local basedness entails that there is $\overline {\nu }$ with $\mathrm {qftp}_{L_{s,\alpha }}(\overline {\nu }) = \mathrm {qftp}_{L_{s,\alpha }}(\overline {\eta })$ such that

$$ \begin{align*}\models \varphi(b;a_{\overline{\nu},i},a). \end{align*} $$

But by mutual s-indiscernibility, $a_{\overline {\nu },i} \equiv _{Aa} a_{\overline {\eta },i}$ and, by (3), $a_{\overline {\eta },i} \equiv _{Aa} a^{\prime }_{\overline {\eta },i}$ and hence $\varphi (x;a_{\overline {\nu },i},a)$ forks over A as well. This contradicts the inductive hypothesis that

This shows that our choice of $(a^{\prime }_{\eta ,i})_{\eta \in \mathcal {T}_{\alpha }}$ satisfies the requirements.

Having constructed our sequence $I^{\prime }_{\gamma } = \langle (a^{\prime }_{\eta ,i})_{\eta \in \mathcal {T}_{\alpha }} : i < \gamma \rangle $ , we have $I^{\prime }_{\gamma } \equiv _{A} I_{\gamma }$ by (3) and $(b^{\prime }_{\eta })_{\eta \in \mathcal {T}_{\alpha }}$ is s-indiscernible over $I^{\prime }_{\gamma }$ by (2). Moreover, each $(a_{\eta ,i})_{\eta \in \mathcal {T}_{\alpha }}$ is indiscernible over $A(b^{\prime }_{\eta })_{\eta \in \mathcal {T}_{\alpha }}(a_{\eta ,j})_{\eta \in \mathcal {T}_{\alpha },j \neq i}$ by (1). Finally, by (4), we have . Choosing $(a_{\eta ,\gamma })$ such that

$$ \begin{align*} I_{\gamma} (a_{\eta,\gamma})_{\eta \in \mathcal{T}_{\alpha}} \equiv_{A} I^{\prime}_{\gamma} (b^{\prime}_{\eta})_{\eta \in \mathcal{T}_{\alpha}} \end{align*} $$

we arrive at $I_{\gamma +1}$ . There is nothing to do at limits, so we have succeeded in constructing our sequence $I_{\kappa }$ . Applying Erdős–Rado to $I_{\kappa }$ , then, we obtain the desired sequence I.

Proof Suppose $\overline {\eta }$ and $\overline {\nu }$ are tuples from $\mathcal {T}_{\alpha +1} \setminus \{\emptyset \}$ with $\mathrm {qftp}_{L_{s,\alpha +1}}(\overline {\eta }) = \mathrm {qftp}_{L_{s,\alpha +1}}(\overline {\nu })$ . After possibly reordering the tuples, there are $i_{0} < \cdots < i_{k-1}$ and $j_{0} < \cdots < j_{k-1}$ such that $\overline {\eta } = (\overline {\eta }_{0}, \ldots , \overline {\eta }_{k-1})$ and $\overline {\nu } = (\overline {\nu }_{0}, \ldots , \overline {\nu }_{k-1})$ where each $\overline {\eta }_{l}$ comes from the tree $\unrhd \langle i_{l} \rangle $ and $\overline {\nu }_{l}$ comes from the tree $\unrhd \langle j_{l} \rangle $ for $l < k$ . Then, in particular, $\mathrm {qftp}_{L_{s,\alpha +1}}(\overline {\eta }_{l}) = \mathrm {qftp}_{L_{s,\alpha +1}}(\overline {\nu }_{l})$ for all $l < k$ . Additionally, for all $l < k$ , let $\overline {\eta }^{\prime }_{l}$ be the element of the tree $\unrhd \langle j_{l} \rangle $ corresponding to $\overline {\eta }_{l}$ (i.e., replace each node $\langle i_{l}\rangle ^{\frown } \xi $ enumerated in $\overline {\eta }_{l}$ with $\langle j_{l} \rangle ^{\frown } \xi $ ). Because I is an A-indiscernible sequence, we have

$$ \begin{align*} (a_{\overline{\eta}_{0}}, \ldots, a_{\overline{\eta}_{k-1}}) \equiv_{A} (a_{\overline{\eta}^{\prime}_{0}}, \ldots, a_{\overline{\eta}_{k-1}'}). \end{align*} $$

