Positive Łoś Theorem 2.10. Let $\prod _{x \in X}M_x / F$ be a prime product. For every positive formula $\varphi (v_1, \dots , v_n)$ and $a_1, \dots , a_n \in S_F$ ,

Consequently, a positive sentence holds in a structure M if and only if it holds in some (equiv. every) prime power of M.

Proof We recall that positive formulas are preserved by homomorphisms. Therefore, the assumption that $a_1, \dots , a_n \in S_F$ guarantees that is an upset of $\mathbb {X}$ , for every positive formula $\psi $ . We will use this fact without further notice.

We reason by induction on the construction of $\varphi $ . In the base case, $\varphi $ is an atomic formula and the result holds by the definition of a prime product. The case where $\varphi = \psi _1 \land \psi _2$ follows from the inductive hypothesis and the fact that F is a filter over $\mathbb {X}$ . The case where $\varphi = \psi _1 \lor \psi _2$ follows from the inductive hypothesis and the fact that F is prime. It only remains to consider the case where $\varphi = \exists w\kern2pt \psi (w, v_1, \dots , v_n)$ . By the induction hypothesis we have

Therefore, it only remains to prove that

For the sake of readability, we will write

. Since E is an upset of $\mathbb {X}$ and

for every $b \in S_F$ , the implication from right to left in the above display is straightforward. To prove the other implication, suppose that $E \in F$ and let Y be the set of minimal elements of E. For every $y \in Y \subseteq E$ there exists $b(y) \in M_y$ such that

(1) $$ \begin{align} M_y \vDash \psi(b(y), a_1(y), \dots, a_n(y)). \end{align} $$

As $\mathbb {X}$ is a wellfounded forest, for each $x \in E$ there exists a unique $y_x \in Y$ such that $y_x \leq x$ . Therefore, we can define an element $b \in \prod _{x \in E}M_x$ as

for each $x \in E$ . As $E \in F$ , we have $b \in S_E \subseteq S_F$ . Furthermore, from Condition (1) and the fact that positive formulas are preserved by homomorphisms, it follows

$$\begin{align*}M_x \vDash \psi(b(x), a_1(x), \dots, a_n(x)) \text{ for every }x \in E. \end{align*}$$

Consequently,

. Since $E \in F$ and

is an upset of $\mathbb {X}$ , we conclude that

as desired.

Proof It suffices to prove the statement for the case where $\varphi $ is basic h-inductive. Then there are two positive formulas $\psi _1(v_1, \dots , v_n)$ and $\psi _2(v_1, \dots , v_n)$ such that $\varphi = \forall v_1, \dots , v_n\kern2pt (\psi _1 \to \psi _2)$ . We will reason by contraposition. Suppose that $\prod _{x \in X} M_x / F \nvDash \varphi $ . By the Positive Łoś Theorem, there are $a_1, \dots , a_n \in S_F$ such that

Since

is an upset of $\mathbb {X}$ , this means that

Consequently, there exists $x \in V_{a_1} \cap \dots \cap V_{a_n}$ such that

$$\begin{align*}M_x \vDash \psi_1(a_1(x), \dots, a_n(x)) \land \lnot \psi_2(a_1(x), \dots, a_n(x)).\\[-34pt] \end{align*}$$

Proof We recall that a class of similar structures is elementary if and only if it is closed under isomorphisms, ultraproducts, and ultraroots [Reference Frayne, Morel and Scott7, Theorem 2.13]. Furthermore, such a class is h-inductive if and only if it is closed under direct limits of chains of structures. Since ultraproducts and direct limits of chains of structures are special cases of prime products (see Examples 2.5 and 2.8), it follows that every class of similar closed under isomorphisms, prime products, and ultraroots is h-inductive and elementary. Conversely, every h-inductive elementary class is closed under isomorphisms and ultraroots because it is elementary and under prime products by Proposition 2.13.

Positive Keisler Isomorphism Theorem 3.2. Under GCH, two structures are positively equivalent if and only if they have isomorphic prime powers of ultrapowers.

