Fact 2.2 ( $\aleph _1$ -saturation)

Let G be a non-standard finite group. Every internal countable cover of an internal set admits a finite sub-covering: given an internal subset X of G and a countable family $(Y_n)_{n\in \mathbb {N}}$ of internal subsets of G such that $X=\bigcup _{n\in \mathbb {N}} Y_n$ , there is some natural number k such that

$$\begin{align*}X=Y_{n_1} \cup \dots \cup Y_{n_k}. \end{align*}$$

Equivalently, a countable intersection of internal sets is non-empty whenever every finite sub-intersection is.

Proof $(a)\Rightarrow (b):$ No finite set lies in a non-principal ultrafilter $\mathfrak {U}$ . Hence, by the way, the non-standard counting measure on the non-standard finite group $G=\prod _{n\to \mathfrak {U}} G_n$ has been defined, it follows from (a) that $\mu _G(X)\le r+\epsilon $ for every $\epsilon>0$ , so $\mu _G(X)\le r$ , as desired.

$(b)\Rightarrow (a):$ If (a) is false, negating quantifiers, there is some value $\epsilon>0$ such that for every natural number $n_0\ge 1$ , we find some $n> n_0$ with $|X_n|> (r+ \epsilon ) |G_n|$ . Hence, the subset

$$\begin{align*}Q_{\epsilon} = \left\{ n\in \mathbb{N} : \ |X_n|> (r+\epsilon) |G_n| \right\} \end{align*}$$

is infinite. Choose a non-principal ultrafilter $\mathfrak {U}$ containing each $Q_\epsilon $ (which exists). By definition of the non-standard counting measure on the non-standard finite group $G=\prod _{n\to \mathfrak {U}} G_n$ , we deduce

$$\begin{align*}\mu_G(X) = \lim_{n\to\mathfrak{U}} \frac{|X(G_n)|}{|G_n|} \ge \lim_{n\to\mathfrak{U}} r+\epsilon = r +\epsilon. \end{align*}$$

Thus, $\mu _G(X)> r$ , which yields the desired contradiction.

Fact 3.4 Given any countable family $\mathcal F_0$ of internal sets, there is a countable tame family $\mathcal F$ of internal sets extending $\mathcal F_0$ .

Proof Write $X=\bigcap _{n\in \mathbb {N}} X_n$ . By $\aleph _1$ -saturation, it suffices to show that

$$\begin{align*}X_1\cap \dots\cap X_m \cap (G\setminus Y_1) \cap \dots \cap (G\setminus Y_m) \neq\emptyset \end{align*}$$

for every $m\ge 1$ , where $Y_1, \dots , Y_m$ are fibers of internal sets in $\mathcal {F}$ defined over B and of density $0$ . If the above intersection were empty, then $ X_1\cap \dots \cap X_m$ would have density $0$ , and X would not be dense.

Proof Assume for a contradiction that $b'$ is not dense over $A\cup \{c'\}$ . By Remark 3.9, there exists a fiber X of an internal set defined over $A\cup \{c'\}$ of density $0$ and containing $b'$ . Write $X=Z_{c'}$ for some internal set $Z\subset G^{k+m}$ defined over A, where $k=|b|=|b'|$ and $m=|c|=|c'|$ . By Remark 3.7 (after taking set-theoretic complements), there is for every natural number $n\ge 1$ an internal set $Y_{\frac {1}{n}}$ defined over A such that

$$\begin{align*}\left\{ y \in G^m : \ \mu_{G^k} (Z_y) < \frac{1}{n} \right\} \subseteq Y _{\frac{1}{n}} \subseteq \left\{ y \in G^m : \ \mu_{G^k} (Z_y)\le \frac{1}{n} \right\}. \end{align*}$$

Now, the internal set $\widetilde Z_n=Z\cap \big (G^{k}\times Y_{\frac {1}{n}}\big )$ is defined over A for every natural number $n\ge 1$ and contains $(b',c')$ , so $(b,c)$ also belongs to the intersection of all the $\widetilde Z_n$ . We conclude that b is not dense over $A\cup \{c\}$ , witnessed by the fiber $\widetilde {Z}_{c}$ , as desired.

Fact 3.12 [Reference Martin-Pizarro and Palacín28, Lemma 1.10 as well as Remarks 1.13 and 1.14 and Lemma 1.15]

Let G be a non-standard finite group and X be an $\omega $ -internal subset of $G^n$ defined over a countable subset A. The following are equivalent:

1. (1) There exists some $(b_1,\ldots ,b_n)$ in X in good position over A, that is, each $b_i$ is dense over $A\cup \{b_j:j<i\}$ .

2. (2) The set X is dense.

Furthermore, every tuple as in (a) is dense over A with respect to the Loeb measure on $G^n$ .

Fact 4.3 Let G be a non-standard finite group and let $\alpha $ be a real number in $[0,1]$ .