Additionally, in the tree $\unrhd \langle j_{l} \rangle $ (naturally viewed as an $L_{s,\alpha }$ -structure), we have $\mathrm {qftp}_{L_{s,\alpha }}(\overline {\eta }_{l}') = \mathrm {qftp}_{L_{s,\alpha }}(\overline {\eta }_{l})$ for all $l < k$ . Thus, mutual s-indiscernibility entails

$$ \begin{align*} (a_{\overline{\eta}^{\prime}_{0}}, \ldots, a_{\overline{\eta}_{k-1}'}) \equiv_{A} (a_{\overline{\nu}_{0}}, \ldots, a_{\overline{\nu}_{k-1}}). \end{align*} $$

Thus we have shown that $a_{\overline {\eta }} \equiv _{A} a_{\overline {\nu }}$ . Therefore, it follows, by local basedness, that $I \equiv _{A} I'$ and the result follows.

Fact 2.19. Suppose T is NSOP $_{1}$ with existence and A is a set of parameters.

1. (1) If , there is an $Ab$ -indiscernible sequence $I = \langle a_{i} : i < \omega \rangle $ over A with $a_{0} = a$ such that I is weak tree Morley over A [Reference Dobrowolski, Kim and Ramsey5, Lemma 4.7].

2. (2) Kim’s lemma for weak tree Morley sequences: the formula $\varphi (x;a_{0})$ Kim-divides over A if and only if $\{\varphi (x;a_{i}) : i < \omega \}$ is inconsistent for some weak tree Morley sequence $\langle a_{i} : i < \omega \rangle $ over A if and only if $\{\varphi (x;a_{i}) : i < \omega \}$ is inconsistent for all weak tree Morley sequences $\langle a_{i} : i < \omega \rangle $ over A [Reference Dobrowolski, Kim and Ramsey5, Corollary 4.8].

3. (3) If $a \equiv ^{L}_{A} b$ and , there is an -Morley sequence over A starting with $(a,b)$ as its first two elements (follows from Fact 2.7 as in [Reference Kaplan and Ramsey7, Corollary 6.6]).

Fact 2.20 [Reference Dobrowolski, Kim and Ramsey5, Lemma 5.7].

Suppose T is NSOP $_{1}$ with existence. If A is a set of parameters, c is an arbitrary tuple, and , then there is $a' \equiv ^{L}_{Ab} a$ such that .

Proof (1) By compactness, we may assume y is finite. Fix $c \models p$ and define $\Gamma (x,y)$ by

$$ \begin{align*} \Gamma(x,y) = p(x) \cup \{\neg \varphi(y;x) : \varphi(y;c) \text{ Kim-divides over } A\}. \end{align*} $$

By symmetry, invariance, and Kim-forking = Kim-dividing, this partial type is as desired.

(2) One can take $\Delta $ to be the partial type that asserts $\langle x_{i} : i < \omega \rangle $ is A-indiscernible, every $x_{i} \models p$ , and (which is type-definable over A by (1)).

Proof By Fact 2.20, there is $c' \equiv ^{L}_{Ab} c$ such that . Let $\sigma \in \mathrm {Autf}(\mathbb {M}/Ab)$ be an automorphism such that $\sigma (c') = c$ and let $c_{0} = \sigma (c)$ . Then we have and $c_{0} \equiv _{Ab}^{L} c$ and hence, in particular, $c_{0}b \equiv _{A}^{L} cb$ . By symmetry and a second application of Fact 2.20 once again, we find $b"c" \equiv ^{L}_{Ac} bc_{0}$ with . Let $\tau \in \mathrm {Autf}(\mathbb {M}/Ac)$ be a strong automorphism with $\tau (b"c") = bc_{0}$ and define $b_{1} = \tau (b)$ . Then by construction, we have $b"c" \equiv _{A} bc_{0}$ and $bc_{0} \equiv ^{L}_{A} bc$ , it follows that $b"c" \equiv ^{L}_{A} bc$ , and hence $\tau (b"c") \equiv ^{L}_{A} \tau (bc)$ , which, after unraveling definitions, gives $bc_{0} \equiv ^{L}_{A} b_{1}c$ . Moreover, since , we obtain by invariance. Let $b_{0} = b$ and $c_{1} = c$ . By Fact 2.19(3), we can extend the sequence $\langle (b_{i},c_{i}) : i < 2 \rangle $ to a weak tree Morley sequence $I = \langle (b_{i},c_{i}) : i \in \mathbb {Z} \rangle $ over A.

