Some remarks on one-basedness

TL;DR

The paper discusses properties of one-based types within simple theories, establishing that types analysable in one-based types are also one-based, thus contributing to the understanding of their structural behavior.

Contribution

It proves that a type analysable in one-based types in a simple theory is itself one-based, clarifying the relationship between analysability and one-basedness.

Findings

Types analysable in one-based types are themselves one-based.

The result applies within simple theories.

Enhances understanding of type structures in model theory.

Abstract

A type analysable in one-based types in a simple theory is itself one-based.

Equations

$$X\subseteq \Sigma \mathrm{cl}(\tilde{a}A)\cap \Sigma \mathrm{cl}(C)\subseteq \Sigma \mathrm{cl}({\bar{a}}_{0}B).$$

$$\mathrm{Cb}({\bar{a}}_0/\Sigma \mathrm{cl}(C))\subseteq Y\subseteq \Sigma \mathrm{cl}(XB)\subseteq \Sigma \mathrm{cl}({\bar{a}}_{0}B).\text{qed}$$

$$\mathrm{Cb}(\bar{a}/\Sigma \mathrm{cl}(A))\subseteq \Sigma \mathrm{cl}(\bar{a})\cap \mathrm{Cb}(\bar{a}\bar{b}/\Sigma \mathrm{cl}(A))=\mathrm{dcl}(\varnothing ),$$

$$\mathrm{Cb}(\bar{a}{b}_1/\Sigma \mathrm{cl}(A))$$

$$\subseteq \Sigma \mathrm{cl}(\bar{a}{b}_1)\subseteq \Sigma \mathrm{cl}(\bar{a}\bar{b}),$$

$$\mathrm{Cb}(\bar{a}\bar{b}/\Sigma \mathrm{cl}(A))$$

$$\subseteq \Sigma \mathrm{cl}(\bar{a}\bar{b})\cap \mathrm{Cb}(\bar{a}\bar{b}/\Sigma \mathrm{cl}(A))=\mathrm{dcl}(\varnothing ),$$

$$\mathrm{Cb}(\bigcup _{i<\beta }{a}_i/\Sigma \mathrm{cl}(A))=\bigcup _{i<\beta }\mathrm{Cb}(\bigcup _{j\le i}{a}_i/\Sigma \mathrm{cl}(A))\subseteq \Sigma \mathrm{cl}(\bigcup _{i<\beta }{a}_i).\text{qed}$$

$$\mathrm{Cb}(\bar{a}/\Sigma \mathrm{cl}(B))=\mathrm{Cb}(\bar{a}/\Sigma \mathrm{cl}(\bar{A}B))\subseteq \Sigma \mathrm{cl}(\bar{a}\bar{A})\cap \Sigma \mathrm{cl}(B)=\Sigma \mathrm{cl}(\bar{a})\cap \Sigma \mathrm{cl}(B),$$

$$C=\mathrm{Cb}(ab/\Sigma \mathrm{cl}(a_i{b}_i:i<\omega )).$$

$$C=C^{\prime }\cap C\subseteq \Sigma \mathrm{cl}(ab)\cap C=\mathrm{dcl}(\varnothing ),$$

