[proofplan] We pass from the combinatorial set $E \subset \mathbb{Z}^d$ to a measure-preserving $\mathbb{Z}^d$-system using the multidimensional correspondence principle. The Furstenberg-Katznelson multidimensional multiple recurrence theorem is then applied to the finite family of commuting transformations indexed by the finite pattern $F$. The correspondence principle transfers the resulting positive-measure recurrence intersection back to a positive upper Banach density intersection of translates of $E$, and any lattice point in that intersection gives the desired translate and dilation. [/proofplan] [step:Transfer the positive-density set to a measure-preserving $\mathbb{Z}^d$-system] Let $d^*(E)$ denote the upper Banach density of $E$ in $\mathbb{Z}^d$. Since $E$ has positive upper Banach density, $d^*(E)>0$. We use the [multidimensional correspondence principle](/theorems/multidimensional-correspondence-principle) in the following form: for every set $E \subset \mathbb{Z}^d$ with $d^*(E)>0$, there exist a probability space $(X,\mathcal{B},\mu)$, a measure-preserving action \begin{align*} T: \mathbb{Z}^d &\to \operatorname{Aut}(X,\mathcal{B},\mu) \\ v &\mapsto T_v, \end{align*} where $T_{v+w}=T_v \circ T_w$ and the transformations $T_v$ are invertible and measure-preserving, and a measurable set $A \in \mathcal{B}$ with $\mu(A)=d^*(E)>0$, such that for every finite set $G \subset \mathbb{Z}^d$, \begin{align*} d^*\left(\bigcap_{g \in G} (E-g)\right) \geq \mu\left(\bigcap_{g \in G} T_g^{-1}A\right). \end{align*} Here $E-g:=\{z \in \mathbb{Z}^d : z+g \in E\}$ and $T_g^{-1}A:=\{x \in X : T_gx \in A\}$. Applying this principle to the given set $E$, we obtain $(X,\mathcal{B},\mu)$, the action $T$, and $A \in \mathcal{B}$ with $\mu(A)>0$. [guided] The purpose of the correspondence principle is to replace a positive-density subset of the lattice by a positive-measure set in a dynamical system. We first name the density: let $d^*(E)$ denote the upper Banach density of $E$ in $\mathbb{Z}^d$. The hypothesis says exactly that $d^*(E)>0$. We use the [multidimensional correspondence principle](/theorems/multidimensional-correspondence-principle) in the precise form needed here. It gives a probability space $(X,\mathcal{B},\mu)$, a measure-preserving action \begin{align*} T: \mathbb{Z}^d &\to \operatorname{Aut}(X,\mathcal{B},\mu) \\ v &\mapsto T_v, \end{align*} meaning that $T_{v+w}=T_v \circ T_w$ and each $T_v$ is invertible and measure-preserving, and it gives a measurable set $A \in \mathcal{B}$ satisfying $\mu(A)=d^*(E)>0$. The transfer inequality says that for every finite set $G \subset \mathbb{Z}^d$, \begin{align*} d^*\left(\bigcap_{g \in G} (E-g)\right) \geq \mu\left(\bigcap_{g \in G} T_g^{-1}A\right), \end{align*} where $E-g:=\{z \in \mathbb{Z}^d : z+g \in E\}$ and $T_g^{-1}A:=\{x \in X : T_gx \in A\}$. This inequality is the bridge we need: if recurrence later proves that the measure on the right is positive for a particular finite set $G$, then the lattice set on the left has positive upper Banach density and is therefore nonempty. [/guided] [/step] [step:Apply multidimensional multiple recurrence to the pattern directions] For each $f \in F$, define the transformation $S_f: X \to X$ by \begin{align*} S_f(x) := T_f x. \end{align*} Because $F$ is finite and $T$ is a $\mathbb{Z}^d$-action, the family $(S_f)_{f \in F}$ is a finite family of commuting invertible measure-preserving transformations of $(X,\mathcal{B},\mu)$. By the [Furstenberg-Katznelson multidimensional multiple recurrence theorem](/theorems/furstenberg-katznelson-multidimensional-multiple-recurrence-theorem), applied to this finite commuting family and to the set $A$ with $\mu(A)>0$, there exists $r \in \mathbb{N}_{>0}$ such that \begin{align*} \mu\left(\bigcap_{f \in F} S_f^{-r}A\right)>0. \end{align*} Since $S_f^r=T_{rf}$ for each $f \in F$, this is \begin{align*} \mu\left(\bigcap_{f \in F} T_{rf}^{-1}A\right)>0. \end{align*} [guided] We now use the recurrence theorem on the finite list of pattern directions. For every $f \in F$, define a transformation $S_f: X \to X$ by \begin{align*} S_f(x) := T_f x. \end{align*} The set $F$ is