1. Ranked lattices and their representations

This section, comprising three subsections, culminates with a description of how to obtain elementary extensions realizing a given finite ranked lattice. Section 1.1 repeats some material from [Reference Kossak and Schmerl1, Chapter 4]. Section 1.2 extends Section 1.1 and puts a new perspective on it. Finally, Section 1.3 extends Section 1.2 from lattices to ranked lattices.

1.1. Representations of lattices

For any set A, let $\operatorname {\mathrm {Eq}}(A)$ be the lattice of equivalence relations on A, ordered in such a way that if $\Theta _1,\Theta _2 \in \operatorname {\mathrm {Eq}}(A)$ , then $\Theta _1 \leq \Theta _2$ iff $\Theta _1 \subseteq \Theta _2$ (that is, $\Theta _1$ refines $\Theta _2$ ). We let $0 {\hspace{-6.3pt} 0}_A$ be the discrete equivalence relation on A (that is, $0 {\hspace{-6.3pt} 0}_A$ is the equality relation on A) and $1 {\hspace {-3.5pt} 1}_A$ be the trivial equivalence relation (that is, $1 {\hspace {-3.5pt} 1}_A = A \times A$ ). Thus, for any $\Theta _1, \Theta _2 \in \operatorname {\mathrm {Eq}}(A)$ , we have that

$$ \begin{align*}0 {\hspace{-6.3pt} 0}_A \leq \Theta_1 \leq 1 {\hspace{-3.5pt} 1}_A \end{align*} $$

and

$$ \begin{align*}\Theta_1 \wedge \Theta_2 = \Theta_1 \cap \Theta_2.\end{align*} $$

If $\Theta \in \operatorname {\mathrm {Eq}}(A)$ and $B \subseteq A$ , then $\Theta \cap B^2 \in \operatorname {\mathrm {Eq}}(B)$ . If f is a function with domain A, then f induces $\Theta \in \operatorname {\mathrm {Eq}}(A)$ if whenever $a,b \in A$ , then $\langle a , b \rangle \in \Theta $ iff $f(a) = f(b)$ .

Let L be a finite lattice. A representation of L is a one-to-one function $\alpha : L \longrightarrow \operatorname {\mathrm {Eq}}(A)$ such that

$$ \begin{align*}\alpha(0_L) &= 1 {\hspace{-3.5pt} 1}_A, \\\alpha(1_L) &= 0 {\hspace{-6.3pt} 0}_A, \end{align*} $$

and

$$ \begin{align*}\alpha(r \vee s) = \alpha(r) \wedge \alpha(s) \end{align*} $$

for each $r,s \in L$ . (It is not required that $\alpha (r \wedge s ) = \alpha (r) \vee \alpha (s)$ .) We say that $\alpha $ is finite if A is a finite set. If $B \subseteq A$ , then $\alpha |B : L \longrightarrow \operatorname {\mathrm {Eq}}(B)$ is such that $(\alpha |B)(r) = \alpha (r) \cap B^2$ for each $r \in L$ . The representation $\beta : L \longrightarrow \operatorname {\mathrm {Eq}}(B)$ is isomorphic to $\alpha $ (in symbols: $\alpha \cong \beta $ ) if there is a bijection $f : A \longrightarrow B$ such that for any $x,y \in A$ and $r \in L$ , $\langle x , y \rangle \in \alpha (r)$ iff $\langle f(x) , f(y)\rangle \in \beta (r)$ . If such is the case, then we say that f demonstrates that $\alpha \cong \beta $ . If $\alpha : L \longrightarrow \operatorname {\mathrm {Eq}}(A)$ is a representation and $\Theta \in \operatorname {\mathrm {Eq}}(B)$ , then $\Theta $ is canonical (for $\alpha $ ) if $B \subseteq A$ and $\Theta = \alpha (r) \cap B^2$ for some $r \in L$ .

Suppose that $\alpha : L \longrightarrow \operatorname {\mathrm {Eq}}(A)$ is a representation of the finite lattice L, and $\mathcal {B}$ is a set of representations of L. Then $\alpha $ arrows ${\mathcal B}$ (in symbols: $\alpha \longrightarrow {\mathcal B}$ ) if whenever $\Theta \in \operatorname {\mathrm {Eq}}(A)$ , then there is $B \subseteq A$ such that $\Theta \cap B^2$ is canonical for $\alpha $ and $\alpha |B \cong \beta $ for some $\beta \in {\mathcal B}$ . We usually write $\alpha \longrightarrow \beta $ instead of $\alpha \longrightarrow \{\beta \}$ .

We next define, by recursion on $n < \omega $ , when the representation $\alpha : L \longrightarrow \operatorname {\mathrm {Eq}}(A)$ of the finite lattice L has the n-canonical partition property (or, briefly, is n-CPP). First, $\alpha $ is $0$ -CPP if for every $r \in L$ , there do not exist exactly $2\ \alpha (r)$ -classes; next, $\alpha $ is $(n+1)$ -CPP if there is a set ${\mathcal B}$ of n-CPP representations of L such that $\alpha \longrightarrow {\mathcal B}$ .

Given ${\mathcal M}$ , we say that a representation $\alpha $ of a finite lattice L is an ${\mathcal M}$ -representation if it is ${\mathcal M}$ -definable. Also, if $A \in \operatorname {\mathrm {Def}}({\mathcal M})$ , then we let $\operatorname {\mathrm {Eq}}^{\mathcal M}(A)$ be the set of those $\Theta \in \operatorname {\mathrm {Eq}}(A)$ that are definable in ${\mathcal M}$ . All the definitions in this subsection up to this point make sense when interpreted in a model ${\mathcal M}$ and are applied just to ${\mathcal M}$ -representations. In particular, it makes sense to refer to an ${\mathcal M}$ -finite ${\mathcal M}$ -representation $\alpha $ as being a-CPP for $a \in M$ . Thus, for every finite lattice L, there is a $\Sigma _1$ formula $cpp_L(x)$ such that for any ${\mathcal M}$ and $a \in M$ , ${\mathcal M} \models cpp_L(a)$ iff there is an ${\mathcal M}$ -finite ${\mathcal M}$ -representation of L that ${\mathcal M}$ thinks is a-CPP. The following theorem can be found in [Reference Kossak and Schmerl1, Chapter 4] or [Reference Schmerl3].

Theorem 1.1. Let L be a finite lattice and ${\mathcal M}$ be a nonstandard, countable model. The following are equivalent:

1. (1) There are ${\mathcal N}_0 \succ {\mathcal M}_0 \equiv {\mathcal M}$ such that $\operatorname {\mathrm {Lt}}({\mathcal N}_0 / {\mathcal M}_0) \cong L$ .

2. (2) For every $n < \omega $ , ${\mathcal M} \models cpp_L(n)$ .

3. (3) There is ${\mathcal N} \succ _{\mathsf {cf}} {\mathcal M}$ such that $\operatorname {\mathrm {Lt}}({\mathcal N} / {\mathcal M}) \cong L$ .

Notice that Corollary 2 follows from Theorem 1 and $(1) \Longrightarrow (3)$ of the previous theorem.

Corollary 1.2. If L is a finite lattice, then the following are equivalent:

1. (1) There is ${\mathcal N}$ such that $\operatorname {\mathrm {Lt}}({\mathcal N}) \cong L$ .

2. (2) There are ${\mathcal M} \prec {\mathcal N}$ such that $\operatorname {\mathrm {Lt}}({\mathcal N} / {\mathcal M}) \cong L$ .

Proof Obviously, $(1) \Longrightarrow (2)$ by letting ${\mathcal M}$ be the prime elementary submodel of ${\mathcal N}$ . The converse $(2) \Longrightarrow (1)$ follows from Theorem 1.1 as long as ${\mathcal M}$ is not a model of True Arithmetic (TA) and so its prime elementary submodel is nonstandard. If ${\mathcal M}$ is a model of TA, then by $(1) \implies (2)$ of Theorem 1.1, ${\mathcal M} \models cpp_L(n)$ for all $n < \omega $ . Since TA is undecidable, there is a prime, nonstandard ${\mathcal M}_0$ such that ${\mathcal M}_0 \models cpp_L(n)$ for all $n < \omega $ . Then by $(1) \implies (2)$ of Theorem 1.1, there is ${\mathcal N}_0 \succ _{\mathsf {cf}} {\mathcal M}_0$ such that $\operatorname {\mathrm {Lt}}({\mathcal N}_0) = \operatorname {\mathrm {Lt}}({\mathcal N}_0 / {\mathcal M}_0) \cong L$ .

1.2. Correct sets of representations

This subsection consists of a definition followed by a theorem generalizing Theorem 1.1.

Definition 1.3. Let ${\mathcal M}$ be a model and L be a finite lattice. We say that ${\mathcal C}$ is an ${\mathcal M}$ -correct set of representations of L if each of the following holds.

1. (1) ${\mathcal C}$ is a nonempty set of $0$ -CPP ${\mathcal M}$ -representations of L.

2. (2) Whenever $\alpha : L \longrightarrow \operatorname {\mathrm {Eq}}(A)$ is in ${\mathcal C}$ and $\Theta \in \operatorname {\mathrm {Eq}}^{\mathcal M}(A)$ , then there is a $B \subseteq A$ such that $\alpha |B \in {\mathcal C}$ and $\Theta \cap B^2$ is canonical for $\alpha $ .

Here is an example. Suppose that ${\mathcal M}$ is nonstandard and that ${\mathcal M} \models cpp_L(n)$ for every $n < \omega $ . Let ${\mathcal C}$ be the set of those ${\mathcal M}$ -finite ${\mathcal M}$ -representations $\alpha $ of L such that, for some nonstandard $n \in M$ , ${\mathcal M}$ thinks that $\alpha $ is n-CPP. Then, ${\mathcal C}$ is ${\mathcal M}$ -correct. With this example, we see that the following theorem generalizes a good portion of Theorem 1.1. It is a consequence of Theorem 1.1 when ${\mathcal M}$ is countable and nonstandard.