Proof Condition (i) follows from [Reference Keisler9, Theorem 2.1] and [Reference Kunen10, Theorem 3.2]. For Condition (ii), see the proof of [Reference Keisler9, Corollary 2.3].

Proof By Theorem 3.8 (i) there exists a $\kappa $ -saturated ultrapower $M^\ast $ of $M_u$ . As $M_u$ is an ultrapower of M and ultrapowers of ultrapowers are still ultrapowers, we may assume that $M^\ast $ is an ultrapower of M. Furthermore, as $M_u$ embeds into $M^\ast $ , we can view f as a homomorphism from N to $M^\ast $ .

Proof Consider two similar structures $M_1$ and $M_2$ . If $M_1$ and $M_2$ have isomorphic prime powers of ultrapowers, then they are positively equivalent by the classical Łoś Theorem and its positive version. Conversely, suppose that $M_1$ and $M_2$ are positively equivalent. By Theorem 3.8, under GCH each $M_i$ has a saturated ultrapower $M_i^\ast $ . Furthermore, by the same theorem $M_i^\ast $ can be assumed to be either finite (if $M_i$ is finite) or of size $\geq \vert L \vert $ (if $M_i$ is infinite). Since $M_1^\ast $ and $M_2^\ast $ are also positively equivalent, we can apply Theorem 3.10 obtaining that there exists .

Proof Since the theory T is h-inductive, from the assumption that f is an immersion and that $M \vDash T$ it follows that $N \vDash T$ . To prove that N is pec, we will use Proposition 3.13. Consider a positive formula $\varphi (\vec {v}\kern2pt)$ and $\vec {a} \in N$ such that $N \nvDash \varphi (\vec {a})$ . Since f is an immersion and $\varphi $ positive, we obtain $M \nvDash \varphi (f(\vec {a}))$ . As M is a pec model of T, there exists $\psi (\vec {v}\kern2pt) \in \operatorname {\mathrm {Res}}_T(\phi )$ such that $M \vDash \psi (f(\vec {a}))$ . Since f is an immersion and $\psi $ positive, this yields $N \vDash \psi (\vec {a})$ as desired.

Proof See the paragraph immediately after the proof of [Reference Poizat and Yeshkeyev14, Proposition 12].

Proof We will define a chain of structures $\{ M_n \mid n \in \omega \}$ such that $M_n \vDash \operatorname {\mathrm {Th}}^+(M)$ for each $n \in \omega $ . First let

. Then suppose that the chain of structures $\{ M_m \mid m \leq n \}$ has already been defined. Since $M_n$ is a model of the h-inductive theory $\operatorname {\mathrm {Th}}^+(M)$ , by Proposition 3.14 (i) there exists a pec model $M_n^\ast $ of $\operatorname {\mathrm {Th}}^+(M)$ with a homomorphism $g_n \colon M_n \to M_n^\ast $ . Furthermore, as $M_n^\ast $ is a model of $\operatorname {\mathrm {Th}}^+(M)$ , the structures $M_n^\ast $ and M have the same positive theory. In particular, M satisfies the positive theory of $M_n^\ast $ . Therefore, we can apply Proposition 3.4 obtaining an ultrapower $M_{n +1}$ of M with a homomorphism $h_n \colon M_n^\ast \to M_{n +1}$ . Since M is a model of $\operatorname {\mathrm {Th}}^+(M)$ , so is the ultrapower $M_{n +1}$ . In addition, the ultrapower $M_{n +1}$ can be assumed to be $\omega $ -saturated in virtue of Corollary 3.9. Then we let $\{ M_m \mid m \leq n + 1 \}$ be the chain of structures obtained by extending $\{ M_m \mid m \leq n \}$ with $M_{n +1}$ and a homomorphism $f_{m \kern2pt\kern2pt\kern2pt n+1} \colon M_m \to M_{n+1}$ for each $m \leq n +1$ defined as follows:

Lastly, let $M_\omega $ be the direct limit of the chain of structures $\{ M_n \mid n \in \omega \}$ . Since $\operatorname {\mathrm {Th}}^+(M)$ is an h-inductive theory and $M_n \vDash \operatorname {\mathrm {Th}}^+(M)$ for each $n \in \omega $ , we can apply Theorem 2.15 obtaining that $M_\omega \vDash \operatorname {\mathrm {Th}}^+(M)$ .