Full text

I should like to thank Zoé Chatzidakis for fruitful discussions

1. Introduction

Recall that a type $p$ over a set $A$ in a simple theory is one-based if for any tuple $\bar{a}$ of realizations of $p$ and any $B\supseteq A$ the canonical base $\mathrm{Cb}(\bar{a}/B)$ is contained in $\mathrm{bdd}(\bar{a}A)$. One-basedness implies that the forking geometry is particularly well-behaved; for instance one-based groups are bounded-by-abelian-by-bounded. Ehud Hrushovski showed in [3, Proposition 3.4.1] that for stable stably embedded types one-basedness is preserved under analyses: If $p$ is stable stably embedded in a supersimple theory, and analysable (in the technical sense defined in the next section) in one-based types, then $p$ is itself one-based. Zoé Chatzidakis then gave another proof for supersimple structures [1, Theorem 3.10], using semi-regular analyses. We shall give an easy direct proof of the theorem stated in the abstract, thus removing the hypotheses of stability, stable embedding, or supersimplicity; it is similar to Hrushovski’s proof, but does not use germs of definable functions (which work less well in simple unstable theories), and has to deal with non-stationarity of types. While we are at it, we shall also generalize the notion of bounded closure and one-basedness to $\Sigma$-closure and $\Sigma$-basedness, where $\Sigma$ is an $\emptyset$-invariant collection of partial types (thought of as small). This may for instance be applied to consider one-basedness modulo types of finite $SU$-rank, or modulo superstable types.

Our notation is standard and follows [5]. Throughout the paper, the ambient theory will be simple, and we shall be working in $\mathfrak{M}^{heq}$, where $\mathfrak{M}$ is a suffiviently saturated model of the ambient theory. Thus tuples are tuples of hyperimaginaries, and $\mathrm{dcl}=\mathrm{dcl}^{heq}$.

2. $\Sigma$-closure

In this section $\Sigma$ will be an $\emptyset$-invariant family of partial types. We first recall the notions of internality and analysability.

Definition 1**.**

Let $\pi$ be a partial type over $A$. Then $\pi$ is

• •

(almost, resp.) internal in $\Sigma$, or (almost, resp.) $\Sigma$-internal, if for every realization $a$ of $\pi$ there is B\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}_{A}a and $\bar{b}$ realizing types in $\Sigma$ based on $B$, such that $a\in\mathrm{dcl}(B\bar{b})$ (or $a\in\mathrm{bdd}(B\bar{b})$, respectively).

• •

analysable in $\Sigma$, or $\Sigma$-analysable, if for any $a\models\pi$ there are $(a_{i}:i<\alpha)\in\mathrm{dcl}(A,a)$ such that $\mathrm{tp}(a_{i}/A,a_{j}:j<i)$ is $\Sigma$-internal for all $i<\alpha$, and $a\in\mathrm{bdd}(A,a_{i}:i<\alpha)$.

A type $\mathrm{tp}(a/A)$ is foreign to $\Sigma$ if a\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}_{AB}\bar{b} for all B\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}_{A}a and $\bar{b}$ realizing types in $\Sigma$ over $B$.

Definition 2**.**

The $\Sigma$-closure $\Sigma\mathrm{cl}(A)$ of a set $A$ is the collection of all hyperimaginaries $a$ such that $\mathrm{tp}(a/A)$ is $\Sigma$-analysable.

We think of $\Sigma$ as a family of small types. For instance, if $\Sigma$ is the family of all bounded types, then $\Sigma\mathrm{cl}(A)=\mathrm{bdd}(A)$. Other possible choices might be the family of all types of $SU$-rank $<\omega^{\alpha}$, for some ordinal $\alpha$, or the family of all superstable types. If $P$ is an $\emptyset$-invariant family of types, and $\Sigma$ is the family of all $P$-analysable types to which all types in $P$ are foreign, then $\Sigma\mathrm{cl}(A)=\mathrm{cl}_{P}(A)$ as defined in [5, Definition 3.5.1]; if $P$ consists of a single regular type $p$, this in turn is the $p$-closure from [2] (see also [4, p. 265]).

Remark 1**.**

In general $\mathrm{bdd}(A)\subseteq\Sigma\mathrm{cl}(A)$; if the inequality is strict, then $\Sigma\mathrm{cl}(A)$ has the same cardinality as the ambient monster model, and hence violates the usual conventions. However, this is usually harmless. Note that $\Sigma\mathrm{cl}(.)$ is a closure operator.