Theorem 1.4. Suppose that ${\mathcal M}$ is a model and L is a finite lattice.

1. (1) If there is ${\mathcal N} \succ {\mathcal M}$ such that $\operatorname {\mathrm {Lt}}({\mathcal N} / {\mathcal M}) \cong L$ , then there is an ${\mathcal M}$ -correct set of representations of L.

2. (2) If ${\mathcal M}$ is countable and there is an ${\mathcal M}$ -correct set of representations of L, then there is ${\mathcal N} \succ {\mathcal M}$ such that $\operatorname {\mathrm {Lt}}({\mathcal N} / {\mathcal M}) \cong L$ .

Proof $(1)$ Suppose that ${\mathcal N} \succ {\mathcal M}$ and that $F : L \longrightarrow \operatorname {\mathrm {Lt}}({\mathcal N} / {\mathcal M})$ is an isomorphism. Let $f : L \longrightarrow N$ be such that for $r \in L$ , $f(r)$ generates $F(r)$ over ${\mathcal M}$ . Let $a = f(1_L)$ .

For each pair of elements $r,s \in L$ , let $g_{r,s} : N \longrightarrow N$ and $h_{r,s} : N^2 \longrightarrow N$ be functions that are ${\mathcal N}$ -definable using parameters only from M such that

• • $g_{r,s}(f(r \vee s)) = f(r)$ ,

• • $h_{r,s}(f(r),f(s)) = f(r \vee s)$ .

The functions $g_{r,s}$ exist since $f(r) \in F(r \vee s)$ ; the functions $h_{r,s}$ exist for a similar reason. Let $g_r = g_{r,1}$ , so that $g_r(a) = f(r)$ . In particular, $g_1(a) = a$ . The two equalities above become

• • $g_{r,s}(g_{r \vee s}(a)) = g_r(a)$ ,

• • $h_{r,s}(g_r(a),g_s(a)) = g_{r \vee s}(a)$ .

For each $X \in \operatorname {\mathrm {Def}}({\mathcal M})$ , let $\alpha _X : L \longrightarrow \operatorname {\mathrm {Eq}}(X)$ be such that whenever $r \in L$ , then $\alpha _X(r)$ is the equivalence relation in $\operatorname {\mathrm {Eq}}(X)$ induced by $g_r \hspace {-4pt} \upharpoonright \hspace {-3pt} X$ . Let B be the set of all $x \in M$ such that

• • $g_{r,s}(g_{r \vee s}(x)) = g_r(x)$ ,

• • $h_{r,s}(g_r(x),g_s(x)) = g_{r \vee s}(x)$ ,

• • $g_1(x) = x$ .

Clearly, $B \in \operatorname {\mathrm {Def}}({\mathcal M})$ and $a \in B^{\mathcal N}$ . We claim that $\alpha _B$ is an ${\mathcal M}$ -representation of L. But even more is true. If $X \subseteq B$ , $X \in \operatorname {\mathrm {Def}}({\mathcal M})$ and $a \in X^{\mathcal N}$ , then $\alpha _X = \alpha _B|X$ .

We now claim that each such $\alpha _X$ is an ${\mathcal M}$ -representation of L.

First, $\alpha _X$ is one-to-one. For, suppose that $r,s \in L$ and $\alpha _X(r) = \alpha _X(s)$ . Then, $g_r \hspace {-4pt} \upharpoonright \hspace {-3pt} X$ and $g_s \hspace {-4pt} \upharpoonright \hspace {-3pt} X$ induce the same equivalence relations on X. It follows that there are ${\mathcal M}$ -definable functions $e_0,e_1 : M \longrightarrow M$ such that for all $x \in X$ , $e_0(g_r(x)) = g_s(x)$ and $e_1(g_s(x)) = g_r(x)$ . But then $e_0^{\mathcal N}(g_r(a)) = g_s(a)$ and $e_1^{\mathcal N}(g_s(a)) = g_r(a)$ , implying that $F(r) = F(s)$ and, therefore, $r = s$ .

Next, to prove that each $\alpha _X$ is a representation of L, it is enough to show that $\alpha _B$ is.

For all $x \in B$ , $g_0(x) = f(0)$ and $g_1(x) = x$ , so $\alpha _B(0)$ is trivial and $\alpha _B(1)$ is discrete. Finally, we show that if $r,s \in L$ , then $\alpha _B(r \vee s) = \alpha _B(r) \wedge \alpha _B(s)$ . To do so, we let $x,y \in X$ , and then show that $\langle x,y \rangle \in \alpha _B(r \vee s) \Longleftrightarrow \langle x,y \rangle \in \alpha _B(r) \cap \alpha _B(s)$ .

$$ \begin{align*} \langle x,y \rangle \in \alpha_B(r \vee s) & \Rightarrow g_{r \vee s}(x) = g_{r \vee s}(y) \nonumber \\ & \Rightarrow g_{r,s}(g_{r \vee s}(x)) = g_{r,s}(g_{r \vee s}(y)) \nonumber \\ & \Rightarrow g_r(x) = g_r(y) \nonumber \\ & \Rightarrow \langle x,y \rangle \in \alpha_B(r). \nonumber \end{align*} $$

Similarly, $\langle x,y \rangle \in \alpha _B(r \vee s) \Rightarrow \langle x,y \rangle \in \alpha _B(s)$ . Conversely,

$$ \begin{align*} \langle x,y \rangle \in \alpha_B(r) \cap \alpha_B(s) & \Rightarrow g_r(x) = g_r(y) \ \& \ g_s(x) = g_s(y) \nonumber \\ & \Rightarrow h_{r,s} (g_r(x),g_s(x)) = h_{r,s} (g_r(y),g_s(y)) \nonumber \\ & \Rightarrow g_{r\vee s}(x) = g_{r\vee s}(y) \nonumber \\ & \Rightarrow \langle x,y \rangle \in \alpha_B(r \vee s). \nonumber \end{align*} $$

Having that each $\alpha _X$ is a representation of L, we easily see that it is $0$ -CPP. For if X is partitioned into $Y,Z \in \operatorname {\mathrm {Def}}({\mathcal M})$ , then either $a \in Y^{\mathcal N}$ or $a \in Z^{\mathcal N}$ , but nodoubling both.

Now let ${\mathcal C}$ be the set of all such $\alpha _X$ ; that is,

$$ \begin{align*}{\mathcal C} = \{\alpha_X : X \subseteq B, X \in \operatorname{\mathrm{Def}}({\mathcal M}), a \in X^{\mathcal N}\}. \end{align*} $$

We have just seen that ${\mathcal C}$ is a nonempty set of $0$ -CPP ${\mathcal M}$ -representations of L, so that $(1)$ of Definition 1.3 is verified. We prove (2) of Definition 1.3. Consider $\alpha _X \in {\mathcal C}$ . Let $\Theta \in \operatorname {\mathrm {Eq}}(X)$ be ${\mathcal M}$ -definable. Define $m : X \longrightarrow X$ so that if $x \in X$ , then $m(x) = \min ([x]_\Theta )$ . Let $r \in L$ be such that $m^{\mathcal N}(a)$ generates $F(r)$ over ${\mathcal M}$ . There are functions $e_0,e_1 : N \longrightarrow N$ that are ${\mathcal N}$ -definable but using parameters only from M such that $e_0(m^{\mathcal N}(a)) = g_r(a)$ and $e_1(g_r(a)) = m^{\mathcal N}(a)$ . Let $Y = \{x \in X : e_0(m^{\mathcal N}(x)) = g_r(x)$ and $e_1(g_r(x)) = m^{\mathcal N}(x)\}$ . Then $m \hspace {-4pt} \upharpoonright \hspace {-3pt} Y$ induces $\alpha _Y(r)$ . This completes the proof of (1).

$(2)$ Since ${\mathcal C} \neq \varnothing $ , let $\alpha : L \longrightarrow \operatorname {\mathrm {Eq}}(A)$ be in ${\mathcal C}$ . Let $\Theta _0,\Theta _1,\Theta _2, \ldots $ enumerate all ${\mathcal M}$ -definable equivalence relations on M. By recursion, obtain a sequence $X_0 \supseteq X_1 \supseteq X_2 \supseteq \cdots $ of sets in $\operatorname {\mathrm {Def}}({\mathcal M})$ as follows. Let $X_0 = A$ . Suppose that we have $X_n$ and that $\alpha |X_n \in {\mathcal C}$ . Let $X_{n+1} \subseteq X_n$ be such that $\alpha |X_{n+1} \in {\mathcal C}$ and $\Theta _n \cap X_{n+1}^2$ is canonical for $\alpha $ . The $X_n$ ’s generate a complete type over ${\mathcal M}$ (using that each $\alpha |X_n$ is 0-CPP). Let ${\mathcal N}$ be an elementary extension of ${\mathcal M}$ generated by an element a realizing this type.

For each $r \in L$ , let $t_r : M \longrightarrow M$ be ${\mathcal M}$ -definable such that whenever $x \in X_0$ , then $t_r(x) = \min ([x]_{\alpha (r)})$ . Define the function F on L so that if $r \in L$ , then $F(r)$ is the elementary substructure of ${\mathcal N}$ generated by $t^{\mathcal N}_r(a)$ over ${\mathcal M}$ . One easily checks that $F : L \longrightarrow \operatorname {\mathrm {Lt}}({\mathcal N} / {\mathcal M})$ is an isomorphism.