Choose $a'$ such that $a_{1}c_{1} \equiv ^{L}_{A} a'c_{0}$ . Then we have $a_{0} \equiv ^{L}_{A} a'$ , as well as , by our assumptions. Additionally, since , $b_{0} = b$ and $c \equiv _{Ab} c_{0}$ , we have . Therefore, by Fact 2.7(6), there is $a_{*}$ with $a_{*} \equiv ^{L}_{Ab_{0}} a_{0}$ , $a_{*} \equiv ^{L}_{Ac_{0}} a'$ , with .

Because I is a weak tree Morley sequence and , by Kim’s lemma, compactness, and an automorphism, there is such that $a_{**}b_{0}c_{0} \equiv ^{L}_{A} a_{*}b_{0}c_{0}$ and such that I is $Aa_{**}$ -indiscernible. Note that, by construction, $a_{**} \equiv ^{L}_{Ab} a_{0}$ , $a_{**} \equiv ^{L}_{Ac} a_{1}$ , and .

Additionally, the sequence $\langle b_{i} : i \in \mathbb {Z}^{\leq 0} \rangle $ is a weak tree Morley sequence over A which is $Aa_{**}c$ -indiscernible and containing $b_{0} = b$ , hence , by Kim’s lemma for weak tree Morley sequences. Similarly, the sequence $\langle c_{i} : i \in \mathbb {Z}^{\geq 1} \rangle $ is a weak tree Morley sequence over A containing $c = c_{1}$ which is $Aa_{**}b$ -indiscernible, yielding . By symmetry, we conclude.

Proof By induction on $\alpha $ , we will construct trees $(a^{\alpha }_{\eta })_{\eta \in \mathcal {T}_{\alpha }}$ so that:

1. (1) $(a^{\alpha }_{\eta })_{\eta \in \mathcal {T}_{\alpha }}$ is s-indiscernible and weakly spread out over B.

2. (2) $a^{\alpha }_{\eta } \models \text {tp}(a/B)$ for all $\eta \in \mathcal {T}_{\alpha }$ .

3. (3) If $\alpha $ is a successor, .

4. (4) If $\alpha < \beta $ , then $a^{\beta }_{\iota _{\alpha \beta }(\eta )} = a^{\alpha }_{\eta }$ for all $\eta \in \mathcal {T}_{\alpha }$ .

For $\alpha = 0$ , we put $a^{0}_{\emptyset } = a$ , and for $\delta $ limit, we will define $(a^{\delta }_{\eta })_{\eta \in \mathcal {T}_{\delta }}$ by setting $a^{\delta }_{\iota _{\alpha \delta }(\eta )} = a^{\alpha }_{\eta }$ for all $\alpha < \delta $ and $\eta \in \mathcal {T}_{\alpha }$ which, by (4) and induction, is well-defined and satisfies the requirements.