1. (1) Given two internal subsets X of $G^{n+r}$ and Y of $G^{n+s}$ in $\mathcal {F}$ , the relation $R^\alpha _{X,Y}$ of $G^r\times G^s$ defined by

   $$\begin{align*}R^\alpha_{X,Y}(a,b) \ \Leftrightarrow \ \mu_{G^n}(X_a \cap Y_b) \le \alpha \end{align*}$$

   is stable. Notice that $R^\alpha _{X,Y}$ is $\omega $ -internal by Definition 3.2(IV), and definable over the same parameters needed to define both X and Y, by Remark 3.7.

2. (2) If the two subsets X of $G^{n+r}$ and Y of $G^{n+s}$ are $\omega $ -internal and defined over A, the relation $R_{X,Y}$ of $G^r\times G^s$ defined by

   $$\begin{align*}R_{X,Y}(a,b) \ \Leftrightarrow \ \text{ the } \omega-\text{internal subset } X_a \cap Y_b \text{ of } G^n \text{ is not dense}\end{align*}$$

   is equational, that is, there is no A-indiscernible sequence $(a_i, b_i)_{i\in \mathbb {N}}$ such that $X_{a_0} \cap Y_{b_0}$ is dense yet $X_{a_0} \cap Y_{b_1}$ is not.

3. (3) Equationality implies the following (see [Reference Martin-Pizarro and Palacín29, Remark 2.1 and Corollary 2.3]). For every $\omega $ -internal subset X of $G^{n+r}$ defined over a countable rich subset M, if the fiber $X_a$ is dense for some a which is itself dense over M, then so is the intersection $\bigcap _{i=0}^m X_{b_i}$ for every m in $\mathbb {N}$ and every tuple $(b_0, \ldots , b_m)$ of elements in $\operatorname {tp}(a/M)$ in good position over M, as in Fact 3.12(a), that is, the element $b_i$ is dense over $M\cup \{b_j:j<i\}$ for $i\le m$ .

Question Is there a model-theoretic account that every almost surely k-stable internal relation is comparable (or $\epsilon $ -comparable) to an internal k-stable relation?

Proof If a does not lie in X, neither does any element of $\operatorname {tp}(a/M)$ , and likewise if b does not lie in Y. In this case, $X\times Y$ (and thus S) is clearly disjoint from $\operatorname {GP}(\operatorname {tp}(a/M),\operatorname {tp}(b/M))$ . So we may assume that $\operatorname {GP}(\operatorname {tp}(a/M),\operatorname {tp}(b/M))$ is a subset of $X\times Y$ .

Suppose now that S is not disjoint from $\operatorname {GP}(\operatorname {tp}(a/M),\operatorname {tp}(b/M))$ . Up to relabeling, we may assume by Remark 4.5(a) that $(a,b)$ lies in S, with $(a, b)$ in good position, that is, the element b is dense over $M\cup \{a\}$ (keeping in mind that a is already dense over M).

Proof of Claim

Suppose to the contrary that there is a pair $(c, d)$ with c dense over $M\cup \{d\}$ not contained in S. By Remark 3.8 (using the fact that d belongs to the type $\operatorname {tp}(b/M)$ ), there is a pair $(a',b)$ in $\operatorname {tp}(c, d/M)$ . So the original tuple $(a, b)$ belongs to S but $(a', b)$ does not lie in S (for S is definable over M). In particular, we have two pairs $(a, b)$ and $(a', b)$ satisfying the following:

• • the type $\operatorname {tp}(a/M)=\operatorname {tp}(a'/M)$ is dense;

• • the element b is dense over $M \cup \{a\}$ and $a'$ is dense over $M \cup \{b\}$ (by Lemma 3.11);

• • the pair $(a, b)$ belongs to S but $(a', b)$ does not.

We will now construct inductively, for every n in $\mathbb {N}$ , a dense sequence $(a_0, b_0, \ldots , a_n, b_n)_{0\le i\le n}$ in good position over M witnessing that S induces a half-graph of height n such that $a_i$ belongs to $\operatorname {tp}(a/M)$ and $b_i$ to $\operatorname {tp}(b/M)$ for each $i\le n$ . In particular, the case $n=k$ contradicts the almost sure stability of S by Remark 4.5(b).

For $n=0$ , set $a_0=a$ and $b_0=b$ , so $(a_0, b_0)$ is in good position over M and lies in S, as desired. Suppose now that the sequence $(a_0, b_0, \ldots , a_r, b_r)$ has already been constructed for some $r< n$ . Consider the $\omega $ -internal set

$$\begin{align*}Z =\{ (x,y) \in X\times Y : \ x\in \operatorname{tp}(a/M) \ \text{and} \ (x,y) \notin S \},\end{align*}$$

which is defined over M. Clearly, $(a', b)$ belongs to Z, so the fiber $Z_b\subseteq X$ is dense. By Fact 4.3(c), so is the intersection $ Z_{b_0}\cap \cdots \cap Z_{b_r}$ , since the subtuple $(b_0,\ldots , b_r)$ is also in good position over M and all coordinates lie in $\operatorname {tp}(b/M)$ . Hence, there is an element $a_{r+1}$ in the above intersection which is dense over $M\cup \{a_i, b_i\}_{0\le i\le r}$ . In particular, the element $a_{r+1}$ lies in $\operatorname {tp}(a/M)$ but $(a_{r+1}, b_i)$ does not lie in S for $i\le r$ .