1.3. Ranked lattices

To refine the notions of end/cofinal/mixed extensions, we appeal to rankings of lattices [Reference Kossak and Schmerl1, Definition 4.2.6]. Suppose that L is a finite lattice. A function $\rho : L \longrightarrow L$ is a ranking of L if for each $r,s \in L$ :

1. (1) $\rho (r) \geq r,$

2. (2) $\rho (\rho (r)) = \rho (r)$ ,

3. (3) $\rho (r) \leq \rho (s)$ or $\rho (s) \leq \rho (r)$ ,

4. (4) $\rho (r \vee s) = \rho (r) \vee \rho (s)$ .

A ranking $\rho $ of L uniquely determines, and is uniquely determined by, its rankset $\{\rho (r) : r \in L\}$ . If L is finite and $R \subseteq L$ , then R is a rankset iff R is linearly ordered and $1_L \in R$ . If $\rho $ is a ranking of L, then $(L,\rho )$ is a ranked lattice.

If ${\mathcal M} \prec {\mathcal N}$ and $\operatorname {\mathrm {Lt}}({\mathcal N} / {\mathcal M})$ is finite, then let $\rho : \operatorname {\mathrm {Lt}}({\mathcal N} / {\mathcal M}) \longrightarrow \operatorname {\mathrm {Lt}}({\mathcal N} / {\mathcal M})$ be such that if ${\mathcal K} \in \operatorname {\mathrm {Lt}}({\mathcal N} / {\mathcal M})$ , then $\rho ({\mathcal K})$ is uniquely defined by

$$ \begin{align*}{\mathcal K} \prec_{\mathsf{cf}} \rho({\mathcal K}) \prec_{\mathsf{end}} {\mathcal N}. \end{align*} $$

One easily verifies that $\rho $ is a ranking of $ \operatorname {\mathrm {Lt}}({\mathcal N} / {\mathcal M})$ . We let $\operatorname {\mathrm {Ltr}}({\mathcal N} / {\mathcal M})$ be the ranked lattice $(\operatorname {\mathrm {Lt}}({\mathcal N} / {\mathcal M}),\rho )$ .

Suppose that $\rho $ is a ranking of the finite lattice L. Then $\rho $ is an end ranking if $\rho (0_L) = 0_L$ , a cofinal ranking if $\rho (0_L) = 1_L$ and a mixed ranking if $0_L < \rho (0_L) < 1_L$ . Obviously, L has a unique cofinal ranking. If $\rho $ is an end, cofinal, or mixed ranking, then $(L,\rho )$ is, respectively, an end, cofinal, or mixed ranked lattice. These definitions are appropriate: if $\operatorname {\mathrm {Ltr}}({\mathcal N} / {\mathcal M})$ is an end, cofinal, or mixed ranked lattice, then ${\mathcal N}$ is, respectively, an end, cofinal, or mixed extension of ${\mathcal M}$ .

Of the $10$ rankings of $\mathbf {N}_5$ , four are depicted in Figure 2 by letting $\bullet $ denote those points in the rankset and $\circ $ those that are not. Of all the ranked pentagons, the four in Figure 2 are the most important for us because of the following.

Figure 2 Four ranked pentagon lattices.

Henceforth, we use the labeling of ${\mathbf N}_5$ as given in Figure 1.

Proposition 1.5. If ${\mathcal M} \prec {\mathcal N}$ and $\operatorname {\mathrm {Ltr}}({\mathcal N} / {\mathcal M}) \cong ({\mathbf N}_5, \rho )$ , then $\rho = \nu _i$ for some $i \leq 3$ .

Proof We first show that $\rho (c) = 1$ . If $\rho (c) \neq 1$ , then $\rho (c) = c$ . We apply the Gaifman Condition [Reference Kossak and Schmerl1, Proposition 4.2.12] by letting $x = a$ , $y = b$ , and $z =c$ , to get the contradiction that $a = b$ .

If $\rho (0) = 1$ , then $\rho = \nu _0$ . So, assume that $\rho (0) < 1$ . Since $\rho (c) = 1$ and $c \wedge b = 0$ , it follows from the Blass Condition [Reference Kossak and Schmerl1, Proposition 4.2.7] that $\rho (b) = b$ . Finally, $\rho (0) \neq b$ by [Reference Kossak and Schmerl1, Theorem 4.6.1]. Thus, $\rho (0) \in \{0,a\}$ , so it must be that $\rho \in \{\nu _1,\nu _2, \nu _3\}$ .

We make some comments about this proposition. First, Proposition 1.5 ${}^*$ is also valid. Theorem 1 can now be restated as: For all countable ${\mathcal M}$ there are $i \in \{1,2\}$ and ${\mathcal N} \succ {\mathcal M}$ such that $\operatorname {\mathrm {Ltr}}({\mathcal N} / {\mathcal M}) \cong ({\mathbf N}_5, \nu _i)$ . In fact, Wilkie’s proof of Theorem 1 yields that $i = 1$ . A similar proof shows that for every countable ${\mathcal M}$ there is ${\mathcal N} \succ _{\mathsf {end}} {\mathcal M}$ such that $\operatorname {\mathrm {Ltr}}({\mathcal N} / {\mathcal M}) \cong ({\mathbf N}_5, \nu _2)$ . Since $\nu _3$ is the only mixed ranking of the four in Figure 2, then in Theorem 4 we get ${\mathcal N}^*$ such that $\operatorname {\mathrm {Ltr}}({\mathcal N}^* / {\mathcal M}^*) \cong ({\mathbf N}_5, \nu _3)$ .

The next order of business is to generalize Definition 1.3 and Theorem 1.4 from lattices to ranked lattices.

First, we need some terminology. Suppose that ${\mathcal M}$ is a model, $A \in \operatorname {\mathrm {Def}}({\mathcal M})$ , and $\Theta \in \operatorname {\mathrm {Eq}}^{\mathcal M}(A)$ . We say that a set ${\mathcal E}$ of $\Theta $ -classes is ${\mathcal M}$ -bounded if there is a bounded $I \in \operatorname {\mathrm {Def}}({\mathcal M})$ such that $I \cap X \neq \varnothing $ for each $X \in {\mathcal E}$ .

If $(L,\rho )$ is a finite ranked lattice, then a representation $\alpha $ of L is a representation of $(L,\rho )$ if whenever $r \leq s \in L$ , then $s \leq \rho (r)$ iff every $\alpha (r)$ -class is the union of a finite set of $\alpha (s)$ -classes. This definition should help motivate the next definition.

Definition 1.6. Let ${\mathcal M}$ be a model and $(L,\rho )$ a finite ranked lattice.

(1) A representation $\alpha : L \longrightarrow \operatorname {\mathrm {Eq}}(A)$ is an ${\mathcal M}$ -representation of $(L,\rho )$ if $\alpha $ is an ${\mathcal M}$ -representation of L and whenever $r \leq s \in L$ , then $s \leq \rho (r)$ iff every $\alpha (r)$ -class is the union of an ${\mathcal M}$ -bounded set of $\alpha (s)$ -classes.

(2) We say that ${\mathcal C}$ is an ${\mathcal M}$ -correct set of representations of $(L,\rho )$ if ${\mathcal C}$ is an ${\mathcal M}$ -correct set of representations of L and each $\alpha \in {\mathcal C}$ is an ${\mathcal M}$ -representation of $(L,\rho )$ .

We next generalize Theorem 1.4 from lattices to ranked lattices.

Theorem 1.7. Suppose that ${\mathcal M}$ is a model and $(L,\rho )$ is a finite ranked lattice.

1. (1) If there is ${\mathcal N} \succ {\mathcal M}$ such that $\operatorname {\mathrm {Ltr}}({\mathcal N} / {\mathcal M}) \cong (L,\rho )$ , then there is an ${\mathcal M}$ -correct set of representations of $(L,\rho )$ .

2. (2) If ${\mathcal M}$ is countable and there is an ${\mathcal M}$ -correct set of representations of $(L,\rho )$ , then there is ${\mathcal N} \succ {\mathcal M}$ such that $\operatorname {\mathrm {Ltr}}({\mathcal N} / {\mathcal M}) \cong (L,\rho )$ .

Proof $(1)$ Obtain ${\mathcal C}$ as in the proof of Theorem 1.4(1), so that ${\mathcal C}$ is an ${\mathcal M}$ -correct set of representations of L. If $\alpha : L \longrightarrow \operatorname {\mathrm {Eq}}(A)$ is in ${\mathcal C}$ , $r \leq s$ but not $s \leq \rho (r)$ , then there is some $\alpha (r)$ -class that is not the union of an ${\mathcal M}$ -bounded set of $\alpha (s)$ -classes. (For, otherwise, there would be an ${\mathcal M}$ -definable function $b : M \longrightarrow M$ such that $b(f(r)) \geq f(s)$ .) However, it could be that $r \leq s \leq \rho (r)$ and some $\alpha (r)$ -class is not the union of an ${\mathcal M}$ -bounded set of $\alpha (s)$ -classes. Let ${\mathcal C}_0$ be the set of those $\alpha \in {\mathcal C}$ that are ${\mathcal M}$ -representations of $(L, \rho )$ . We will show that this ${\mathcal C}_0$ is an ${\mathcal M}$ -correct set of representations of $(L,\rho )$ . To see this, it suffices to show that for each $\alpha : L \longrightarrow \operatorname {\mathrm {Eq}}(A)$ in ${\mathcal C}$ , there is $B \subseteq A$ such that $\alpha |B \in {\mathcal C}_0$ .

Suppose that we have $\alpha : L \longrightarrow \operatorname {\mathrm {Eq}}(A)$ in ${\mathcal C}$ and that $r \leq s \leq \rho (r)$ . Partition A into two sets $A_0,A_1$ , so that $A_0$ is the union of those $\alpha (r)$ -classes that are the union of an ${\mathcal M}$ -bounded set of $\alpha (s)$ -classes. Since ${\mathcal C}$ is ${\mathcal M}$ -correct, then either $\alpha |A_0 \in {\mathcal C}$ or $\alpha |A_1 \in {\mathcal C}$ . By what was previously said, the latter option is impossible, so we have that $\alpha |A_0 \in {\mathcal C}$ . Repeating this for all such $r,s \in L$ , finally yields $B \subseteq A$ as required. This completes the proof of (1).