We will prove that $M_\omega $ is a pec model of $\operatorname {\mathrm {Th}}^+(M)$ that is positively $\omega $ -saturated and belongs to . To this end, observe that $M_\omega $ is also the direct limit of:

1. (i) a chain of structures whose members are of the form $M_n^\ast $ for $n \in \omega $ , and of

2. (ii) a chain of structures whose members are of the form $M_{n + 1}$ for $n \in \omega $ .

On the one hand, each $M_n^\ast $ is a pec model of $\operatorname {\mathrm {Th}}^+(M)$ by construction. Therefore, from Condition (i) it follows that $M_\omega $ is the direct limit of a chain of pec models of $\operatorname {\mathrm {Th}}^+(M)$ . Since the theory $\operatorname {\mathrm {Th}}^+(M)$ is h-inductive, from Proposition 3.14 (ii) it follows that $M_\omega $ is a pec model of $\operatorname {\mathrm {Th}}^+(M)$ . On the other hand, each $M_{n + 1}$ is an ultrapower of M. Therefore, from Condition (ii) it follows .

It only remains to prove that $M_\omega $ is positively $\omega $ -saturated. Recall that $M_\omega $ is the direct limit of the chain of structures $\{ M_n \mid n \in \omega \}$ . For each $n \in \omega $ we will denote the canonical homomorphism from $M_n$ to $M_\omega $ associated with the direct limit $M_\omega $ by $f_n \colon M_n \to M_\omega $ . Then consider $a_1, \dots , a_n \in M_\omega $ and let $p(v_1, \dots , v_n, a_1, \dots , a_n)$ be a set of positive formulas that is finitely satisfiable in $M_\omega $ . Since $M_\omega $ is the direct limit of $\{ M_n \mid n \in \omega \}$ , there exist $m \in \omega $ and $\hat {a}_1, \dots , \hat {a}_n \in M_{m}$ such that $f_m(\hat {a}_1) = a_1, \dots , f_m(\hat {a}_n) = a_n$ .

By the universal property of the direct limit we have

$$\begin{align*}f_m = f_{m+1} \circ f_{m \kern2pt\kern2pt\kern2pt m+1} = f_{m+1} \circ h_m \circ g_m. \end{align*}$$

Thus,

(2) $$ \begin{align} \begin{split} p(v_1, \dots, v_n, a_1, \dots, a_n) &= p(v_1, \dots, v_n, f_m(\hat{a}_1), \dots, f_m(\hat{a}_n))\\ &=p(v_1, \dots, v_n, f_{m+1}(h_m(g_m(\hat{a}_1))), \dots, f_{m+1}(h_m(g_m(\hat{a}_n)))). \end{split} \end{align} $$

Since $M^\ast _m$ is a pec model of $\operatorname {\mathrm {Th}}^+(M)$ and $M_{\omega }$ a model of $\operatorname {\mathrm {Th}}^+(M)$ , the homomorphism $f_{m+1} \circ h_m \colon M^\ast _m \to M_\omega $ is an immersion. Consequently, from the assumption that the set $p(v_1, \dots , v_n, a_1, \dots , a_n)$ is finitely satisfiable in $M_\omega $ and the above display it follows that $p(v_1, \dots , v_n, g_m(\hat {a}_1), \dots , g_m(\hat {a}_n))$ is finitely satisfiable in $M^\ast _m$ . As positive formulas are preserved by homomorphisms, the set $p(v_1, \dots , v_n, h_{m}(g_m(\hat {a}_1)), \dots , h_{m}(g_m(\hat {a}_n)))$ is finitely satisfiable in $M_{m+1}$ . Since $M_{m+1}$ is $\omega $ -saturated, there exist $b_1, \dots , b_n \in M_{m+1}$ such that

$$\begin{align*}M_{m+1} \vDash p(b_1, \dots, b_n, h_{m}(g_m(\hat{a}_1)), \dots, h_{m}(g_m(\hat{a}_n))). \end{align*}$$