Fact 2**.**

The following are equivalent:

1. (1)

$\mathrm{tp}(a/A)$* is foreign to $\Sigma$.* 2. (2)

a\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}_{A}\Sigma\mathrm{cl}(A). 3. (3)

a\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}_{A}\mathrm{dcl}(aA)\cap\Sigma\mathrm{cl}(A). 4. (4)

$\mathrm{dcl}(aA)\cap\Sigma\mathrm{cl}(A)\subseteq\mathrm{bdd}(A)$.

*Proof: *This follows immediately from [5, Proposition 3.4.12]; see also [5, Lemma 3.5.3].∎

$\Sigma$-closure is well-behaved with respect to independence.

Lemma 3**.**

Suppose A\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}_{B}C. Then \Sigma\mathrm{cl}(A)\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}_{\Sigma\mathrm{cl}(B)}\Sigma\mathrm{cl}(C). More precisely, for any $A_{0}\subseteq\Sigma\mathrm{cl}(A)$ we have A_{0}\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}_{B_{0}}\Sigma\mathrm{cl}(C), where $B_{0}=\mathrm{dcl}(A_{0}B)\cap\Sigma\mathrm{cl}(B)$. In particular, $\Sigma\mathrm{cl}(AB)\cap\Sigma\mathrm{cl}(BC)=\Sigma\mathrm{cl}(B)$.

*Proof: *Let $B_{1}=\Sigma\mathrm{cl}(B)\cap\mathrm{dcl}(BC)$. Then C\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}_{B}A implies C\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}_{B_{1}}A, and $\mathrm{tp}(C/B_{1})$ is foreign to $\Sigma$ by Fact 2 $(3\Rightarrow 1)$. Hence C\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}_{B_{1}}\Sigma\mathrm{cl}(A), and C\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}_{B_{1}}A_{0}.

Since $\mathrm{tp}(A_{0}/B_{0})$ is foreign to $\Sigma$ by Fact 2, we obtain A_{0}\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}_{B_{0}}\Sigma\mathrm{cl}(B_{0}). But $\Sigma\mathrm{cl}(B_{0})=\Sigma\mathrm{cl}(B)\supseteq B_{1}$, whence A_{0}\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}_{B_{0}}C by transitivity, and finally A_{0}\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}_{B_{0}}\Sigma\mathrm{cl}(C) by foreignness to $\Sigma$ again.∎

3. $\Sigma$-basedness

Again, $\Sigma$ will be an $\emptyset$-invariant family of partial types.

Definition 3**.**

A type $p$ over $A$ is $\Sigma$-based if $\mathrm{Cb}(\bar{a}/\Sigma\mathrm{cl}(B))\subseteq\Sigma\mathrm{cl}(\bar{a}A)$ for any tuple $\bar{a}$ of realizations of $p$ and any $B\supseteq A$.

Remark 4**.**

Equivalently, $p\in S(A)$ is $\Sigma$-based if \bar{a}\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}_{\Sigma\mathrm{cl}(\bar{a}A)\cap\Sigma\mathrm{cl}(B)}\Sigma\mathrm{cl}(B) for any tuple $\bar{a}$ of realisations of $p$ and any $B\supseteq A$.

Lemma 5**.**

Suppose $\mathrm{tp}(a/A)$ is $\Sigma$-based, $A\subseteq B$, and $a_{0}\in\Sigma\mathrm{cl}(\bar{a}B)$, where $\bar{a}$ is a tuple of realizations of $\mathrm{tp}(a/A)$. Then $\mathrm{tp}(a_{0}/B)$ is $\Sigma$-based.

*Proof: *Let $\bar{a}_{0}$ be a tuple of realizations of $\mathrm{tp}(a_{0}/B)$, and $C\supseteq B$. There is a tuple $\tilde{a}$ of realizations of $\mathrm{tp}(a/A)$ such that $\bar{a}_{0}\in\Sigma\mathrm{cl}(\tilde{a}B)$; we may choose it such that \tilde{a}\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}_{\bar{a}_{0}B}C. Then $\Sigma\mathrm{cl}(\tilde{a}B)\cap\Sigma\mathrm{cl}(C)\subseteq\Sigma\mathrm{cl}(\bar{a}_{0}B)$ by Lemma 3.