(2) Let ${\mathcal C}$ be an ${\mathcal M}$ -correct set of representations of $(L,\rho )$ . Then ${\mathcal C}$ is an ${\mathcal M}$ -correct set of representations of L, so we can obtain ${\mathcal N} \succ {\mathcal M}$ as in the proof of Theorem 1.4(2). Then $\operatorname {\mathrm {Lt}}({\mathcal N} / {\mathcal M}) \cong L$ .

We use the notation from the proof of Theorem 1.4(2). Thus, $F : L \longrightarrow \operatorname {\mathrm {Lt}}({\mathcal N} / {\mathcal M})$ is an isomorphism and $F(r)$ is generated by $t_r(a)$ over ${\mathcal M}$ . We prove that F is also an isomorphism of the ranked lattices $(L,\rho )$ and $\operatorname {\mathrm {Ltr}}({\mathcal N} / {\mathcal M})$ . It suffices to prove: whenever $r < s \in L$ , then $s \leq \rho (r)$ iff $F(r) \prec _{\mathsf {cf}} F(s)$ . So, let $r < s \in L$ .

$(\Longrightarrow )$ : Suppose that $s \leq \rho (r)$ . Consider $\alpha \in {\mathcal C}$ . Every $\alpha (r)$ -class is the union of an ${\mathcal M}$ -bounded set of $\alpha (s)$ -classes. Let $g : M \longrightarrow M$ be an ${\mathcal M}$ -definable function such that if $x \in X$ , then $g(x) = \max \{t_s(y) : \langle x,y \rangle \in \alpha (r)\}$ . Clearly, $g(x)$ is well defined for $x \in X$ , so there is such an ${\mathcal M}$ -definable g. Thus, $g(t_r(x)) \geq t_s(x)$ for all $x \in X$ , so that $g^{\mathcal N}(t_r^{\mathcal N}(a)) \geq t_s^{\mathcal N}(a)$ . Therefore, $F(r) \prec _{\mathsf {cf}} F(s)$ .

$(\Longleftarrow )$ : Suppose that $F(r) \prec _{\mathsf {cf}} F(s)$ . Then there is an ${\mathcal M}$ -definable $g : M \longrightarrow M$ such that $g^{\mathcal N}(t_r^{\mathcal N}(a)) \geq t_s^{\mathcal N}(a)$ . There is $X_i$ such that $g(t_r(x)) \geq t_s(x)$ for all $x \in X_i$ . Let $\alpha _i = \alpha |X_i \in {\mathcal C}$ . Thus, each $\alpha _i(r)$ -class is the union of an ${\mathcal M}$ -bounded set of $\alpha _i(s)$ -classes. Then $s \leq \rho (r)$ .

Wilkie’s proof of Theorem 1 made implicit use of Theorem 1.7(2).

2. Proving Theorem 3

This section is devoted to proving Theorem 3.

With the idea of obtaining a contradiction, assume that ${\mathcal M} \prec _{\mathsf {mix}} {\mathcal N}$ and that $\operatorname {\mathrm {Lt}}({\mathcal N} / {\mathcal M}) \cong \mathbf {N}_5$ . Proposition 1.4 implies that $\operatorname {\mathrm {Ltr}}({\mathcal N} / {\mathcal M})\ \cong (\mathbf {N}_5, \nu _3)$ . Following Theorem 1.7(1), we let ${\mathcal C}$ be an ${\mathcal M}$ -correct set of representations of $({\mathbf N}_5, \nu _3)$ . In the course of this proof, we will see that ${\mathcal C}$ must have certain properties. We will also see that there are other properties that ${\mathcal C}$ possibly could have, and we will then assume that ${\mathcal C}$ does have these properties.

Since ${\mathcal C} \neq \varnothing $ , fix some $\alpha \in {\mathcal C}$ . Thus, $\alpha : {\mathbf N}_5 \longrightarrow \operatorname {\mathrm {Eq}}(A)$ . We can assume:

1. (C1) For every $\beta \in {\mathcal C}$ , there is $B \subseteq A$ such that $\beta = \alpha | B$ .

Since $a \vee c = 1$ , then $\alpha (a) \cap \alpha (c) = 0 {\hspace{-6.3pt} 0}_A$ ; therefore, whenever X is an $\alpha (a)$ -class and Z is an $\alpha (c)$ -class, then $|X \cap Z| \leq 1$ . Since $0 < a = \nu _3(0)$ , then, according to Definition 1.6, the set of $\alpha (a)$ -classes is ${\mathcal M}$ -bounded; let $n +1\in M$ be the number of $\alpha (a)$ -classes according to ${\mathcal M}$ . Then, we can assume:

1. (C2) $\alpha : {\mathbf N}_5 \longrightarrow \operatorname {\mathrm {Eq}}(A)$ , where $A = [0,n] \times M$ and n is nonstandard, is such that if $\langle i,j \rangle , \langle i',j' \rangle \in A$ , then

   $$ \begin{align*} \big\langle \langle i,j \rangle, \langle i',j' \rangle \big\rangle \in \alpha(a) \mbox{ iff } i = i', \end{align*} $$
   $$ \begin{align*}\big\langle \langle i,j \rangle, \langle i',j' \rangle \big\rangle \in \alpha(c) \mbox{ iff } j = j'. \end{align*} $$

At first, it may look as if we can only assume that $A \subseteq [0,n] \times M$ . But it is always possible to enlarge the set A so as to get $[0,n] \times M$ .

For just this proof, let us say that the ${\mathcal M}$ -representation $\beta $ of ${\mathbf N}_5$ is rectangular if $|X \cap Z| = 1$ for each $\beta (a)$ -class X and $\beta (c)$ -class Z. We see from (C2) that $\alpha $ is rectangular. We can even assume:

1. (C3) Every $\beta \in {\mathcal C}$ is a rectangular representation.

To see why, let ${\mathcal C}_0$ be the set of those rectangular ${\mathcal M}$ -representations $\beta $ , where $B \subseteq A$ and $\beta = \alpha |B$ , for which there is $A_0 \subseteq B$ such that $\alpha |A_0 \in {\mathcal C}$ . To prove that this ${\mathcal C}_0$ is an ${\mathcal M}$ -correct set of representations of $({\mathbf N}_5,\nu _3)$ , it suffices to prove that if $A_1 \subseteq A$ and $\alpha | A_1\in {\mathcal C}$ , then there is $B \subseteq A_1$ such that $\alpha |B \in {\mathcal C}_0$ . To prove this, consider some $\alpha _1 = \alpha |A_1 \in {\mathcal C}$ . Define $\Theta \in \operatorname {\mathrm {Eq}}(A_1)$ so that if $y,z \in A_1$ , then $\langle y,z \rangle \in \Theta $ iff the following holds for each $\alpha _1(a)$ -class X: there is $u \in X$ such that $\langle u,y \rangle \in \alpha _1(c)$ iff there is $v \in X$ such that $\langle v,z \rangle \in \alpha _1(c)$ . Clearly, $\alpha _1(c) \subseteq \Theta \in \operatorname {\mathrm {Eq}}(A_1)$ . Since ${\mathcal C}$ is ${\mathcal M}$ -correct, there are $A_0 \subseteq A_1$ and $r \in \{0,c\}$ such that $\alpha |A_0 \in {\mathcal C}$ and $\alpha _1(r) \cap A_0^2 = \Theta \cap A_0^2$ . The number of $\Theta $ -classes is at most $2^{n+1}$ , so it must be that $r = 0$ . Let B be the union of those $\alpha _1(c)$ -classes that have a nonempty intersection with $A_0$ . Then $A_0 \subseteq B \subseteq A_1$ and $\beta = \alpha |B \in {\mathcal C}_0$ . This proves that ${\mathcal C}_0$ is an ${\mathcal M}$ -correct set of rectangular representations of $({\mathbf N}_5, \nu _3)$ , so we can assume (C3).

Moreover, we can also assume:

1. (C4) If $I \subseteq I' \subseteq [0,n]$ , $J \subseteq J' \subseteq M$ , and $\alpha |(I \times J) \in {\mathcal C}$ , then $\alpha |(I' \times J') \in {\mathcal C}$ .

Working in ${\mathcal M}$ , let $\langle B_k : k \in M \rangle $ be a one-to-one enumeration of all $\alpha (b)$ -classes. Thus, we let $\psi (u,v)$ be an ${\mathcal L}(M)$ -formula such that

$$ \begin{align*}{\mathcal M} \models \forall u,v[\psi(u,v) \leftrightarrow v \in B_u]. \end{align*} $$

We also let $q : [0,n] \times M \longrightarrow M$ be such that if $\langle i,j \rangle \in A$ , then $q(i,j) = k$ , where $\langle i,j \rangle \in B_k$ .

If $\beta : {\mathbf N}_5 \longrightarrow \operatorname {\mathrm {Eq}}(X)$ is in ${\mathcal C}$ and $p \in M$ , then there is $X' \subseteq X$ such that $\beta | X' \in {\mathcal C}$ and if $B_k \cap X' \neq \varnothing $ , then $k> p$ . To see why, just consider $\Theta \in \operatorname {\mathrm {Eq}}(X)$ such that $\bigcup \{B_k \cap X : k> p\}$ is a $\Theta $ -class and then apply Definitions 1.3 and 1.6.