Now suppose we are given $(a^{\beta }_{\eta })_{\eta \in \mathcal {T}_{\beta }}$ satisfying the requirements for all $\beta \leq \alpha $ . Let $\langle (a^{\alpha }_{\eta ,i})_{\eta \in \mathcal {T}_{\alpha }} : i < \omega \rangle $ be a mutually s-indiscernible Morley sequence over B with $a^{\alpha }_{\eta ,0} = a^{\alpha }_{\eta }$ for all $\eta \in \mathcal {T}_{\alpha }$ , which exists by Lemma 2.15. Apply extension to find $a_{*} \equiv _{B} a$ so that . Define a tree $(b_{\eta })_{\eta \in \mathcal {T}_{\alpha +1}}$ by setting $b_{\emptyset }= a_{*}$ and $b_{\langle i \rangle \frown \eta } = a^{\alpha }_{\eta ,i}$ for all $i < \omega $ and $\eta \in \mathcal {T}_{\alpha }$ . We may define $(a^{\alpha +1}_{\eta })_{\eta \in \mathcal {T}_{\alpha +1}}$ to be a tree which is s-indiscernible over B and locally based on $(b_{\eta })_{\eta \in \mathcal {T}_{\alpha +1}}$ over B. By an automorphism, we may assume $a^{\alpha +1}_{\iota _{\alpha \alpha +1}(\eta )} = a^{\alpha }_{\eta }$ for all $\eta \in \mathcal {T}_{\alpha }$ , hence conditions (2), and (4) are clearly satisfied. Moreover, by Lemma 2.16, we have $\langle (a_{\eta ,i})_{\eta \in \mathcal {T}_{\alpha }} : i < \omega \rangle \equiv _{B} \langle a^{\alpha +1}_{\unrhd \langle i \rangle } : i < \omega \rangle $ so $\langle a^{\alpha +1}_{\unrhd \langle i \rangle } : i < \omega \rangle $ is a Morley sequence over B. Then by (4) and induction, it follows that $(a^{\alpha +1}_{\eta })_{\eta \in \mathcal {T}_{\alpha +1}}$ is s-indiscernible and spread out over B, which shows (1).

For (3), we just note that, by symmetry, if , there is some formula $\varphi (x;a_{\emptyset }^{\alpha +1}) \in \text {tp}(Ba^{\alpha +1}_{\vartriangleright \emptyset }/Aa^{\alpha +1}_{\emptyset })$ that Kim-divides over A. As the tree $(a^{\alpha +1}_{\eta })_{\eta \in \mathcal {T}_{\alpha +1}}$ is locally based on $(b_{\eta })_{\eta \in \mathcal {T}_{\alpha +1}}$ , it follows that some tuple from $Bb_{\vartriangleright \emptyset }$ also realizes $\varphi (x;a_{*})$ and $a_{*} \equiv _{B} a^{\alpha +1}_{\emptyset }$ , so $\varphi (x;a_{*})$ Kim-divides over A as well, contradicting the choice of $a_{*}$ . This contradiction establishes (3), completing the induction.

By considering $(a^{\kappa }_{\eta })_{\eta \in \mathcal {T}_{\kappa }}$ for $\kappa $ sufficiently large, we may apply Erdős–Rado, as in [Reference Kaplan and Ramsey7, Lemma 5.10], to find the desired sequence.

Proof Define a partial type $\Gamma (x;b,a)$ over $Aab$ as follows:

$$ \begin{align*} \Gamma(x;b,a) = \text{tp}(c/Ab) \cup \{\neg \varphi(x,a;b) : \varphi(x,y;b) \in L(Ab) \text{ Kim-divides over }A\}. \end{align*} $$

Next we define a partial type $\Delta (x;b,a)$ as follows:

$$ \begin{align*} \Delta(x;b,a) = \Gamma(x;b,a) \cup \{\neg\psi(x;b,a) : \psi(x;b,a) \in L(Aab) \text{ Kim-divides over }Ab\}. \end{align*} $$

Proof of claim

Suppose not. Then because Kim-forking and Kim-dividing are the same in NSOP $_{1}$ with existence, there is some formula $\psi (x;b,a) \in L(Aab)$ such that

$$ \begin{align*} \Gamma(x;b,a) \vdash \psi(x;b,a) \end{align*} $$

and $\psi (x;b,a)$ Kim-divides over $Ab$ . As , we know by Lemma 3.1 that there is a sequence $\langle a_{i} : i < \omega \rangle $ with $a_{0} = a$ which is a weak tree Morley sequence over $Ab$ and satisfies for all $i < \omega $ . Then by Claim 1, $\bigcup _{i < \omega } \Gamma (x;b,a_{i})$ is consistent. However, we have

$$ \begin{align*} \bigcup_{i < \omega} \Gamma(x;b,a_{i}) \vdash \{\psi(x;b,a_{i}) : i < \omega\} \end{align*} $$

and $\{\psi (x;b,a_{i}) : i < \omega \}$ is inconsistent, because weak tree Morley sequences witness Kim-dividing. This contradiction proves the claim.