Put $X=\mathrm{Cb}(\tilde{a}/\Sigma\mathrm{cl}(C))$. By $\Sigma$-basedness of $\mathrm{tp}(a/A)$ we have

[TABLE]

As \tilde{a}\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}_{X}\Sigma\mathrm{cl}(C) we get \tilde{a}B\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}_{XB}\Sigma\mathrm{cl}(C), and hence \bar{a}_{0}\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}_{Y}\Sigma\mathrm{cl}(C) by Lemma 3, where $Y=\Sigma\mathrm{cl}(XB)\cap\mathrm{dcl}(\bar{a}_{0}XB)$. As $Y\subseteq\Sigma\mathrm{cl}(C)$, we have

[TABLE]

Lemma 6**.**

If $\mathrm{tp}(a)$ and $\mathrm{tp}(b)$ are $\Sigma$-based, so is $\mathrm{tp}(ab)$.

*Proof: *Let $\bar{a}$ and $\bar{b}$ be tuples of realizations of $\mathrm{tp}(a)$ and $\mathrm{tp}(b)$, respectively, and consider a set $A$ of parameters. We add $\Sigma\mathrm{cl}(\bar{a}\bar{b})\cap\mathrm{Cb}(\bar{a}\bar{b}/\Sigma\mathrm{cl}(A))$ to the language. By $\Sigma$-basedness of $\mathrm{tp}(a)$ we get

[TABLE]

whence \bar{a}\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}\Sigma\mathrm{cl}(A); similarly \bar{b}\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}\Sigma\mathrm{cl}(A).

Put $b_{1}=\mathrm{Cb}(\bar{b}/\Sigma\mathrm{cl}(\bar{a}A))$, and choose $\bar{a}^{\prime}A^{\prime}\models\mathrm{tp}(\bar{a}A/b_{1})$ with \bar{a}^{\prime}A^{\prime}\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}_{b_{1}}\bar{a}\bar{b}A. Then $b_{1}\in\Sigma\mathrm{cl}(\bar{a}^{\prime}A^{\prime})$; by $\Sigma$-basedness of $\mathrm{tp}(a)$ and Lemma 5 applied to $\bar{a}b_{1}\in\Sigma\mathrm{cl}(\bar{a}\bar{a}^{\prime}A^{\prime})$ we have $\mathrm{Cb}(\bar{a}b_{1}/\Sigma\mathrm{cl}(AA^{\prime}))\subseteq\Sigma\mathrm{cl}(\bar{a}b_{1}A^{\prime})$.

If $Y=\Sigma\mathrm{cl}(\emptyset)\cap\mathrm{dcl}(b_{1})$, then A\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}_{Y}b_{1} by Lemma 3, as $b_{1}\in\Sigma\mathrm{cl}(\bar{b})$ by $\Sigma$-basedness of $\mathrm{tp}(b)$ and because \bar{b}\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}\Sigma\mathrm{cl}(A); since $\mathrm{tp}(A^{\prime}/b_{1})=\mathrm{tp}(A/b_{1})$ we also have A^{\prime}\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}_{Y}b_{1}, whence A^{\prime}\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}_{Y}\bar{a}b_{1}A, and A^{\prime}\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}_{YA}\bar{a}b_{1}. As $\Sigma\mathrm{cl}(YA)=\Sigma\mathrm{cl}(A)$, Lemma 3 implies