For each $j \in M$ , there is a (unique) permutation $\pi _j$ of $[0,n]$ defined by the following condition: if $i,i' \leq n$ , then $\pi _j(i) \leq \pi _j(i')$ iff $q(i,j) \leq q(I',j)$ . Using these permutations, we define $\Psi \in \operatorname {\mathrm {Eq}}(A)$ so that $\big \langle \langle i,j \rangle , \langle i',j' \rangle \big \rangle \in \Psi $ iff $\pi _{j} = \pi _{j'}$ . Clearly, $\beta (c) \subseteq \Psi $ and the set of $\Psi $ -classes is ${\mathcal M}$ -bounded as ${\mathcal M}$ thinks that there are no more than $(n+1)!\ \Psi $ -classes. Thus, there are $I \times J \subseteq [0,n] \times M = A$ and $\pi $ such that $\alpha |(I \times J) \in {\mathcal C}$ and $\pi _j = \pi $ whenever $\langle i,j \rangle \in I \times J$ . Without loss of generality, we assume that $J = M$ and that $\pi $ is the identity permutation. Thus, we can assume:

1. (C5) $A = I \times J = [0,n] \times M$ , and if $i,i' \leq n$ and $j \in M$ , then

   $$ \begin{align*}i \leq i' \mbox{ iff } q(i,j) \leq q(i',j). \end{align*} $$

We now have that ${\mathcal C}$ is an ${\mathcal M}$ -correct set of representations of $({\mathbf N}_5,\nu _3)$ satisfying (C1)–(C5).

With (C5) in mind, we make a couple of definitions concerning a $\beta \in {\mathcal C}$ , where $\beta : {\mathbf N}_5 \longrightarrow \operatorname {\mathrm {Eq}}(I \times J)$ . Suppose that X and Y are $\beta (b)$ -classes. We say that X is below Y if there are $i,i' \in I$ and $j \in J$ such that $\langle i,j \rangle \in X$ , $\langle i',j \rangle \in Y$ , and $i < i'$ . If X is below Y, then Y is above X. Thus, (C5) says: If $B_k$ is below $B_{k'}$ (as $\alpha (b)$ -classes), then $k < k'$ . The following is a consequence of (C5):

1. (C6) For each $\alpha (b)$ -class, the set of $\alpha (b)$ -classes below it is ${\mathcal M}$ -bounded.

We next show that it can be assumed that $\alpha : {\mathbf N}_5 \longrightarrow \operatorname {\mathrm {Eq}}(A)$ has a very specific form. Recall that if $d \in M$ , then $M^d$ is the set (of codes) of all definable sequences of length d. For a given $d \in M$ , with $d> 0$ , we will think of $M^d = M$ . Also, $t \in M^d$ and $i < d$ , then $t_i$ is the i-th element of the sequence coded by t. If $t \in M^d$ and $e \leq d$ , then $t \hspace {-4pt} \upharpoonright \hspace {-3pt} e$ is the (code of) the sequence of length e whose i-th element is the same as the i-th element of t.

We can assume the following, which improves on (C2):

1. (C7) $\alpha : {\mathbf N}_5 \longrightarrow \operatorname {\mathrm {Eq}}(A)$ , where $A = [0,n] \times M^n$ , such that, in addition to (C2), we have if $\langle i,t \rangle , \langle i',t' \rangle \in A$ , then

   $$ \begin{align*}\big\langle \langle i,t \rangle, \langle i',t' \rangle \big\rangle\in \alpha(b) {\mbox{ iff }} i = i' {\mbox{ and }} t\hspace{-4pt} \upharpoonright \hspace{-3pt} i = t' \hspace{-4pt} \upharpoonright \hspace{-3pt} i'. \end{align*} $$

Let $\alpha $ and A be, for now, as in (C2). Let G be a function on A such that if $\langle i,j \rangle \in A$ , then

$$ \begin{align*}G(\langle i,j \rangle) = \{q(i',j) : i' \leq i\}. \end{align*} $$

Then G is definable in ${\mathcal M}$ . Let $\Theta \in \operatorname {\mathrm {Eq}}(A)$ be induced by G. We easily see that $\Theta \subseteq \alpha (b)$ . Thus, there is a definable $X \subseteq A$ such that $X = I \times J$ , $\alpha \hspace {-4pt} \upharpoonright \hspace {-3pt} X \in {\mathcal C}$ and $\alpha (r) \cap X^2 = \Theta \cap X^2$ for some $r \in {\mathbf N}_5$ , with $r \geq b$ . From (C6) we get that $r = b$ . We then see that if $j \in J$ , $i,i' \in I$ and $i < i'$ , then $B_{q( i,j )} \cap X \supseteq B_{q( i',j )} $ . It then follows that we can also assume (C7).

A consequence of (C7) is that $\alpha $ and q are defined by $\Sigma _0\ {\mathcal L}(\{n\})$ -formulas.

We saw in (C3) that every $\beta \in {\mathcal C}$ is rectangular. We also can assume the following:

1. (C8) If $\beta = \alpha |(I \times J) \in {\mathcal C}$ , then $\{\min (I)\} \times J$ is a $\beta (b)$ -class.

To see why, suppose that $\beta = \alpha |(I \times J) \in {\mathcal C}$ and $\{\min (I)\} \times J$ is not a $\beta (b)$ -class. Let $\Theta \in \operatorname {\mathrm {Eq}}(I \times J)$ be such that and if $\langle i,j \rangle , \langle i',j' \rangle \in I \times J$ , then $\big \langle \langle i,j \rangle , \langle i',j' \rangle \big \rangle \in \Theta $ iff $\big \langle \langle \min (I),j \rangle , \langle \min (I),j' \rangle \big \rangle \in \beta (b)$ . Clearly, $\Theta \in \operatorname {\mathrm {Eq}}^{\mathcal M}(I \times J)$ . Let $X \subseteq I \times J$ be such that $\beta |X \in {\mathcal C}$ and for some $r \in {\mathbf N}_5$ , $\Theta \cap X^2 = \beta (r) \cap X^2$ . It must be that $r = 0$ , so there is a $\beta (b)$ -class B such that $X \subseteq B \times J$ . Then, we can assume that $\beta \not \in {\mathcal C}$ and that $\beta |(B \times J) \in {\mathcal C}$ .

We will refer to the $\beta (b)$ -class from (C8) as the root of $\beta $ .

Consider some $\beta : {\mathbf N}_5 \longrightarrow \operatorname {\mathrm {Eq}}(X)$ in ${\mathcal C}$ , where $X = I \times J \subseteq [0,n] \times M$ . For each $m < \omega $ , we say that $\beta $ is m-thick if whenever $f : M \longrightarrow M$ is definable, then there are $i_0,i_1, \ldots ,i_m \in I$ and $j \in J$ such that $f(q(i_k,j)) < q(i_{k+1},j)$ for each $k < m$ .

Proof Notice that every $\beta \in {\mathcal C}$ is vacuously $0$ -thick.

First, we show that every $\beta : {\mathbf N}_5 \longrightarrow \operatorname {\mathrm {Eq}}(B)$ in ${\mathcal C}$ is $1$ -thick. By Corollary 2.1, there is a $\beta (b)$ -class $B \cap B_{k_0}$ having an ${\mathcal M}$ -unbounded set of $\beta (b)$ -classes above it. Thus, for any ${\mathcal M}$ -definable $f : M \longrightarrow M$ , there is a $\beta (b)$ -class $B \cap B_{k_1}$ above $B \cap B_{k_0}$ such that $f(k_0) < k_1$ . Let $i_0,i_1,j$ be such that $\langle i_0,j \rangle \in B \cap B_{k_0}$ and $\langle i_1,j \rangle \in B \cap B_{k_1}$ . Then $f(q(i_0,j)) = k_0 < k_1 = q(i_1,j)$ .

Next, we assume that $1 < m < \omega $ . We will prove that every $\beta \in {\mathcal C}$ is m-thick. Actually, we will prove something even stronger:

If $\beta : {\mathbf N}_5 \longrightarrow \operatorname {\mathrm {Eq}}(B)$ is in ${\mathcal C}$ and $f : M \longrightarrow M$ is definable, then there is $I \times J \subseteq B$ such that $\beta | (I \times J) \in {\mathcal C}$ and whenever $i,i' \in I$ , $j \in J$ and $i < i'$ , then $f(q(i,j)) < q(i',j)$ .

To prove this, suppose that $\beta : {\mathbf N}_5 \longrightarrow \operatorname {\mathrm {Eq}}(B)$ is in ${\mathcal C}$ and $f : M \longrightarrow M$ is definable. We assume that $B = I_0 \times J_0$ and, without loss, that whenever $i < i' \in M$ , then, then $\max (i',f(i)) < f(i')$ .

We will obtain I and J in two steps.

In the first step, for each $j \in J_0$ , let $R_j \subseteq I_0^2$ be such that if $\langle i,i' \rangle \in I_0^2$ , then

$$ \begin{align*}\langle i,i' \rangle \in R_j {\mbox{ iff }} f(q(i,j)) < q(i',j). \end{align*} $$

Let $\Theta _0 \in \operatorname {\mathrm {Eq}}(B)$ be such that if $i,i' \in I_0$ and $j,j' \in J_0$ , then

$$ \begin{align*}\big\langle \langle i,j \rangle, \langle i',j' \rangle \big\rangle \in \Theta_0 \mbox{ iff } R_j = R_{j'}. \end{align*} $$

Obviously, $\beta (c) \subseteq \Theta _0 \in \operatorname {\mathrm {Eq}}^{\mathcal M}(B)$ and the set of $\Theta _0$ -classes is ${\mathcal M}$ -bounded. Thus, there is $B_1 = I_1 \times J_1 \subseteq B$ such that $\beta | B_1 \in {\mathcal C}$ and $\Theta _0\cap B_1^2 $ is trivial. Notice that if $j,j' \in J_1$ , then $R_j \cap I_1^2 = R_{j'} \cap I_1^2$ . For some (or all) $j \in J_1$ , let $R = R_j \cap I_1^2$ .