Similarly, the $\omega $ -internal set

$$\begin{align*}W =\{ (x,y) \in X\times Y : \ y\in \operatorname{tp}(b/M) \ \text{and}\ (x,y) \in S \}\end{align*}$$

is also defined over M and the fiber $W_a\subseteq Y$ is dense, since $(a, b)$ lies in W. Thus, there exists an element $b_{r+1}$ in $\operatorname {tp}(b/M)$ contained in $W_{a_0}\cap \cdots \cap W_{a_{r+1}}$ dense over $M\cup \{a_i, b_i\}_{0\le i\le r} \cup \{a_{r+1}\}$ , which yields that the tuple

$$\begin{align*}(a_0, b_0, \ldots, a_{r+1}, b_{r+1})\end{align*}$$

is in good position over M by Fact 3.12 and satisfies the desired properties.

By the previous claim, choose any pair $(a',b')$ in $S\cap \operatorname {GP}(\operatorname {tp}(a/M), \operatorname {tp}(b/M))$ such that $a'$ is dense over $M\cup \{b'\}$ . The pair $(b',a')$ is in good position over M and it belongs to the inverse relation $S^{\mathrm {opp}}$ , which is again almost surely k-stable, by Remark 4.5(a). We conclude by the Claim (applied to $S^{\mathrm {opp}}$ inverting the roles of X and Y) that all pairs $(b", a")$ in $\operatorname {GP}(\operatorname {tp}(b/M), \operatorname {tp}(a/M))$ with $b"$ dense over $M\cup \{a"\}$ must be contained in $S^{\mathrm {opp}}$ . Hence, all pairs $(a", b")$ in $\operatorname {GP}(\operatorname {tp}(a/M), \operatorname {tp}(b/M))$ in good position over M belong to S, so the set $\operatorname {GP}(\operatorname {tp}(a/M), \operatorname {tp}(b/M))$ is fully contained in S, as desired.

Proof If the set-theoretic difference $S_a \setminus S_{a'}$ were dense for some choice a and $a'$ , both in the same dense type over M, then it would contain some element b which is dense over $M\cup \{a, a '\}$ . In particular, both $(a, b)$ and $(a', b)$ would lie in $\operatorname {GP}(\operatorname {tp}(a/M), \operatorname {tp}(b/M))$ , yet $(a,b)$ belongs to S but $(a', b)$ does not, which contradicts Theorem 4.7.

Proof If S is not dense in $G^{n+m}$ , then it suffices to take U to be the empty union of boxes. Thus, we may assume that the internal subset S of $G^{n+m}$ is dense.

Claim 1 Consider an arbitrary subset Z of S in the $\sigma $ -algebra of internal sets defined over M. (Note that Z need not be $\omega $ -internal, but it is measurable with respect to the extension of the Loeb measure.) If Z has positive measure, then there exists a dense element a in $G^n$ such that the fiber $Z_a$ is dense and satisfies the following property ( $\star $ ): for every internal subset W of Y defined over M, if $\mu _{G^m}(W\setminus S_a)=0$ , then

$$\begin{align*}\mu_{G^n}\big( \{x\in G^n : \ \mu_{G^m}(W\setminus S_x)=0\} \big)> 0.\end{align*}$$

Proof of Claim 1

By Fubini–Tonelli (Remark 2.6), the measurable set

$$\begin{align*}Z_1=\{ x\in G^n : \ \mu_{G^m}(Z_x)>0\}\end{align*}$$

has positive measure (since $\mu _{G^{n+m}} (Z)$ is not zero). For every internal subset W of Y defined over M, set

$$\begin{align*}D_W=\{ x\in G^n : \ \mu_{G^m}(W \setminus S_x)=0 \}.\end{align*}$$

Notice that there are only countably many such sets $D_W$ . Consider the subfamily $\mathcal D$ of all sets $D_W$ of measure $0$ . Similarly, let $\mathcal E$ be the countable family of all internal subsets $X'$ of $G^n$ defined over M of measure $0$ . By $\sigma $ -additivity of the (extension of the) Loeb measure, the set

$$\begin{align*}Z_1 \setminus \Big( \bigcup_{D_W\in\mathcal D} D_W \ \cup \ \bigcup_{X'\in \mathcal E} X' \Big) \end{align*}$$

has positive measure, so it is non-empty. Now choose some element a in the above set-theoretic difference. By the choice of $\mathcal E$ , the type $\operatorname {tp}(a/M)$ is dense and so is the fiber $Z_a$ (since a lies in $Z_1$ ). Furthermore, the element a satisfies ( $\star $ ) by our choice of the family $\mathcal D$ .