[TABLE]

by Lemma 3 since A^{\prime}\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}_{\bar{a}b_{1}Y}A. On the other hand, put $C=\mathrm{Cb}(\bar{a}b_{1}/\Sigma\mathrm{cl}(A))$. Then \bar{b}\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}_{b_{1}}\Sigma\mathrm{cl}(\bar{a}A) by definition of $b_{1}$, whence \bar{a}\bar{b}\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}_{\bar{a}b_{1}}\Sigma\mathrm{cl}(A); as \bar{a}b_{1}\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}_{C}\Sigma\mathrm{cl}(A) we get \bar{a}\bar{b}\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}_{C}\Sigma\mathrm{cl}(A), whence $\mathrm{Cb}(\bar{a}\bar{b}/\Sigma\mathrm{cl}(A))\subseteq C$. So

[TABLE]

whence \bar{a}\bar{b}\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}\Sigma\mathrm{cl}(A).∎

Corollary 7**.**

If $\mathrm{tp}(a_{i})$ is $\Sigma$-based for all $i<\alpha$, so is $\mathrm{tp}(\bigcup_{i<\alpha}a_{i})$.

*Proof: *We use induction on $\beta$ to show that $\mathrm{tp}(\bigcup_{i<\beta}a_{i})$ is $\Sigma$-based, for $\beta\leq\alpha$. This is clear for $\beta=0$; it follows from Lemma 6 for successor ordinals. And if $\beta$ is a limit ordinal, then for any set $A$

[TABLE]

Lemma 8**.**

If $\mathrm{tp}(a/A)$ is $\Sigma$-based and a\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}A, then $\mathrm{tp}(a)$ is $\Sigma$-based.

*Proof: *Let $\bar{a}$ be a tuple of realizations of $\mathrm{tp}(a)$, and consider a set $B$ of parameters. For every $a_{i}\in\bar{a}$ choose $A_{i}$ with $\mathrm{tp}(a_{i}A_{i})=\mathrm{tp}(aA)$ and A_{i}\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}_{a_{i}}(\bar{a},B,A_{j}:j<i). As A_{i}\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}a_{i} we obtain A_{i}\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}(\bar{a},B,A_{j}:j<i), whence A_{i}\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}_{(A_{j}:j<i)}\bar{a}B, and inductively (A_{j}:j\leq i)\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}\bar{a}B. Put $\bar{A}=\bigcup_{a_{i}\in\bar{a}}A_{i}$; we just saw that \bar{A}\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}\bar{a}B. Now $\mathrm{tp}(a_{i}/\bar{A})$ is $\Sigma$-based for all $a_{i}\in\bar{a}$, and so is $\mathrm{tp}(\bar{a}/\bar{A})$ by Corollary 7. As \bar{a}\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}_{B}\bar{A}, Lemma 3 implies

[TABLE]

where the last equality follows from \bar{a}A\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}_{\bar{a}}B and Lemma 3.∎

Corollary 9**.**

If $p$ is almost internal in $\Sigma$-based types, then $p$ is $\Sigma$-based.

*Proof: *Suppose $p=\mathrm{tp}(a/A)$, and choose B\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}_{A}a and $\bar{b}$ such that $a\in\mathrm{bdd}(B\bar{b})$ and $\mathrm{tp}(b/B)$ is $\Sigma$-based for all $b\in\bar{b}$. Then $\mathrm{tp}(\bar{b}/AB)$ is $\Sigma$-based by Lemma 7, as is $\mathrm{tp}(a/AB)$ by Lemma 5, and $\mathrm{tp}(a/A)$ by Lemma 8.∎

Lemma 10**.**

If $\mathrm{tp}(a)$ and $\mathrm{tp}(b/a)$ are $\Sigma$-based, so is $\mathrm{tp}(ab)$.

*Proof: *Consider a tuple $\bar{a}\bar{b}$ of realizations of $\mathrm{tp}(ab)$, and a set $A$ of parameters. As $\mathrm{tp}(\bar{a})$ and $\mathrm{tp}(\bar{b}/\bar{a})$ are both $\Sigma$-based, we may suppose $a=\bar{a}$ and $b=\bar{b}$. Put $C=\mathrm{Cb}(ab/\Sigma\mathrm{cl}(A))$; again we add $\Sigma\mathrm{cl}(ab)\cap C$ to the language. By $\Sigma$-basedness of $\mathrm{tp}(a)$ we get a\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}\Sigma\mathrm{cl}(A).