For the second step, define $\Theta _1 \in \operatorname {\mathrm {Eq}}(B_1)$ so that if $i,i' \in I_1$ and $j,j' \in J_1$ , then

$$ \begin{align*}\big\langle \langle i,j \rangle, \langle i',j' \rangle \big\rangle \in \Theta_1 \mbox{ iff } \{k \in I_1 : \langle i,k \rangle \in R\} = \{k \in I_1 : \langle i',k \rangle \in R\}. \end{align*} $$

Obviously, $(\beta | B_1)(a) \subseteq \Theta _1 \in \operatorname {\mathrm {Eq}}^{\mathcal M}(B_1)$ . Thus, there is $B_2 = I \times J \subseteq B_1$ such that $\beta | B_2 \in {\mathcal C}$ and $\Theta _1 \cap B_2^2 \in \{\beta (a) \cap B_2^2, \beta (0) \cap B_2^2\}$ .

We show that $\Theta _1 \cap B_2^2 \neq \beta (0) \cap B_2^2$ . Assume to the contrary that $\Theta _1 \cap B_2^2 = \beta (0) \cap B_2^2$ . Then, whenever $i,i' \in I$ and $i < i'$ , then $f(q(i,j)) \geq f(q(i',j))$ . But then $\beta |((I \backslash \max (I)) \times J)$ is in ${\mathcal C}$ but is not $1$ -thick, which is a contradiction.

Therefore, $\Theta _1 \cap B_2^2 = \beta (0) \cap B_2^2$ . It then follows that $I \times J$ has the desired property.

Lemma 2.3 implies a strengthening of itself via Corollary 2.2.

We need some more notation and terminology. Recall that a cut K (of ${\mathcal M}$ ) is a subset of M such that $0 \in K \neq M$ and that $x + 1 \in K$ whenever $x \leq y \in K$ . If $m < \omega $ , then the cut K is $\Sigma _m$ -closed iff whenever $\varphi (x)$ is a $\Sigma _m\ {\mathcal L}(K)$ -formula and ${\mathcal M} \models \exists x \varphi (x)$ , then there is $d \in K$ such that ${\mathcal M} \models \varphi (d)$ . If K is a $\Sigma _0$ -closed cut and $\varphi $ is an ${\mathcal L}(K)$ -formula, then $\ulcorner \varphi \urcorner $ , the Gödel number of $\varphi $ , is in K.

Also, recall that $\alpha : {\mathbf N}_5 \longrightarrow \operatorname {\mathrm {Eq}}(A)$ , $A = [0,n] \times M$ and $\langle B_k : k \in M \rangle $ is a definable, one-to-one enumeration of the $\alpha (b)$ -classes (as defined a few lines after (C3)).

We work in ${\mathcal M}$ . For each $k \leq n$ , let $\Lambda _k$ be the set of all prenex ${\mathcal L}(M)$ -sentences $\sigma $ having length at most n and having the form

(*) $$ \begin{align} \sigma = \mathsf Q_1 x_1 \mathsf Q_2 x_2 \cdots {\mathsf Q}_ix_i \cdots \mathsf Q_{\ell} x_{\ell} \varphi(\overline x), \end{align} $$

where $\ell < k$ , each ${\mathsf Q}_i$ is either $\exists $ or $\forall $ and $\varphi (\overline x)$ is a $\Sigma _0$ formula. If $\ell < k \leq n$ , then $\Lambda _\ell \subseteq \Lambda _k$ . Let $\Lambda = \Lambda _n = \bigcup _{k \leq n}\Lambda _k$ . If $ \ell < n$ , $t = \langle t_0,t_1, \ldots ,t_\ell \rangle \in [0,n]^{\ell +1}$ , $j \in M$ , and $\sigma \in \Lambda $ as in $(*)$ , let

(**) $$ \begin{align} \sigma^{(t,j)} = \mathsf Q_1 x_1 \leq q(t_1,j)\mathsf Q_2 x_2 \leq q(t_2,j) \cdots \mathsf Q_{\ell} x_{\ell} \leq q(t_{\ell},j) \varphi(\overline x). \end{align} $$

There is an ${\mathcal L}(\{n\})$ -formula, call it $\varphi _0(x)$ , such that for every standard $\sigma \in \Lambda $ and each t and j as in $(**)$ ,

(1) $$ \begin{align} {\mathcal M} \models \varphi_0(\ulcorner \sigma^{(t,j)} \urcorner) \leftrightarrow \sigma^{(t,j)}. \end{align} $$

Let $g : M \longrightarrow M$ be an increasing, ${\mathcal L}(\{n\})$ -definable function that is sufficiently fast-growing in the following sense: whenever $a \in M$ , then there is a $\Sigma _0$ -closed cut K such that $\{n,a\} \subseteq K \subseteq [0,g(a)]$ . (There is such a g since n is nonstandard.) Notice that $g(0)> n^n$ ; in particular, $g(0)> n^k$ for every $k < \omega $ .

Continue working in ${\mathcal M}$ . Define $F : A \longrightarrow M$ so that if $\langle i,j \rangle \in A$ , then $F(\langle i,j \rangle )$ is the set of all pairs $\langle t,\sigma \rangle $ such that for some $\ell < n$ and $t \in [0,n]^{\ell +1}$ , $t_0 = i$ , $\sigma $ is an ${\mathcal L}([0,g(q(i,j)])$ in $\Lambda $ (as in $(*)$ ) and that $\sigma ^{(t,j)}$ is true. Let $\Theta \in \operatorname {\mathrm {Eq}}(A)$ be induced by F. It is clear that if $\big \langle \langle i,j \rangle , \langle i',j' \rangle \big \rangle \not \in \alpha (b)$ , then $F(\langle i,j \rangle ) \neq F(\langle i,j \rangle )$ so that $\big \langle \langle i,j \rangle , \langle i',j' \rangle \big \rangle \not \in \Theta $ . Thus, $\Theta \subseteq \alpha (b)$ . On the other hand, if $B_k$ is an $\alpha (b)$ -class, then the set of possible $F(\langle i,j \rangle )$ for $\langle i,j \rangle \in B_k$ is ${\mathcal M}$ -bounded. Thus, there is $B = I \times J \subseteq A$ such that $\alpha |B$ is in ${\mathcal C}$ and $\alpha (b) \cap B^2 = \Theta \cap B^2$ . Let $\beta = \alpha |B$ .

By Lemma 2.3, there are $j \in J$ , a sufficiently large $\ell < \omega $ , $t = \langle t_0,t_1, \ldots ,t_\ell \rangle \in I^{\ell + 1}$ and a sufficiently fast-growing, definable $f : M \longrightarrow M$ such that $f(q(t_k,j)) < q(t_{k+1},j)$ . By “sufficiently,” we mean that for every standard $\Sigma _\ell \ {\mathcal L}([0,g(q(t_0,j))])$ -sentence in $\Lambda $ , we have that

(2) $$ \begin{align} {\mathcal M} \models \sigma \leftrightarrow \sigma^{(t,j)}. \end{align} $$

Thus, it follows from (1) and (2) that for each such $\sigma $ , that

(3) $$ \begin{align} {\mathcal M} \models \varphi_0(\ulcorner \sigma^{(t,j)} \urcorner) \leftrightarrow \sigma. \end{align} $$

Now let K be the smallest $\Sigma _0$ -closed cut such that $n,q(t_0,j) \in K$ . Thus we have that $K \subseteq [0,g(q(t_0,j))]$ . From (3), we get that there is a $\Sigma _\ell \ {\mathcal L}(K)$ -formula $\varphi (x)$ such that for every $\Sigma _{\ell -1}\ {\mathcal L}(K)$ -sentence $\sigma $ ,

(4) $$ \begin{align} {\mathcal M} \models \varphi(\ulcorner \sigma \urcorner) \leftrightarrow \sigma. \end{align} $$

But the existence of $\varphi (x)$ in (4) contradicts the following version of Tarski’s Theorem on the undefinability of truth, which is an immediate consequence of Gödel’s Diagonalization Lemma.

This contradiction completes the proof of Theorem 3.

3. Representations of ${\mathbf N}_5$

For almost all of this section, we ignore PA and concentrate just on representations of ${\mathbf N}_5$ . Only in the first and last paragraphs is PA considered.

Caveat lector: In the next definition, and throughout this paper, $\omega ^n$ is not an ordinal but is the set of n-tuples of natural numbers. If $s \in \omega ^n$ and $i < n$ , then $s_i$ is the i-th element of s. Also, remember that if $n < \omega $ , then $n = \{0,1, \ldots , n-1\}$ . If $s \in \omega ^n$ and $i < m \leq n$ , then $s \hspace {-4pt} \upharpoonright \hspace {-3pt} m \in \omega ^m$ and $(s \hspace {-4pt} \upharpoonright \hspace {-3pt} m)_i = s_i$ .

Proof Suppose that $m,n,I$ are as given. Let $i_{m+1} = \max (I)$ and $I \backslash \{i_{m+1}\} = \{i_0,i_1, \ldots , i_m\}$ , where $i_0 < i_1 < \cdots < i_m$ . We consider separately the two cases: ${i_{m+1} < n+1}$ and $i_{m+1} = n+1$ .

First, suppose that $i_{m+1} < n+1$ .

$$ \begin{align*}D = \{t \in \omega^{n+1} : t_i = 0 {\mbox{ whenever }} i = i_{m+1} {\mbox{ or }} i \in (n+1) \backslash I \}. \end{align*} $$

We show that $\alpha _m \cong \alpha _n|(I \times D)$ . Let $h : (m+2) \times \omega ^{m+1} \longrightarrow (n+2) \times \omega ^{n+1}$ be such that if $\langle j, s \rangle \in (m+2) \times \omega ^{m+1}$ , then $h(\langle j,s \rangle ) = \langle i_j,t \rangle $ , where $t \in \omega ^{n+1}$ and

$$ \begin{align*} t_k = \begin{cases} s_j, {\mbox{ if }} j \leq m {\mbox{ and }} i_j = k, \\ 0, {\mbox{ otherwise, }} \end{cases} \end{align*} $$

for all $k \leq n$ . One easily verifies that $h : (m+2) \times \omega ^{m+1} \longrightarrow I \times D$ is a bijection and that whenever $\langle j,s \rangle , \langle j',s' \rangle \in (m+2) \times \omega ^{m+1}$ and $r \in {\mathbf N}_5$ , then

$$ \begin{align*}\big\langle \langle j,s \rangle, \langle j',s' \rangle \big\rangle \in \alpha_m(r) \Longleftrightarrow \big\langle \langle h(\langle j,s \rangle), h(\langle j',s' \rangle) \big\rangle \in \alpha_n(r). \end{align*} $$

Thus, $\alpha _m \cong \alpha _n|(I \times D)$ .