Since positive formulas are preserved by homomorphisms, from Condition (2) it follows that $M_\omega \vDash p(f_{m+1}(b_1), \dots , f_{m+1}(b_n), a_1, \dots , a_n)$ . Hence, we conclude that $M_\omega $ is positively $\omega $ -saturated.

To prove the second part of the statement, suppose that M is positively saturated. If M is finite, we have $M_n \cong M$ for each $n \in \omega $ because $M_n$ is an ultrapower of M. Therefore, the above construction yields as desired. Then we consider the case where M is infinite and $\vert L \vert \leq \vert M \vert $ . Since M is a model of $\operatorname {\mathrm {Th}}^+(M)$ , by Proposition 3.14 (i) there exists a pec model $M^\ast $ of $\operatorname {\mathrm {Th}}^+(M)$ with a homomorphism $g \colon M \to M^\ast $ . As M is infinite and such that $\vert L \vert \leq \vert M \vert $ , by the downward Löwenheim–Skolem Theorem there exists an elementary substructure N of $M^\ast $ containing $g[M]$ such that $\vert N \vert \leq \vert M \vert $ . Since N is an elementary substructure of $M^\ast $ and $M^\ast $ is a pec model of the h-inductive theory $\operatorname {\mathrm {Th}}^+(M)$ , we can apply Proposition 3.15 obtaining that N is also a pec model of $\operatorname {\mathrm {Th}}^+(M)$ . Therefore, we may assume without loss of generality that $M^\ast = N$ and, therefore, that $\vert M^\ast \vert \leq \vert M \vert $ . As $M^\ast $ is a model of $\operatorname {\mathrm {Th}}^+(M)$ , we know that M and $M^\ast $ have the same positive theory. Together with $\vert M^\ast \vert \leq \vert M \vert $ and the assumption that M is positively saturated, this implies that there exists a homomorphism $h \colon M^\ast \to M$ by Proposition 3.7. Then we consider the endomorphism of M. Then let $M_\omega $ be the direct limit of the chain of structures

$$\begin{align*}M \xrightarrow{f} M \xrightarrow{f} M \xrightarrow{f} \cdots. \end{align*}$$

The argument detailed above shows that $M_\omega $ is an $\omega $ -saturated pec model of $\operatorname {\mathrm {Th}}^+(M)$ . Furthermore, by construction.

Proof From the classical Łoś Theorem and its positive version it follows that if there exists

, then $M_1$ and $M_2$ are positively equivalent. Conversely, suppose that $M_1$ and $M_2$ are positively equivalent. Then let

By Proposition 3.18 there are

positively $\omega $ -saturated pec models of T. Furthermore, $M^\ast _1$ and $M^\ast _2$ have the same positive theory by the classical Łoś Theorem and its positive version. Consequently, $M_1^\ast $ and $M_2^\ast $ are elementarily equivalent by Proposition 3.16. In view of the Keisler–Shelah Isomorphism Theorem, there exists

. Together with the above display, this yields

By Proposition 2.9 this simplifies to

as desired.

To prove the second part of the statement, suppose that $M_1$ and $M_2$ are positively saturated (in addition to positively equivalent) and that each $M_i$ is either finite or of size $\geq \vert L \vert $ . By Proposition 3.18 we can take

and repeat the argument above obtaining that

.

Proof of Proposition 2.9

We detail the proof of the inclusion , since the proof of the other inclusion regarding reduced and filter products is analogous (in fact, simpler).