Consider a Morley sequence $(a_{i}b_{i}:i<\omega)$ in $\mathrm{lstp}(ab/C)$; we may assume that (a_{i}b_{i}:i<\omega)\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}_{C}abA. Since (a_{i}:i<\omega)\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}C we get ab\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}(a_{i}:i<\omega). Moreover, as $\mathrm{tp}(ab/C)$ is foreign to $\Sigma$, we have ab\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}_{C}\Sigma\mathrm{cl}(a_{i}b_{i}:i<\omega). On the other hand $C\in\mathrm{dcl}(a_{i}b_{i}:i<\omega)$, whence

[TABLE]

Put $b^{\prime}=\mathrm{Cb}(ab/\Sigma\mathrm{cl}(a,a_{i}b_{i}:i<\omega))$. Then $a\in b^{\prime}$, and $b^{\prime}\in\Sigma\mathrm{cl}(ab)$ by $\Sigma$-basedness of $\mathrm{tp}(b/a)$. Put $X=\Sigma\mathrm{cl}(\emptyset)\cap\mathrm{dcl}(b^{\prime})$. Then b^{\prime}\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}_{X}(a_{i}:i<\omega) by Lemma 3; as $\mathrm{tp}(b^{\prime}/a_{i}:i<\omega)$ is $\Sigma$-based by Lemma 5 and Corollary 7 applied to $b^{\prime}\in\Sigma\mathrm{cl}(a,a_{i}b_{i}:i<\omega)$, so is $\mathrm{tp}(b^{\prime}/X)$ by Lemma 8. Put $C^{\prime}=\mathrm{Cb}(b^{\prime}/\Sigma\mathrm{cl}(a_{i}b_{i}:i<\omega))$, then $C^{\prime}\subseteq\Sigma\mathrm{cl}(b^{\prime})\subseteq\Sigma\mathrm{cl}(ab)$ by $\Sigma$-basedness.

Now ab\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}_{b^{\prime}}\Sigma\mathrm{cl}(a_{i}b_{i}:i<\omega) by definition of $b^{\prime}$; as b^{\prime}\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}_{C^{\prime}}\Sigma\mathrm{cl}(a_{i}b_{i}:i<\omega) by definition, we get ab\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}_{C^{\prime}}\Sigma\mathrm{cl}(a_{i}b_{i}:i<\omega), whence $C\subseteq C^{\prime}$. We obtain

[TABLE]

whence ab\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}\Sigma\mathrm{cl}(A).∎

Theorem 11**.**

Let $p$ be analysable in $\Sigma$-based types. Then $p$ is $\Sigma$-based.

*Proof: *Suppose $p=\mathrm{tp}(a/A)$. Then there is a sequence $(a_{i}:i<\alpha)\subseteq\mathrm{dcl}(aA)$ such that $a\in\mathrm{bdd}(A,a_{i}:i<\alpha)$ and $\mathrm{tp}(a_{i}/A,a_{j}:j<i)$ is internal in $\Sigma$-based types for all $i<\alpha$. So $\mathrm{tp}(a_{i}/A,a_{j}:j<i)$ is $\Sigma$-based for all $i<\alpha$ by Corollary 9; we use induction on $i$ to show that $\mathrm{tp}(a_{j}:j<i/A)$ is $\Sigma$-based. This is clear for $i=0$ and $i=1$; by Lemma 7 it is true for limit ordinals, and by Lemma 10 it holds for successor ordinals.∎

Corollary 12**.**

If $p$ is analysable in one-based types, then $p$ is itself one-based.∎