Next, suppose that $i_{m+1} = n+1$ . In this case, let

$$ \begin{align*}D = \{t \in \omega^{n+1} : t_i = 0 {\mbox{ whenever }} i \in (n+1) \backslash I \}. \end{align*} $$

Showing that $\alpha _m \cong \alpha _n|(I \times D)$ is much like in the first case.

Our primary goal in this section is to prove the following theorem, which will be given a more precise formulation in Theorem 3.9.

To prove this theorem, we will take a detour and visit some other lattices and their representations. These lattices are introduced in Definition 3.4 and their representations in Definition 3.5.

Definition 3.4. Suppose that $1 \leq m \leq n < \omega $ . Let $G_{m,n}$ be the set consisting of all pairs $\langle \theta , f \rangle $ , where $\theta \in \operatorname {\mathrm {Eq}}(n+1)$ and $f : n+1 \longrightarrow m+1$ are such that if $i,j \leq n$ and $\langle i,j \rangle \in \theta $ , then $f(i) = f(j)$ . Let $\unlhd $ be the partial ordering of $G_{m,n}$ such that if $\langle \theta , f \rangle , \langle \psi , g \rangle \in G_{m,n}$ , then

$$ \begin{align*} \langle \theta, f \rangle \unlhd \langle \psi, g \rangle {\mbox{ iff }} \theta \supseteq \psi {\mbox{ and }} f(i) \leq g(i) {\mbox{ for all }} i \leq n. \end{align*} $$

Clearly, $\unlhd $ really is a partial ordering. It should be observed that $G_{m,n}$ with $\unlhd $ , as in Definition 3.4, is a lattice in which

$$ \begin{align*}0_{G_{m,n}} = \langle 1 {\hspace{-3.5pt} 1}_{n+1},0 \rangle, \end{align*} $$

$$ \begin{align*}1_{G_{m,n}} = \langle 0 {\hspace{-6.3pt} 0}_{n+1},m \rangle, \end{align*} $$

$$ \begin{align*}\langle \theta,f \rangle \vee \langle \psi, g \rangle = \langle \theta \cap \psi,\sup(f,g) \rangle, \end{align*} $$

where $\sup (f,g) = h$ iff $h(i) = \max (f(i),g(i))$ for all $i \leq n$ . In the above equalities, we are identifying $k \leq m$ with the function that is constantly k on $n+1$ . We will continue to do so.

Our real concern is with the lattices $G_n = G_{n,n}$ . The more general $G_{m,n}$ are introduced in order to be able to do an inductive proof. One of the reasons for introducing the lattices $G_n$ is that there is an embedding $e_n : {\mathbf N}_5 \longrightarrow G_n$ defined by:

$$ \begin{align*} e_n(0) & = 0_{G_n}, \\ e_n(a) & = \langle 0 {\hspace{-6.3pt} 0}_{n+1}, 0 \rangle, \\ e_n(b) & = \langle 0 {\hspace{-6.3pt} 0}_{n+1}, \mathsf{id}_{n+1}\rangle, \\ e_n(c) & = \langle 1 {\hspace{-3.5pt} 1} _{n+1},n \rangle, \\ e_n(1) & = 1_{G_n}. \end{align*} $$

As usual, $\mathsf {id}_X$ is the identity function on X.

It is routine to verify that each $e_n$ is an embedding. Figure 3 depicts the lattice $G_1$ with ${\mathbf N}_5$ embedded in it. If $r \in {\mathbf N}_5$ , then $e_1(r)$ is labeled with r. The unlabeled point is $\langle 0 {\hspace{-6.3pt} 0}_2, 1-\mathsf {id}_2 \rangle $ , where $1-\mathsf {id}_2$ is the function $f : 2 \longrightarrow 2$ such that $f(0) = 1$ and $f(1) = 0$ .

Figure 3 Embedding ${\mathbf N}_5$ into $G_1$ .

Next, we define representations of the $G_{m,n}$ .

Observe that $\gamma _{m,n}$ is indeed a representation of $G_{m,n}$ . However, no $\gamma _{m,n}$ is $0$ -CPP since if $\theta $ has exactly 2 equivalence classes, then $\gamma _{m,n}(\langle \theta , 0 \rangle )$ has exactly 2 equivalence classes. In fact, if E is a $\theta $ -class, then $E \times \omega ^m$ is a $\gamma _{m,n}(\langle \theta , 0 \rangle )$ -class. Thus, the number of $\gamma _{m,n}(\langle \theta , 0 \rangle )$ -classes is equal to the number of $\theta $ -classes. On the other hand, if $f : n+1 \longrightarrow m+1$ is not constantly $0$ , then there are infinitely many $\gamma _{m,n}(\langle \theta ,f \rangle )$ -classes; specifically, if $f(i)> 0$ and $s_0 \neq t_0$ , where $s,t \in \omega ^m$ , then $\big \langle \langle i,s \rangle , \langle i,t \rangle \big \rangle \not \in \gamma _{m,n}(\langle \theta ,f \rangle )$ .

Let $\gamma _n = \gamma _{n,n}$ . Note that for $ n < \omega $ , both of the representations $\gamma _{n+1}$ and $\alpha _n$ are into $\operatorname {\mathrm {Eq}}\big ((n+2) \times \omega ^{n+1}\big )$ . In fact, even more is true.

We next come to the main result about the representations $\gamma _{m,n}$ .

Proof What this lemma says is that if $1 \leq m \leq n < \omega $ and $\Theta \in \operatorname {\mathrm {Eq}}\big ((n+1) \times \omega ^m\big )$ , then there is $X \subseteq (n+1) \times \omega ^m$ and $r \in G_{m,n}$ such that $\gamma _{m,n}|X \cong \gamma _{m,n}$ and $\Theta \cap X^2 = \gamma _{m,n}(r) \cap X^2$ . Observe that if $X \subseteq (n+1) \times \omega ^m$ and $\gamma _{m,n}|X \cong \gamma _{m,n}$ , then there is $D \subseteq \omega ^m$ such that $X = (n+1) \times D$ . This follows from the fact that each $\gamma _{m,n}(\langle 1 {\hspace {-3.5pt} 1}_{n+1},m \rangle )$ -class has the form $(n+1) \times \{s\}$ for some $s \in \omega ^m$ .

The proof of the lemma is by induction on $m \geq 1$ with $n \geq m$ being fixed. The basis step is for $m = 1$ and the inductive step for $m> 1$ . Both steps start out the same way. So for now, consider $n \geq m \geq 1$ and let $\Theta \in \operatorname {\mathrm {Eq}}\big ((n+1) \times \omega ^m\big )$ .

Consider an arbitrary $s \in \omega ^{m-1}$ . (Of course, if $m = 1$ , then $s = \varnothing $ is the only choice.) With the idea of invoking Infinite Ramsey’s Theorem for pairs, we define $F_s : [\omega ]^2 \longrightarrow \operatorname {\mathrm {Eq}}\big (\{0,1\} \times (n+1)\big )$ so that whenever $\{k,\ell \} \in [\omega ]^2$ , $k < \ell $ , $e,e' \in \{0,1\}$ and $i,j \leq n$ , then $\big \langle \langle e,i \rangle , \langle e',j \rangle \big \rangle \in F_s(\{k,\ell \})$ iff there are $t,t' \in \omega ^m$ such that ${t_{m-1} = k}$ , $t^{\prime }_{m-1} = \ell $ , $t \hspace {-4pt} \upharpoonright \hspace {-3pt} (m-1) = t' \hspace {-4pt} \upharpoonright \hspace {-3pt} (m-1) = s$ and one of the following:

• • $e = e' = 0$ and $\big \langle \langle i,t \rangle , \langle j,t \rangle \big \rangle \in \Theta $ ;

• • $e = 0$ , $e' = 1$ and $\big \langle \langle i,t \rangle , \langle j,t' \rangle \big \rangle \in \Theta $ ;

• • $e = 1$ , $e' = 0$ and $\big \langle \langle i,t' \rangle , \langle j,t \rangle \big \rangle \in \Theta $ ;

• • $e = e' = 1$ and $\big \langle \langle i,t' \rangle , \langle j,t' \rangle \big \rangle \in \Theta $ .

Now we apply Ramsey to get an infinite $H_s \subseteq \omega $ such that $F_s \hspace {-4pt} \upharpoonright \hspace {-3pt} [H_s]^2$ is constant. Let

$$ \begin{align*}Y_s = \{t \in \omega^m : t \supseteq s \mbox{ and } t_{m-1} \in H_s\}. \end{align*} $$

Each of the following is true for each $s \in \omega ^{m-1}$ :

1. (1) If $i \leq n$ , then $\Theta \cap (\{i\} \times Y_s)^2$ is either trivial or discrete.

2. (2) If $i, j \leq n$ , $\Theta \cap ( \{i\} \times Y_s)^2$ is trivial, $\Theta \cap (\{j\} \times Y_s)^2$ is discrete, and ${t,t' \in Y_s}$ , then $\big \langle \langle i,t \rangle , \langle j,t' \rangle \big \rangle \not \in \Theta $ .