Let $\kappa $ be a cardinal and for each $\alpha < \kappa $ let $M_\alpha $ be the direct limit of a chain of structures $\{ M_x \mid x \in X_\alpha \}$ in $\mathsf {K}$ indexed by a well-ordered poset $\mathbb {X}_\alpha $ . Moreover, let U be a ultrafilter over $\kappa $ . We need to prove that

Without loss of generality, we may assume that the members of $\{\mathbb {X}_\alpha \}_{\alpha < \kappa }$ are pairwise disjoint. Then let $\mathbb {X}$ be the poset obtained as the disjoint union of $\{\mathbb {X}_\alpha \}_{\alpha < \kappa }$ and define

Proof of the Claim

Clearly, F is an upset of $\mathsf {Up}(\mathbb {X})$ . Moreover, F is nonempty as it contains X. To prove that F is closed under binary intersections, consider $V, W \in F$ . Then

(3) $$ \begin{align} \{ \alpha < \kappa \mid X_\alpha \cap V \ne \emptyset \} \cap \{ \alpha < \kappa \mid X_\alpha \cap W \ne \emptyset \} \in U. \end{align} $$

We will show that

(4) $$ \begin{align} \{ \alpha < \kappa \mid X_\alpha \cap V \ne \emptyset \} \cap \{ \alpha < \kappa \mid X_\alpha \cap W \ne \emptyset \} \subseteq \{ \alpha < \kappa \mid X_\alpha \cap V \cap W \ne \emptyset \}. \end{align} $$

To this end, consider $\alpha < \kappa $ such that $X_\alpha \cap V \ne \emptyset $ and $X_\alpha \cap W \ne \emptyset $ . Then there are $v \in X_\alpha \cap V$ and $w \in X_\alpha \cap W$ . Since $\mathbb {X}_\alpha $ is linearly ordered, we may assume that $v \leq w$ . Since $V \cap X_\alpha $ is an upset of $\mathbb {X}_\alpha $ (because V is an upset of $\mathbb {X}$ ), this implies $w \in X_\alpha \cap V$ . Thus, $w \in X_\alpha \cap V \cap W$ and, therefore, $X_\alpha \cap V \cap W \ne \emptyset $ . From Conditions (3) and (4) it follows that

$$\begin{align*}\{ \alpha < \kappa \mid X_\alpha \cap V \cap W \ne \emptyset \} \in U. \end{align*}$$

Thus, $V \cap W \in F$ as desired. We conclude that F is a filter over $\mathbb {X}$ .

The definition of F guarantees that $\emptyset \notin F$ . Therefore, F is proper. To prove that it is prime, consider $V, W \in \mathsf {Up}(\mathbb {X})$ such that $V \cup W \in F$ . Then

$$\begin{align*}\{ \alpha < \kappa \mid X_\alpha \cap V \ne \emptyset \} \cup \{ \alpha < \kappa \mid X_\alpha \cap W \ne \emptyset \} = \{ \alpha < \kappa \mid X_\alpha \cap (V \cup W) \ne \emptyset \} \in U. \end{align*}$$

Since U is an ultrafilter over $\kappa $ , it is also a prime filter over $\kappa $ ordered under the identity relation (Remark 2.2). Therefore, from the above display it follows that

$$\begin{align*}\text{either }\{ \alpha < \kappa \mid X_\alpha \cap V \ne \emptyset \} \in U\text{ or } \{ \alpha < \kappa \mid X_\alpha \cap W \ne \emptyset \} \in U. \end{align*}$$

This, in turn, implies that either V or W belongs to F.

Now, recall that $\mathbb {X}$ is the disjoint union of the well-ordered posets $\mathbb {X}_\alpha $ . Therefore, the union $\{ M_x \mid x \in X \}$ of the ordered systems $\{ M_x \mid x \in X_\alpha \}$ is a well-defined ordered system indexed by a wellfounded forest. Furthermore, F is a prime filter over $\mathbb {X}$ by the Claim. Consequently, we can form the associated prime product $\prod _{x \in X}M_x / F$ .