To conclude, we may take $c'$ to be any realization of $\Delta (x;b,a)$ .

Proof By induction on $\alpha $ , we will construct trees $(a^{\alpha }_{\eta })_{\eta \in \mathcal {T}_{\alpha }}$ satisfying the following:

1. (1) For all $\eta \in \mathcal {T}_{\alpha }$ , $a^{\alpha }_{\eta } \models \text {tp}(a/Ab)$ .

2. (2) The tree $(a^{\alpha }_{\eta })_{\eta \in \mathcal {T}_{\alpha }}$ is s-indiscernible over $Ab$ and weakly spread out over A.

3. (3) If $\alpha $ is a successor, then .

4. (4) .

5. (5) If $\alpha < \beta $ , then $a^{\beta }_{\iota _{\alpha \beta }(\eta )} = a^{\alpha }_{\eta }$ for all $\eta \in \mathcal {T}_{\alpha }$ .

Put $a^{0}_{\emptyset } = a$ and for $\delta $ limit, if we are given $(a^{\alpha }_{\eta })_{\eta \in \mathcal {T}_{\alpha }}$ for every $\alpha < \delta $ , we can define $(a^{\delta }_{\eta })_{\eta \in \mathcal {T}_{\delta }}$ by setting $a^{\delta }_{\iota _{\alpha \delta }(\eta )} = a^{\alpha }_{\eta }$ for all $\alpha < \delta $ and $\eta \in \mathcal {T}_{\alpha }$ , which is well-defined by (5) and is easily seen to satisfy the requirements.

Suppose now we are given $(a^{\alpha }_{\eta })_{\eta \in \mathcal {T}_{\alpha }}$ . Let $J = \langle (a^{\alpha }_{\eta ,i})_{\eta \in \mathcal {T}_{\alpha }} : i < \omega \rangle $ be a mutually s-indiscernible Morley sequence over A with $(a^{\alpha }_{\eta ,0})_{\eta \in \mathcal {T}_{\alpha }} = (a^{\alpha }_{\eta })_{\eta \in \mathcal {T}_{\alpha }}$ , which exists by Lemma 2.15. By (4), symmetry, and the chain condition, Fact 2.7(5) we may assume J is $Ab$ -indiscernible and . By Lemma 3.2, there is $a_{*} \equiv _{Ab} a$ such that and . After defining a tree $(c_{\eta })_{\eta \in \mathcal {T}_{\alpha +1}}$ by $c_{\emptyset } = a_{*}$ and $c_{\langle i \rangle \frown \eta } = a^{\alpha }_{\eta ,i}$ for all $\eta \in \mathcal {T}_{\alpha }$ , these conditions on $a_{*}$ imply that $(c_{\eta })_{\eta \in \mathcal {T}_{\alpha }}$ satisfy (3) and (4), respectively. Let $(a^{\alpha +1}_{\eta })_{\eta \in \mathcal {T}_{\alpha +1}}$ be any tree s-indiscernible over $Ab$ locally based on $(c_{\eta })_{\eta \in \mathcal {T}_{\alpha }}$ over $Ab$ . This still satisfies (2) by Lemma 2.16. Moreover, as $c_{\unrhd \langle i \rangle } \models \mathrm {tp}((a^{\alpha }_{\eta })_{\eta \in \mathcal {T}_{\alpha }}/Ab)$ for all $i < \omega $ , it follows from local basedness that $a^{\alpha +1}_{\unrhd \langle i \rangle } \models \mathrm {tp}((a^{\alpha }_{\eta })_{\eta \in \mathcal {T}_{\alpha }}/Ab)$ for all $i < \omega $ as well. Hence, by an automorphism over $Ab$ , we may assume $a^{\alpha +1}_{0 \frown \eta } = a^{\alpha }_{\eta }$ for all $\eta \in \mathcal {T}_{\alpha }$ , which ensures the constructed tree satisfies (5), and (1)–(4) are easy to verify.