3. (3) If $i < j \leq n$ and both $\Theta \cap (\{i\} \times Y_s)^2$ and $\Theta \cap (\{j\} \times Y_s)^2$ are discrete, then one of the following:

   1. (3a) if $t,t' \in Y_s$ , then $\big \langle \langle i,t \rangle , \langle j,t' \rangle \big \rangle \not \in \Theta $ ;

   2. (3b) if $t,t' \in Y_s$ , then $\big \langle \langle i,t \rangle , \langle j,t' \rangle \big \rangle \in \Theta $ iff $t = t'$ .

A consequence of (1)–(3) is:

1. (4) If $i,j \leq n$ and $t,t' \in Y_s$ , then

   $$ \begin{align*}\big\langle \langle i,t \rangle, \langle j,t \rangle \big\rangle \in \Theta \Longleftrightarrow \big\langle \langle i,t' \rangle, \langle j,t' \rangle \big\rangle \in \Theta. \end{align*} $$

Let $T_s = \{i \leq n : \Theta \cap ( \{i\} \times Y_s)^2$ is trivial $\}$ . Because of (1), $(n+1) \backslash T_s = \{i \leq n : \Theta \cap ( \{i\} \times Y_s)^2$ is discrete $\}$ . With (4) in mind, we can let $\theta _s \in \operatorname {\mathrm {Eq}}(n+1)$ be such that

$$ \begin{align*}\theta_s = \{\langle i,j \rangle \in (n+1) \times (n+1) : \big\langle \langle i,t \rangle, \langle j,t \rangle \big\rangle \in \Theta\} \end{align*} $$

for each $t \in Y_s$ . Clearly, $T_s$ is the (possibly empty) union of some $\theta _s$ -classes.

Let

$$ \begin{align*}D_0 = \bigcup\{Y_s : s \in \omega^{m-1}\} \end{align*} $$

and

$$ \begin{align*}X_0 = (n+1) \times D_0. \end{align*} $$

It is readily seen that $\gamma _{m,n} | X_0 \cong \gamma _{m,n}$ .

Basis step $m = 1$ : Since $m=1$ , it must be that $s = \varnothing $ . Thus, $Y_s$ is an infinite subset of $\omega ^1$ and $X_0 = (n+1) \times Y_{\varnothing }$ . We have already noted that $\gamma _{1,n}|X_0 \cong \gamma _{1,n}$ . To complete this step, we need to show that there is $r_0 \in G_{1,n}$ such that $\Theta \cap X_0^2 = \gamma _{1,n}(r_0) \cap X_0^2$ .

Let $f : n+1 \longrightarrow 2$ be such that $f(i) = 0$ iff $i \in T_{\varnothing }$ . Then we can take ${r_0 = \langle \theta _\varnothing ,f \rangle }$ . One easily verifies that $\Theta \cap X_0^2 = \gamma _{1,n}(r_0) \cap X_0^2$ .

Inductive step $m> 1$ : Thus, we are assuming $\gamma _{m-1,n} \longrightarrow \gamma _{m-1,n}$ . We already have infinite $Y_s$ for each $s \in \omega ^{m-1}$ and that (1)–(4) hold. Also, we have $D_0 \subseteq \omega ^m$ and $X_0 = (n+1) \times D_0$ and that $\gamma _{m,n} | X_0 \cong \gamma _{m,n}$ . Without loss of generality, we assume, for each $s \in \omega ^{m-1}$ , that $H_s = \omega $ and then $Y_s = \{t \in \omega ^m : t \supseteq s\}$ . Thus, $D_0 = \omega ^m$ and $X_0 = (n+1) \times \omega ^m$ .

Let $\Theta _1 \in \operatorname {\mathrm {Eq}}\big ((n+1) \times \omega ^{m-1}\big )$ be such that if $i,j \leq n$ and $s,s' \in \omega ^{m-1}$ , then $\big \langle \langle i,s \rangle , \langle j,s' \rangle \big \rangle \in \Theta _1$ iff $T_s = T_{s'}$ and $\theta _s = \theta _{s'}$ . By the inductive hypothesis, there are $r_1 \in G_{m-1,n}$ , $D_1 \subseteq \omega ^{m-1}$ , and $X_1 = (n+1) \times D_1$ such that $\gamma _{m-1,n} | X_1 \cong \gamma _{m-1,n}$ and $\Theta _1 \cap X_1^2 = \gamma _{m-1,n}(r_1) \cap X_1^2$ . Since there are only finitely many $\Theta _1$ -classes, it must be that $r_1 = \langle \psi ,0 \rangle $ for some $\psi \in \operatorname {\mathrm {Eq}}(n+1)$ . Thus, we have, for $i,j \leq n$ and $s,s' \in D_1$ , that

$$ \begin{align*} \big\langle \langle i,s \rangle, \langle j,s' \rangle \big\rangle \in \Theta_1 & \Longleftrightarrow T_s = T_{s'} {\mbox{ and }} \theta_s = \theta_{s'} \\ & \Longleftrightarrow \langle i,j \rangle \in \psi. \end{align*} $$

This implies that $\psi $ is trivial and that there are T and $\theta $ such that $T_s = T$ and $\theta _s = \theta $ whenever $s \in D_1$ .

Without loss of generality, we will assume that $D_1 = \omega ^{m-1}$ so that $X_1 = (n+1) \times \omega ^{m-1}$ . Notice that (1)–(4) remain true and, in addition, the following hold:

1. (5) If $i \leq n$ , then $\Theta \cap (\{i\} \times Y_s)^2$ is trivial iff $i \in T$ .

2. (6) If $i,j \leq n$ and $t \in \omega ^m$ , then $\langle i,j \rangle \in \theta $ iff $\big \langle \langle i,t \rangle , \langle j,t \rangle \big \rangle \in \Theta $ .

Let $\Theta _2 \in \operatorname {\mathrm {Eq}}\big ((n+1) \times \omega ^{m-1}\big )$ be such that if $i,j \leq n$ and $s,s' \in \omega ^{m-1}$ , then $\big \langle \langle i,s \rangle , \langle j,s' \rangle \big \rangle \in \Theta _2$ iff one of the following:

• • $i,j \not \in T$ , $\langle i,j \rangle \in \theta $ and $s = s'$ ;

• • $i,j \in T$ and for some (or, equivalently, all) $t \in Y_s$ and $t' \in Y_{s'}$ , $\big \langle \langle i,t \rangle , \langle j,t' \rangle \big \rangle \in \Theta $ .

One easily verifies that, indeed, $\Theta _2 \in \operatorname {\mathrm {Eq}}\big ((n+1) \times \omega ^{m-1}\big )$ . By the inductive hypothesis, there are $D_2 \subseteq \omega ^{m-1}$ , $X_2 = (n+1) \times D_2$ and $r_2 \in G_{m-1,n}$ such that $\gamma _{m-1,n} | X_2 \cong \gamma _{m-1,n}$ and $\Theta _2 \cap X_2^2 = \gamma _{m-1,n}(r_2) \cap X_2^2$ . Let $r_2 = \langle \varphi , f' \rangle $ . It must be that $\varphi = \theta $ .

Now let $D_3 = \{t \in \omega ^m : t \hspace {-4pt} \upharpoonright \hspace {-3pt} (m-1) \in D_2\}$ and $X_3 = (n+1) \times D_3$ . Then $\gamma _{m,n} | X_3 \cong \gamma _{m,n}$ . Let $r_3 = \langle \theta ,f \rangle \in G_{m,n}$ , where $f(i) = f'(i)$ if $i \in T$ and $f(i) = m$ if $i \not \in T$ . Although it is not clear if $\Theta \cap X^2_3 = \gamma _{m,n}(r_3) \cap X_3^2$ , we do have

$$ \begin{align*}\Theta \cap (T \times D_3)^2 = \gamma_{m,n}(r_3) \cap (T \times D_3)^2. \end{align*} $$

Without loss of generality, let $D_3 = \omega ^m$ , so that $X_3 = (n+1) \times \omega ^m$ . Thus, we have, in addition to (1)–(6), that:

1. (7) $\Theta \cap (T \times \omega ^m)^2 = \gamma _{m,n}(r_3) \cap (T \times \omega ^m)^2.$

To complete this inductive step,1 we proceed with what might be called a “thinning” of $\omega ^m$ . The object is to get $D \subseteq \omega ^m$ and $X = (n+1) \times D$ such that $\gamma _{m,n} | X \cong \gamma _{m,n}$ and $\gamma _{m,n}(r_3) \cap X^2 = \Theta \cap X^2$ .

Let $\langle s^k : k < \omega \rangle $ be a one-to-one enumeration of $\omega ^m$ . By recursion on k, choose $t^k \in \omega ^m$ so that:

1. (T1) $t^k \not \in \{t^0,t^1, \ldots ,t^{k-1}\}$ ,

2. (T2) $t^k \hspace {-4pt} \upharpoonright \hspace {-3pt} (m-1) = s^k \hspace {-4pt} \upharpoonright \hspace {-3pt} (m-1)$ ,

3. (T3) $\Theta $ and $\gamma _{m,n}(r_3)$ agree on $(n+1) \times \{t^0,t^1, \ldots , t^k\}$ .

Clearly, $t^0 = s^0$ . If $k> 0$ , then there are only finitely many $t \in \omega $ such that (T2) holds but (T3) fails, so it is always possible to get $t^k$ .

Let $r = r_3$ , $D = \{t^k : k < \omega \}$ , and $X = (n+1) \times D$ . Then, X and r are as required.

Let $R : \omega \longrightarrow \omega $ be the Ramsey function such that if $m < \omega $ , then $R(m)$ is the least $k < \omega $ such that whenever $\chi : [k]^2 \longrightarrow 35$ , then there is $I \subseteq k$ such that $|I| = m$ and $\chi $ is constant on $[I]^2$ . (It seems to be right that 35 is large enough for the following proof to work. But if it isn’t, replace it with something that is.)