In order to conclude the proof, it suffices to prove that

$$\begin{align*}\prod_{\alpha < \kappa} M_\alpha / U \cong \prod_{x \in X} M_x / F. \end{align*}$$

To this end, observe that for every $\alpha < \kappa $ and $a \in M_\alpha $ there exist $z_{a} \in X_\alpha $ and $m_a \in M_{z_{a}}$ such that $f_{z_a}(m_a) = a$ , where $f_{z_a} \colon M_{z_a} \to M_\alpha $ is the canonical homomorphism associated with the direct limit $M_\alpha $ . For every each $a \in \prod _{\alpha < \kappa } M_\alpha $ , let

Notice that for each $x \in Y_{a}$ there exists exactly one $\alpha < \kappa $ such that $z_{a(\alpha )} \leq x$ because $\mathbb {X}$ is the disjoint union of the various $\mathbb {X}_\alpha $ and these are linearly ordered. We will denote this $\alpha $ by $\beta _{a x}$ . Bearing this in mind, let $g(a)$ be the only element of $\prod _{x \in Y_{a}} M_x$ defined for every $x \in Y_{a}$ as

Proof of the Claim

It suffices to show that $Y_a \in F$ and that for every $x, y \in Y_a$ ,

$$\begin{align*}x \leq y \text{ implies }f_{xy}(g(a)(x)) = g(a)(y). \end{align*}$$

By definition $Y_a$ is an upset of $\mathbb {X}$ which, moreover, is nondisjoint with every $X_\alpha $ because $z_{a(\alpha )} \in X_\alpha \cap Y_a$ . Together with the definition of F, this yields $Y_a \in F$ . Then consider $x, y \in Y_a$ such that $x \leq y$ . From $z_{a(\beta _{ax})} \leq x \leq y$ and the fact that $\beta _{ay}$ is the unique $\alpha < \kappa $ such that $z_{a(\alpha )} \leq y$ it follows $\beta _{ax} = \beta _{ay}$ . Therefore, by the definition of $g(a)$ we obtain

$$\begin{align*}g(a)(x) = f_{z_{a(\beta_{ax})} x}(m_{a(\beta_{ax})}) \, \, \text{ and } \, \, g(a)(y) = f_{z_{a(\beta_{ax})} y}(m_{a(\beta_{ax})}). \end{align*}$$

Furthermore, from $z_{a(\beta _{ax})} \leq x \leq y$ it follows $f_{z_{a(\beta _{ax})} y} = f_{xy} \circ f_{z_{a(\beta _{ax})} x}$ . Together with the above display, this yields

$$\begin{align*}f_{xy}(g(a)(x)) = f_{xy}(f_{z_{a(\beta_{ax})} x}(m_{a(\beta_{ax})})) = f_{z_{a(\beta_{ax})} y}(m_{a(\beta_{ax})}) = g(a)(y).\\[-34pt] \end{align*}$$

Then we turn to prove the following:

Claim 3.23. For every atomic formula $\varphi (v_1, \dots , v_n)$ and $a_1, \dots , a_n \in \prod _{\alpha < \kappa } M_\alpha $ ,

(5) $$ \begin{align} &\prod_{\alpha < \kappa} M_\alpha / U \vDash \varphi(a_1/ \mathord{\equiv_{U}}, \dots, a_n/ \mathord{\equiv_{U}}) \, \, \text{ if and only if } \nonumber\\ &\prod_{x \in X} M_x / F \vDash \varphi(g(a_1)/ \mathord{\equiv_{F}}, \dots, g(a_n)/ \mathord{\equiv_{F}}). \end{align} $$

Proof of the Claim

We begin by showing that

The first of the equalities above holds by the definition of $z_{a_i(\alpha )}$ and $m_{a_i(\alpha )}$ , the second because $M_\alpha $ is the direct limit of the chain of structures $\{ M_x \mid x \in X_\alpha \}$ , the third by the definition of $g(a_i)$ , and the fifth by the definition of