As the internal set S is dense, it has positive measure. By the previous claim, there is an element a which is dense over M satisfying $(\star )$ with respect to $Z=S$ such that the fiber $S_a$ is dense. Thus, the fiber $S_a$ contains an element b which is dense over $M\cup \{a\}$ . By Fact 3.12, the dense pair $(a,b)$ is in good position over M and clearly lies in S.

Proof of Claim 2

By Theorem 4.7, there is no element in $\operatorname {tp}(b/M)$ dense over $M\cup \{a\}$ outside of $S_a$ , so the $\omega $ -internal subset $\operatorname {tp}(b/M) \setminus S_a$ of Y is not dense. By $\aleph _1$ -saturation, there is some internal subset $Y'$ of Y defined over M containing b with $\mu _{G^m}(Y'\setminus S_a)=0$ . We deduce by $(\star )$ that $\mu _{G^n}(X')>0$ , where

$$\begin{align*}X'= \{x \in X : \ \mu_{G^m}(Y'\setminus S_x)=0\}\end{align*}$$

is an $\omega $ -internal subset (by Definition 3.2(IV)) containing a (this set is $D_{Y'}$ with $Z=S$ in the notation of Claim 1).

A straightforward computation using Fubini–Tonelli yields

$$\begin{align*}\mu_{G^{n+m}} ((X'\times Y')\setminus S) = \int_{X'} \mu_{G^m}(Y'\setminus S_x) \, \mathrm{d}\mu_{G^n} = 0, \end{align*}$$

as desired.

Consider now the countable collection $\mathcal B_S$ of all subsets $X'\times Y'$ as in Claim 2, that is, the set $Y'$ is internal while $X'$ is $\omega $ -internal, both are defined over M and

$$\begin{align*}X'= \{x \in X : \ \mu_{G^m}(Y'\setminus S_x)=0\},\end{align*}$$

so $\mu _{G^{n+m}}((X'\times Y')\setminus S)=0$ . By $\sigma $ -additivity, we have that

$$\begin{align*}\left(\bigcup_{X'\times Y'\in \mathcal B_S} (X'\times Y') \right)\setminus S \end{align*}$$

has measure $0$ (with respect to the extension of the Loeb measure). Observe further that the set

$$\begin{align*}Z=S\setminus \bigcup_{X'\times Y'\in \mathcal B_S} (X'\times Y') \end{align*}$$

belongs to the $\sigma $ -algebra of internal sets defined over M. If Z has positive measure, we deduce as in the discussion right after Claim 1 that Z contains a pair $(a,b)$ in good position over M with a satisfying $(\star )$ . It follows from Claim 2 that $(a,b)$ lies in some box $X'\times Y'$ in $\mathcal B_S$ , contradicting the choice of Z. Thus, S and $\bigcup _{X'\times Y'\in \mathcal B_S} (X'\times Y')$ are comparable with respect to the extension of the Loeb measure. We need only show that we can replace the latter union, modulo an $\epsilon $ -error, by an actual finite union of internal boxes defined over M.

Given $\epsilon>0$ , by continuity of the Loeb measure, there is some integer $\ell $ and subsets $X^{\prime }_1\times Y_1,\ldots , X^{\prime }_\ell \times Y_\ell $ in $\mathcal B_S$ such that

$$\begin{align*}\mu_{G^{n+m}}\left( S\setminus \bigcup\limits_{i=1}^\ell (X^{\prime}_i\times Y_i)\right)<\frac{\epsilon}{2}.\end{align*}$$

Note that $\ell>0$ , since $\mathcal B_S$ is non-empty, for S is assumed to be dense. For each $1\le i\le \ell $ , the $\omega $ -internal subset $(X^{\prime }_i\times Y_i)\setminus S$ has measure $0$ , so we can find internal subsets $X_i$ , with each $X_i$ containing $X^{\prime }_i$ , such that

$$\begin{align*}\mu_{G^{n+m}}\big((X_i\times Y_i)\setminus S\big) <\frac{\epsilon}{2\ell}.\end{align*}$$

Note that $X_i$ is again internal and defined over M. Now set $U=\bigcup _{i=1}^\ell (X_i\times Y_i)$ , so U contains $\bigcup _{i=1}^\ell (X^{\prime }_i\times Y_i)$ , and it follows that

$$ \begin{align*} \mu_{G^{n+m}}\left(S \triangle U\right) & = \mu_{G^{n+m}} (S \setminus U) + \mu_{G^{n+m}}(U\setminus S) \\ & \le \mu_{G^{n+m}} \left(S \setminus \bigcup_{i=1}^\ell (X_i'\times Y_i) \right) + \sum_{i=1}^{\ell} \mu_{G^{n+m}}\left((X_i\times Y_i)\setminus S \right) \\ & < \frac{\epsilon}{2} + \ell\cdot \frac{\epsilon}{2\ell} = \epsilon. \end{align*} $$

This completes the proof of Proposition 4.9.