Given $(a^{\kappa }_{\eta })_{\eta \in \mathcal {T}_{\kappa }}$ for $\kappa $ sufficiently large, we may, by Erdős–Rado (see, e.g., [Reference Kaplan and Ramsey7, Lemma 5.10]), obtain a weak Morley tree $(b_{\eta })_{\eta \in \mathcal {T}_{\omega }}$ over A satisfying (1)–(4). Then the sequence $I = \langle a_{i} : i < \omega \rangle $ defined by $a_{i} = b_{\zeta _{i}}$ for all $i < \omega $ is a weak tree Morley sequence over A, as it is a path in a weak Morley tree over A, but by (3), we have for all i, so I is -Morley over $Ab$ as well.

Proof By Proposition 3.3 and the assumption that , there is a sequence $I = \langle a_{i} : i < \omega \rangle $ with $a_{0} = a$ such that I is an -Morley sequence over B and a weak tree Morley sequence over A. As , by symmetry, and I is -Morley over B, there is $I' \equiv _{Ba}I$ such that $I'$ is $Bc$ -indiscernible. Because $I'$ is also a weak tree Morley sequence over A, it follows by Kim’s lemma that . By symmetry, we conclude.

Proof (1) $\Rightarrow $ (2) Since , there is a Morley sequence $I=\langle a_i : i<\omega \rangle $ over A with $a_0=a$ such that I is $Ab$ -indiscernible. By [Reference Dobrowolski, Kim and Ramsey5, Lemma 2.17], there is a model N containing A such that and I is a coheir sequence over N. By compactness and extension we can clearly assume the length of I is arbitrarily large, and . Hence by the pigeonhole principle, there is an infinite subsequence J of I such that all the tuples in J have the same type over $Nb$ . Thus, for $a'\in J$ , we have and . Hence $M=f(N)$ is a desired model, where f is an $Ab$ -automorphism sending $a'$ to a.

(2) $\Rightarrow $ (3) Clear.

(3) $\Rightarrow $ (1) follows from transitivity and symmetry of .

Proof (1) $\implies $ (2) It suffices to show that, given any increasing continuous sequence $\langle A_{i} : i < \kappa ^{+} \rangle $ of parameter sets and tuple d such that $|A_{i}| \leq \kappa $ and for all $i < \kappa ^{+}$ , there is a continuous increasing sequence of models $\langle M_{i} : i < \kappa ^{+} \rangle $ and a finite tuple $d'$ such that $|M_{i}| \leq \kappa $ and for all $i < \kappa ^{+}$ . This follows from Fact 2.8, since the existence of such a sequence of models implies T has SOP $_{1}$ . Moreover, after naming constants, we may assume $\kappa = |T|$ .

So suppose we are given $\langle A_{i} : i < |T|^{+} \rangle $ , an increasing continuous sequence of sets of parameters with $|A_{i}| \leq |T|$ for all $i < |T|^{+}$ . Let $A = \bigcup _{i < |T|^{+}} A_{i}$ , and suppose further that we are given some tuple d such that for all $i < |T|^{+}$ . By induction on $i < |T|^{+}$ we will build increasing and continuous sequences $\langle A^{\prime }_{i} : i < |T|^{+} \rangle $ and $\langle M_{i} : i < |T|^{+} \rangle $ satisfying the following for all $i < |T|^{+}$ :

1. (1) $A^{\prime }_{0} = A_{0}$ and $A^{\prime }_{\leq i} \equiv A_{\leq i}$ .

2. (2) $M_{i}\models T$ with $|M_{i}| = |T|$ and $A^{\prime }_{i} \subseteq M_{i}$ .

3. (3) .