. To prove the fourth, it suffices to show that

$$\begin{align*}X_\alpha \cap V_{g(a_1)} \cap \dots \cap V_{g(a_n)} = \{ x \in X \mid x \geq z_{a_1(\alpha)}, \dots, z_{a_n(\alpha)}\}. \end{align*}$$

The above equality, in turn, holds because $V_{g(a_i)} = Y_{a_i}$ by Claim 3.22 and $\mathbb {X}$ is the disjoint union of the various $\mathbb {X}_\beta $ .

Observe that

is an upset of $\mathbb {X}$ (because positive formulas are preserved by homomorphisms). Therefore, from the above series of equalities and the definition of F it follows that

By the classical Łos Theorem and its positive version this yields the desired result.

In view of Claim 3.23, the map

$$\begin{align*}\hat{g} \colon \prod_{\alpha < \kappa} M_\alpha / U \to \prod_{x \in X} M_x / F \end{align*}$$

defined by the rule is a well-defined embedding. Therefore, to prove that $\hat {g}$ is an isomorphism, it only remains to show that it is surjective.

To this end, consider $a \in S_F$ . For each $\alpha < \kappa $ and $x \in X_\alpha \cap V_a$ let $f_x \colon M_x \to M_\alpha $ be the canonical homomorphism associated with the direct limit $M_\alpha $ . For each $\alpha < \kappa $ such that $X_\alpha \cap V_a \ne \emptyset $ there exists some $y_\alpha \in X_\alpha \cap V_a$ such that

(6) $$ \begin{align} z_{f_{y_{\alpha}}(a(y_\alpha))} \leq y_a \, \, \text{ and } \, \, a(y_\alpha) = f_{z_{f_{y_{\alpha}}(a(y_\alpha))} \kern2pt\kern2pt y_\alpha}(m_{f_{y_{\alpha}}(a(y_\alpha))}). \end{align} $$

This is a consequence of the definition of the direct limit $M_\alpha $ and of the assumption that $X_\alpha \cap V_a$ is an upset of the linearly ordered poset $\mathbb {X}_\alpha $ .

We define an element $b \in \prod _{\alpha < \kappa } M_\alpha $ as follows: for each $\alpha < \kappa $ ,

Our aim is to prove that $\hat {g}(b / {\equiv }_U) = a / {\equiv }_F$ , i.e.,

. Since $g(b) \in S_F$ , we know that

is an upset of $\mathbb {X}$ . Therefore, by the definition of F it suffices to show that

As $a \in S_F$ , we know that $V_a \in F$ and, therefore, that $\{ \alpha < \kappa \mid X_\alpha \cap V_a \ne \emptyset \} \in U$ . Therefore, our task reduces to that of proving the inclusion

Accordingly, let $\alpha < \kappa $ be such that $X_\alpha \cap V_a \ne \emptyset $ . Then $y_\alpha \in X_a \cap V_a$ . Furthermore, $b(\alpha ) = f_{y_{\alpha }}(a(y_\alpha ))$ by the definition of b. Therefore, from Condition (6) it follows $z_{b(\alpha )} \leq y_\alpha $ . By the definition of $Y_{b}$ this amounts to $y_\alpha \in Y_b$ . Since $Y_b = V_{g(b)}$ by Claim 3.22, we conclude that $y_\alpha \in V_{g(b)}$ . Thus, $y_\alpha \in V_{g(b)} \cap V_a$ . In addition, from the definition of $g(b)$ and Condition (6) it follows

$$\begin{align*}g(b)(y_\alpha) = f_{z_{f_{y_{\alpha}}(a(y_\alpha))} \kern2pt\kern2pt y_\alpha}(m_{f_{y_{\alpha}}(a(y_\alpha))}) = a(y_\alpha). \end{align*}$$

Therefore,

. Hence, we conclude that

as desired.