Fact 5.1 ([Reference Hrushovski20, Theorem 3.5] & [Reference Montenegro, Onshuus and Simon31, Theorem 2.12])

Let M be a countable rich subset of a non-standard finite group G. Then, there exists an $\omega $ -internal normal subgroup $G_{M}^{00}$ defined over M with the following properties:

1. (1) The $\omega $ -internal subset $G_{M}^{00}$ equals a countable intersection of internal generic symmetric neighborhoods $Z_n$ of the identity, each defined over M, that is,

   • • each $Z_n=Z_n ^{-1}$ is symmetric and contains the identity element $1_G$ ;

   • • finitely many translates of each $Z_n$ cover the group G;

   • • $Z_{n+1}\cdot Z_{n+1}\subseteq Z_n$ for all n in $\mathbb {N}$ .

2. (2) Whenever $(a, b)$ is in good position over M with b in $\operatorname {tp}(a/M)$ , the dense element $b\cdot a ^{-1}$ over M belongs to $G_{M}^{00}$ . In particular, the $\omega $ -internal set $G_{M}^{00}$ is dense.

3. (3) Whenever a is dense over M and g in $G_{M}^{00}$ is dense over M, the $\omega $ -internal set

   $$\begin{align*}\operatorname{tp}(a/M)\cap \operatorname{tp}(a/M)\cdot g\end{align*}$$
   is again dense, that is, there exists some $a'$ in $\operatorname {tp}(a/M)$ which is dense over $M\cup \{g\}$ such that $a'\cdot g$ belongs to $\operatorname {tp}(a/M)$ .

   Similarly, the $\omega $ -internal set $\operatorname {tp}(a/M)\cap g\cdot \operatorname {tp}(a/M)$ is also dense, so there exists some $a"$ in $\operatorname {tp}(a/M)$ which is dense over $M\cup \{g\}$ such that $g\cdot a"$ belongs to $\operatorname {tp}(a/M)$ .

4. (4) If N is a countable rich subset of G containing M, then $G_{N}^{00}$ is a subgroup of $G_{M}^{00}$ .

Furthermore, it follows that if G has finite exponent, then $G_{M}^{00}$ is a countable intersection of internal subgroups, each defined over M and of finite index: Indeed, we can equip the quotient group $G/ G_{M}^{00}$ with the so-called logic topology, according to which a subset in the quotient is closed if its preimage is $\omega $ -internal and defined over M. The quotient group $G/ G_{M}^{00}$ is a compact Hausdorff topological group of finite exponent, and thus profinite by [Reference Iltis22, Theorem 4.5]. Since M and $\mathcal F$ are countable, the logic topology is second countable, so there is a fundamental system of open normal subgroups $H_n/ G_{M}^{00}$ with $G_{M}^{00}=\bigcap _n H_n$ . Now, each quotient $H_n/G_{M}^{00}$ is closed of finite index, by topological compactness, and thus the preimage $H_n$ is internal and defined over M, as desired.

Proof Fix some element g in $G_{M}^{00}$ which is dense over M. In order to show that the internal set $\Lambda _\Box (S)_g$ has positive density, we need only show that $\Lambda _\Box (S)_g$ contains a pair $(a, b)$ dense over $M\cup \{g\}$ . We do so with the help of the following auxiliary claims.

By Claim 4 and Fact 3.12, there is a pair $(a_0, b)$ in good position over $M\cup \{g\}$ contained in $S\cap S\cdot (1_G, g ^{-1})$ . That is, both pairs $(a_0, b)$ and $(a_0, b\cdot g)$ belong to S.

Note that we have not yet used that S is almost surely stable. The idea is to replace $a_0$ by a suitable realization a such that the product $ a\cdot g$ lies again in $\operatorname {tp}(a/M)$ .

Indeed, Fact 5.1(c) yields that there is some a in $\operatorname {tp}(a_0/M)$ dense over $M\cup \{g\}$ such that $a\cdot g$ lies again in $\operatorname {tp}(a_0/M)$ . We may assume that a is dense over $M\cup \{b,g\}$ , by Lemma 3.10. Clearly, each of the pairs $(b, a)$ , $(b, a\cdot g)$ , $(b\cdot g, a)$ , and $(b\cdot g, a\cdot g)$ is in good position over M.

Theorem 4.7 implies that

$$\begin{align*}\operatorname{GP}(\operatorname{tp}(a_0/M), \operatorname{tp}(b/M))\subseteq S \end{align*}$$

since the pair $(a_0, b)$ lies in S and is in good position over M. Thus, both pairs $(a, b)$ and $(a\cdot g, b)$ lie in S.

Now, the pair $(a_0, b\cdot g)$ also lies in S and is in good position over M. Again by Theorem 4.7, the pairs $(a, b\cdot g)$ and $(a\cdot g, b\cdot g)$ of $\operatorname {GP}(\operatorname {tp}(a'/M), \operatorname {tp}(b\cdot g/M))$ must lie in S. In particular, the pair $(a, b)$ is dense over $M\cup \{g\}$ and belongs to the internal set $\Lambda _\Box (S)_g$ , as desired.