To begin, we define $A^{\prime }_{0} = A_{0}$ and take $M_{0}$ be any model containing $A^{\prime }_{0}$ of size $|T|$ . Given $A^{\prime }_{\leq i}$ and $M_{\leq i}$ satisfying the requirements, we pick $A^{\prime \prime }_{i+1}$ such that $A^{\prime }_{\leq i} A^{\prime \prime }_{i+1} \equiv A_{\leq i+1}$ . Then we apply extension, to obtain $A^{\prime }_{i+1} \equiv _{A^{\prime }_{\leq i}} A^{\prime \prime }_{i+1}$ such that . Note that $A^{\prime }_{\leq i+1} \equiv A_{\leq i+1}$ . We define $M_{i+1}$ to be any model containing $A^{\prime }_{i+1}M_{i}$ of size $|T|$ . This satisfies the requirements.

At limit $\delta $ , we define $A^{\prime }_{\delta } = \bigcup _{i < \delta } A^{\prime }_{i}$ and $M_{\delta } = \bigcup _{i < \delta } M_{i}$ . This clearly satisfies (1) and (2) and (3) is trivial. Therefore this completes the construction.

Let $M = \bigcup _{i < |T|^{+}} M_{i}$ . Choose $d'$ such that $d\langle A_{i} : i < |T|^{+} \rangle \equiv d' \langle A^{\prime }_{i} : i < |T|^{+} \rangle $ , which is possible by (1). Then we have for all $i < |T|^{+}$ .

Towards contradiction, suppose that there is some $i < |T|^{+}$ with the property that . Then, in particular, we have . Additionally, because , we know, by symmetry and transitivity, that . By symmetry once more, we get , a contradiction. This shows that for all $i < |T|^{+}$ , completing the proof of this direction.

(2) $\implies $ (3) Suppose (3) fails, i.e., we are given A of size $\kappa ^{+}$ , $p \in S(A)$ , and an increasing continuous sequence of sets $\langle A_{i} : i < \omega \rangle $ such that $|A_{i}| \leq \kappa $ and p Kim-divides over $A_{i}$ for all $i < \kappa ^{+}$ . We will define an increasing continuous sequence of ordinals $\langle \alpha _{i} : i < \kappa ^{+} \rangle $ such that $\alpha _{i} \in \kappa ^{+}$ and $p \upharpoonright A_{\alpha _{i+1}}$ Kim-divides over $A_{\alpha _{i}}$ for all $i < \kappa ^{+}$ . We set $\alpha _{0} = 0$ and given $\langle \alpha _{j} : j \leq i \rangle $ , we know that there is some formula $\varphi (x;a_{i+1}) \in p$ that Kim-divides over $A_{\alpha _{i}}$ , by our assumption on p. Let $\alpha _{i+1}$ be the least ordinal $< \kappa ^{+}$ such that $a_{i+1}$ is contained in $A_{\alpha _{i+1}}$ . For limit i, if we are given $\langle \alpha _{j} : j < i \rangle $ , we put $\alpha _{i} = \sup _{j < i} \alpha _{j}$ . Then we define $\langle A_{i} : i < \kappa ^{+} \rangle $ by $A^{\prime }_{i} = A_{\alpha _{i}}$ for all $i < \kappa $ , and let $d \models p$ be any realization. By construction, we have for all $i < \kappa ^{+}$ , which witnesses the failure of (2).

(3) $\implies $ (1) This was established in [Reference Kaplan, Ramsey and Shelah9, Theorem 3.9].

Proof Suppose towards contradiction that $\varphi (x;a_{0})$ Kim-divides over A and $I = \langle a_{i} : i < \omega \rangle $ is an -Morley sequence over A such that $\{\varphi (x;a_{i}) : i < \omega \}$ is consistent. By naming A as constants, we may assume $|A| \leq |T|$ . We may stretch I such that $I = \langle a_{i} : i < |T|^{+} \rangle $ . Define $A_{i} = Aa_{<i}$ . Then $\langle A_{i} : i < |T|^{+} \rangle $ is increasing and continuous and $|A_{i}| \leq |T|$ . Let $d \models \{\varphi (x;a_{i}) : i < |T|^{+} \}$ . We claim for all $i < |T|^{+}$ . If not, then for some $i < |T|^{+}$ , we have , or, in other words . Since I is an -Morley sequence, we also have , hence , by transitivity (Theorem 3.4). This entails, in particular, that , which is a contradiction, since $\varphi (x;a_{i})$ Kim-divides over A. This completes the proof.