finite by hypothesis, so $(S_f)_{f \in F}$ is a finite family. Each $S_f$ is invertible and measure-preserving because each $T_v$ in the $\mathbb{Z}^d$-action has those properties. The transformations commute because the acting group $\mathbb{Z}^d$ is abelian: for $f,h \in F$, \begin{align*} S_f \circ S_h = T_f \circ T_h = T_{f+h}=T_{h+f}=T_h \circ T_f = S_h \circ S_f. \end{align*} The [Furstenberg-Katznelson multidimensional multiple recurrence theorem](/theorems/furstenberg-katznelson-multidimensional-multiple-recurrence-theorem) applies exactly to a probability space, a finite commuting family of measure-preserving transformations, and a measurable set of positive measure. These hypotheses have now been verified, and $\mu(A)>0$ came from the correspondence principle. Therefore there exists $r \in \mathbb{N}_{>0}$ such that \begin{align*} \mu\left(\bigcap_{f \in F} S_f^{-r}A\right)>0. \end{align*} Because $S_f=T_f$ and $T$ is an action of $\mathbb{Z}^d$, iterating $S_f$ exactly $r$ times gives \begin{align*} S_f^r = T_f^r = T_{rf}. \end{align*} Thus the recurrence conclusion can be rewritten in the lattice-indexed form \begin{align*} \mu\left(\bigcap_{f \in F} T_{rf}^{-1}A\right)>0. \end{align*} This is the form compatible with the correspondence principle, because the same vectors $rf \in \mathbb{Z}^d$ appear as translation parameters for $E$. [/guided] [/step] [step:Transfer the recurrence intersection back to a lattice intersection] Define the finite set $G_r \subset \mathbb{Z}^d$ by \begin{align*} G_r := \{rf : f \in F\}. \end{align*} The correspondence inequality applied to $G_r$ gives \begin{align*} d^*\left(\bigcap_{f \in F} (E-rf)\right) = d^*\left(\bigcap_{g \in G_r} (E-g)\right) \geq \mu\left(\bigcap_{g \in G_r} T_g^{-1}A\right) = \mu\left(\bigcap_{f \in F} T_{rf}^{-1}A\right)>0. \end{align*} Thus \begin{align*} \bigcap_{f \in F} (E-rf) \neq \varnothing. \end{align*} [guided] We now transfer the positive-measure recurrence statement back to the original set $E$. Define the finite set $G_r \subset \mathbb{Z}^d$ by \begin{align*} G_r := \{rf : f \in F\}. \end{align*} This set is finite because $F$ is finite. The correspondence principle applies to every finite subset of $\mathbb{Z}^d$, so it applies to $G_r$. It gives \begin{align*} d^*\left(\bigcap_{g \in G_r} (E-g)\right) \geq \mu\left(\bigcap_{g \in G_r} T_g^{-1}A\right). \end{align*} By the definition of $G_r$, the left-hand intersection is the same as $\bigcap_{f \in F}(E-rf)$, and the right-hand intersection is the same as $\bigcap_{f \in F}T_{rf}^{-1}A$. Therefore \begin{align*} d^*\left(\bigcap_{f \in F} (E-rf)\right) = d^*\left(\bigcap_{g \in G_r} (E-g)\right) \geq \mu\left(\bigcap_{g \in G_r} T_g^{-1}A\right) = \mu\left(\bigcap_{f \in F} T_{rf}^{-1}A\right)>0. \end{align*} A subset of $\mathbb{Z}^d$ with positive upper Banach density cannot be empty, since the empty set has upper Banach density $0$. Hence \begin{align*} \bigcap_{f \in F} (E-rf) \neq \varnothing. \end{align*} [/guided] [/step] [step:Choose a lattice point in the intersection and form the desired copy of $F$] Choose \begin{align*} a \in \bigcap_{f \in F} (E-rf). \end{align*} For every $f \in F$, the membership $a \in E-rf$ means, by definition of translate, that $a+rf \in E$. Hence \begin{align*} a+rF=\{a+rf:f\in F\}\subset E. \end{align*} The element $a \in \mathbb{Z}^d$ and the integer $r \in \mathbb{N}_{>0}$ are therefore the required translate and dilation. [guided] The previous step proved that the set \begin{align*} \bigcap_{f \in F} (E-rf) \end{align*} is nonempty. Choose an element \begin{align*} a \in \bigcap_{f \in F} (E-rf). \end{align*} This membership means that $a$ belongs to $E-rf$ for every $f \in F$. By the definition of the translate $E-rf=\{z \in \mathbb{Z}^d : z+rf \in E\}$, we therefore have $a+rf \in E$ for every $f \in F$. Collecting these inclusions over all $f \in F$ gives \begin{align*} a+rF=\{a+rf:f\in F\}\subset E. \end{align*} The point $a$ lies in $\mathbb{Z}^d$, and the recurrence step produced $r \in \mathbb{N}_{>0}$. Thus $a$ and $r$ are exactly the translate and positive dilation required by the theorem statement. [/guided] [/step]