Proof As in Claim 3 of Theorem 5.2, write $g=c ^{-1} \cdot d$ , where $(c, d)$ are in good position over M and both lie in the same type $\operatorname {tp}(g/M)$ . Suppose toward a contradiction that $\mu _G(A\triangle (g\cdot A))>0$ , or equivalently, that

$$\begin{align*}\mu_G\big((c\cdot A) \triangle (d\cdot A))>0.\end{align*}$$

Without loss of generality, we may assume that $(c\cdot A)\setminus (d\cdot A)$ is dense, so it contains some element b which is dense over $M\cup \{c,d\}$ . This means that $c ^{-1}\cdot b$ belongs to A but $d ^{-1} \cdot b$ does not, or equivalently, the pair $(c, b)$ belongs to the almost surely k-stable relation $\mathrm {Cay}(G,A)$ , but $(d, b)$ does not. Since both pairs belong to $\operatorname {GP}(\operatorname {tp}(g/M), \operatorname {tp}(b/M) )$ , this contradicts Theorem 4.7.

To prove the final part, observe that since A and $g\cdot A$ are comparable, so are A and $g ^{-1} \cdot A$ . By continuity, we need only show that each intersection

$$\begin{align*}\bigcap_{-m\le i\le m} g^i\cdot A \end{align*}$$

has constant measure $\mu _G(A)$ for all m in $\mathbb {Z}$ . Otherwise, the internal set

$$\begin{align*}A \setminus \bigcap_{-m\le i\le m} g^i\cdot A \end{align*}$$

is dense, so by Lemma 3.10, we may choose an element a in A which is dense over $M\cup \{g\}$ . Remark 3.13 implies that a must belong to both $g\cdot A$ and $g ^{-1} \cdot A$ , and inductively, we conclude that a belongs to $g^{i} \cdot A$ for all $-m\le i\le m$ , which gives the desired contradiction.

Proof The proof proceeds by contradiction. Suppose there are $k\ge 1$ and $\delta>0$ such that for every $\ell \ge 1$ (setting $\theta =\epsilon =1/\ell $ ), we can find a finite group $G_\ell $ of order at least $\ell $ and a relation $S_\ell \subseteq G_\ell \times G_\ell $ of density $\delta $ such that $\mathcal {H}_k(S_\ell )\le \frac {1}{\ell } |G_\ell |^{2k}$ , yet $|\Lambda _\Box (S_\ell )|<|G_\ell ^{3}|/\ell $ .

In particular, the family $(G_\ell , S_\ell )_{\ell \in \mathbb {N}}$ is almost surely k-stable, as in Remark 4.6. Take a non-principal ultrafilter $\mathfrak {U}$ on $\mathbb {N}$ and consider the non-standard finite group $G=\prod _{\ell \to \mathfrak {U}} G_\ell $ as well as the internal set $ S=\prod _{\ell \to \mathfrak {U}} S_\ell $ . By Remark 2.5, the set S has Loeb measure $\mu _{G^{2}}(S)\ge \delta $ and is almost surely k-stable by Remark 4.6. Choose any countable rich subset M. Corollary 5.3 yields that $\mu _{G^{3}}(\Lambda _\Box (S))$ is at least $\eta $ for some $\eta>0$ .

Remark 2.4 applied to $r=\eta $ yields the desired contradiction, since $\Lambda _\Box (S_\ell )$ has size at most $\eta |G_\ell |^{3}/2$ for sufficiently large $\ell $ .

Proof As in the proof of Corollary 5.5, negating quantifiers, we deduce that for a fixed choice of $k, r, \delta ,$ and $\epsilon $ , for every $\ell $ in $\mathbb {N}$ (setting $\theta =\eta =1/\ell $ ), there is a finite group $G_\ell $ of exponent at most r and a relation $S_\ell \subseteq G_\ell \times G_\ell $ of density $\delta $ such that

$$\begin{align*}|\mathcal{H}_k(S_\ell)|\le \frac{1}{\ell} |G_\ell|^{2k},\end{align*}$$

yet for every subgroup $H\leqslant G_\ell $ of index at most $\ell $ , the subset of elements g in H with

$$\begin{align*}|\Lambda_\Box(S)_g| <\frac{|S|}{\ell} \end{align*}$$

has size at least $\epsilon |H|$ . In particular, the group $G_\ell $ has size at least $\ell $ , by taking H the trivial subgroup of $G_\ell $ .

Take a non-principal ultrafilter $\mathfrak {U}$ on $\mathbb {N}$ and consider the non-standard finite group $G=~\prod _{\ell \to \mathfrak {U}} G_\ell $ as well as the internal set $ S=\prod _{\ell \to \mathfrak {U}} S_\ell $ . By Remark 2.5, the set S has Loeb measure $\mu _{G^{2}}(S)\ge \delta $ and is almost surely k-stable by Remark 4.6. Choose any countable rich subset M. Notice that the exponent of the non-standard finite group G is bounded by r by Łoś’s Theorem (Theorem 2.1). Fact 5.1 yields that the subgroup $G_{M}^{00}$ is a countable intersection of internal subgroups $H_n$ , each of finite index and defined over M. We may assume that $H_{n+1}\subseteq H_n$ for every n in $\mathbb {N}$ .

Definition 3.2 tells us that the set

$$\begin{align*}Z=\{g \in G_{M}^{00} : \ \mu_{G^{2}}(\Lambda_\Box(S)_g)=0\} = \bigcap\limits_{\substack{n\in \mathbb{N} \\ 0\ne m\in \mathbb{N}}} \left\{g \in H_n : \ \mu_{G^{2}}(\Lambda_\Box(S)_g)<\frac{1}{m}\right\} \end{align*}$$

is $\omega $ -internal.

Fact 3.12 and Theorem 5.2 imply that Z is not dense, so by $\aleph _1$ -saturation and Remark 3.9, we deduce that for some $0\ne m$ in $\mathbb {N}$ and some finite-index internal subgroup $H=H_n$ defined over M, the set $\{g \in H : \ \mu _{G^{2}}(\Lambda _\Box (S))_g) < \frac {1}{m}\}$ is contained in some internal set of density $0$ .

Choose some $\ell _0$ in $\mathbb {N}$ with

$$\begin{align*}0<\frac{\mu_{G^{2}}(S)}{\ell_0}< \frac{1}{2m}\end{align*}$$

such that for $\mathfrak {U}$ -almost all $\ell \ge \ell _0$ , the subgroup $H(G_{\ell })$ has index $[G:H]\le \ell _0$ . Now, Definition 3.2 yields an internal set W with

$$\begin{align*}\left\{g\in H : \ \mu_{G^{2}}(\Lambda_\Box(S))_g) < \frac{1}{2m} \right\}\subseteq W\subseteq \left\{g\in H : \ \mu_{G^{2}}(\Lambda_\Box(S))_g) \le \frac{1}{2m}\right\},\end{align*}$$

whence $\mu _G(W)=0$ and $ \mu _G(W)<\epsilon \cdot \mu _G(H)=\epsilon /[G:H]$ . Therefore, for $\mathfrak {U}$ -almost all $\ell \ge \ell _0$ , we have that

$$\begin{align*}\left|\left\{g \in H(G_\ell) : \ |\Lambda_\Box(S_\ell)_g| < \frac{|S_\ell|}{\ell} \right\} \right| \le |W(G_\ell)|< \frac{\epsilon}{[G:H]} |G_\ell| = \epsilon |H(G_\ell)|.\end{align*}$$

This contradicts our choice of $G_\ell $ , as desired.

Proof Set $\delta =\mu _{G^{2}}(S)/2>0$ . By Definition 3.2, there is an internal set $Y_\delta $ defined over M such that

$$\begin{align*}\{y\in G : \ \mu_G(S_y)> \delta \}\subseteq Y_\delta\subseteq \{y\in G : \ \mu_G(S_y)\ge \delta \},\end{align*}$$

where $S_y=\{x\in G : \ (x, y) \in S\}$ denotes the fiber of S over y. By Fubini–Tonelli (Remark 2.6), the internal set $Y_\delta $ is dense. Hence, choose some c in G dense over M with $\mu _G(S_{c})\ge \delta $ . By Remark 3.6, there is a countable rich subset N containing $M\cup \{c\}$ . Notice that the internal dense subset $S_{c}\subseteq G$ is defined over N.

Proof of Claim

It suffices to find a triple satisfying (i) and (ii) in the statement of the claim with $(a_1,a_3)$ in good position over N, or equivalently, with $(a_1,a_2)$ in good position over N, since squaring preserves dense elements.

Write $G_{N}^{00}$ as a countable intersection of internal generic symmetric neighborhoods $Z_j$ , with j in $\mathbb {N}$ , defined over N as in Fact 5.1(a). In particular, for each $Z_j$ , there are finitely many elements $t_1(j),\ldots , t_{n_j}(j)$ in G such that

$$\begin{align*}G=Z_j\cdot t_1(j)\cup\cdots\cup Z_j\cdot t_{n_j}(j).\end{align*}$$

Note that we may find such elements $t_r(j)$ in N, since N is a rich subset of G. Moreover, the product set $Z_{j+1}\cdot Z_{j+1}$ is a subset of $Z_j$ . Assume toward a contradiction that there is no triple as required, and consider the $\omega $ -internal set

$$\begin{align*}\mathcal Z= \bigcap\limits_{\substack{j\in \mathbb{N} \\ \mathcal X \in \mathcal F}} \left\{ (x,y,z) \in S_c^{3} : \ x\cdot y= z^2 \text{ with} x ^{-1} \cdot z=y\cdot z ^{-1} in Z_j \text{ and } (x,y) \notin \mathcal X \right\}, \end{align*}$$

where $\mathcal F$ is the countable collection of all internal subsets $\mathcal X$ of $G^2$ defined over N of density $0$ . Notice that the projection $\pi $ of $\mathcal Z$ onto the first two coordinates is again $\omega $ -internal by Remark 3.1 and empty, for otherwise, it would contain a pair $(a_1,a_2)$ in good position over N, by Fact 3.12, but such a pair would yield a triple as in the statement. Therefore, the $\omega $ -internal set $\mathcal Z$ must be empty as well, since $\pi (\mathcal Z)=\emptyset $ . By Fact 2.2, there are finitely many sets $\mathcal X_1, \ldots , \mathcal X_r$ in $\mathcal F$ and some j in $\mathbb {N}$ such that the internal set

$$\begin{align*}\widetilde{\mathcal Z}=\left\{ (x,y,z) \in S_c^{3} : \ x\cdot y= z^2 \text{ with } x ^{-1} \cdot z=y\cdot z ^{-1} \text{ in } Z_j \right\}\end{align*}$$

is covered by $\bigcup _{i=1}^r \pi ^{-1}(\mathcal X_i)$ . Now, the internal set $S_c$ is dense, so there is some $t_r(j+1)$ in N such that $S_c\cap (Z_{j+1}\cdot t_r(j+1))$ is dense as well. It follows from the suitability of G that the corresponding set $\mathcal {E}(S_c\cap (Z_{j+1}\cdot t_r(j+1))$ must have positive density. Hence, there is a dense pair $(u_1,u_2)$ over N with each $u_i$ in $S_c\cap (Z_{j+1}\cdot t_r(j+1))$ such that

$$\begin{align*}u_1\cdot u_2= u_3^2 \text{ for some } u_3 \text{ in } S_c\cap(Z_{j+1}\cdot t_r(j+1)).\end{align*}$$

Now, the pair $(u_1,u_2)$ avoids all $\mathcal X_i$ and the common difference

$$\begin{align*}u_1 ^{-1}\cdot u_3 &= u_2\cdot u_3 ^{-1}=u_2\cdot (t_r(j+1) ^{-1} \cdot t_r(j+1) ) \cdot u_3 ^{-1}\\ & =( u_2\cdot t_r(j+1) ^{-1} ) \cdot (u_3\cdot t_r(j+1) ^{-1}) ^{-1} \end{align*}$$

belongs to $Z_{j+1}\cdot Z_{j+1}\subseteq Z_j$ . Thus, the triple $(u_1,u_2, u_3)$ belongs to $\widetilde {\mathcal Z}$ and thus to some $\pi ^{-1}(\mathcal X_j)$ . This implies that the dense pair $(u_1,u_2)$ over N belongs to the internal set $\mathcal X_j$ defined over N of density $0$ , which yields the desired contradiction.

Choose now a tuple $(a_1,a_2,a_3)$ in $S_c^{3}$ as in the Claim with $a_1\cdot a_2 = a_3^2$ , where $g=a_1 ^{-1} \cdot a_3=a_2\cdot a_3 ^{-1}$ lies in $G_{N}^{00}$ and $(a_1,g)$ in good position over N. By Fact 5.1(d), we have that g, viewed as an element of $G_{M}^{00}$ , is dense over M. Fact 5.1(c) yields that for some b in $\operatorname {tp}(c/M)$ dense over $M\cup \{a_1, a_2,a_3\}$ , the product $b\cdot g$ also belongs to $\operatorname {tp}(c/M)$ (and is dense over $M\cup \{a_1, a_2, a_3\}$ ). Hence, the triple $(a_1,g, b)$ is in good position over M. We need only verify that $(a_1, b, g)$ belongs to $\Lambda _{3\times 2}(S)$ .

By construction, the points

$$\begin{align*}(a_1, c), (a_1\cdot g,c) \text{ and } (g\cdot a_1\cdot g, c) \end{align*}$$

all lie in the almost surely k-stable internal relation S. As each pair (with the order of coordinates reversed) is in good position over M, Theorem 4.7 implies that for each $i=1,2,3$ ,

$$\begin{align*}\operatorname{GP}(\operatorname{tp}(a_i/M), \operatorname{tp}(c/M)) \subseteq S,\end{align*}$$

so the pairs

$$\begin{align*}(a_1, b), (a_1\cdot g,b) \text{ and } (g\cdot a_1\cdot g, b) \end{align*}$$

all lie in the almost surely k-stable internal relation S. Analogously, using now that the element $b\cdot g$ of $\operatorname {tp}(b/M)$ is also dense over $M\cup \{a_1, a_1\cdot g, g\cdot a_1\cdot g\}$ , we conclude again by Theorem 4.7 that each of the pairs

$$\begin{align*}(a_1, b\cdot g), (a_1\cdot g, b\cdot g) \text{ and } (g\cdot a_1\cdot g, b\cdot g)\end{align*}$$

lies in the almost surely k-stable internal relation S, as desired.