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Drike
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$T$$G$ is a group with a simple first order theory $T$ as defined by Shelah, hence equiped with a "nice" notion of independence. $M$ is a model of $T$$G$ also has generic elements. I write $acl^n(A)$ for the set of elements in $M$ lying in a finite $A$-definable set of size at most $n$, and $acl(A)$ for the union of $acl^n(A)$ when $n$ ranges over all natural numbers. Let $a$ and $b$ be independent elements over $0$. Assume that they are not in $acl(0)$. One has

$$acl(a)\cap acl(b)=acl(\emptyset)$$

In particular, this holds if $a$ and $b$ are independent generic elements of $G$.

Question 1 : DoesIf $a$ and $b$ are generic, does the following equality hold? $$acl^n(a)\cap acl^n(b)=acl^n(\emptyset)$$

Am I allowed a second question? As I expect the answer to question 1 to be no, let me ask :

Question 2 : Is there a constant $k$ (depending only on $T$ and $n$) such that for all independent generics $a$ and $b$ over $0$, one would have $$acl^n(a)\cap acl^n(b)\subset acl^{k.n}(\emptyset)$$

$T$ is a simple first order theory as defined by Shelah, hence equiped with a "nice" notion of independence. $M$ is a model of $T$. I write $acl^n(A)$ for the set of elements in $M$ lying in a finite $A$-definable set of size at most $n$, and $acl(A)$ for the union of $acl^n(A)$ when $n$ ranges over all natural numbers. Let $a$ and $b$ be independent elements over $0$. Assume that they are not in $acl(0)$. One has

$$acl(a)\cap acl(b)=acl(\emptyset)$$

Question 1 : Does the following equality hold? $$acl^n(a)\cap acl^n(b)=acl^n(\emptyset)$$

Am I allowed a second question? As I expect the answer to question 1 to be no, let me ask :

Question 2 : Is there a constant $k$ (depending only on $T$ and $n$) such that for all independent $a$ and $b$ over $0$, one would have $$acl^n(a)\cap acl^n(b)\subset acl^{k.n}(\emptyset)$$

$G$ is a group with a simple first order theory $T$ as defined by Shelah, hence equiped with a "nice" notion of independence. $G$ also has generic elements. I write $acl^n(A)$ for the set of elements in $M$ lying in a finite $A$-definable set of size at most $n$, and $acl(A)$ for the union of $acl^n(A)$ when $n$ ranges over all natural numbers. Let $a$ and $b$ be independent elements over $0$. One has

$$acl(a)\cap acl(b)=acl(\emptyset)$$

In particular, this holds if $a$ and $b$ are independent generic elements of $G$.

Question 1 : If $a$ and $b$ are generic, does the following equality hold? $$acl^n(a)\cap acl^n(b)=acl^n(\emptyset)$$

As I expect the answer to question 1 to be no, let me ask :

Question 2 : Is there a constant $k$ (depending only on $T$ and $n$) such that for all independent generics $a$ and $b$ over $0$, one would have $$acl^n(a)\cap acl^n(b)\subset acl^{k.n}(\emptyset)$$

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Drike
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$T$ is a simple first order theory as defined by Shelah, hence equiped with a "nice" notion of independence. $M$ is a model of $T$. I write $acl^n(A)$ for the set of elements in $M$ lying in a finite $A$-definable set of size at most $n$, and $acl(A)$ for the union of $acl^n(A)$ when $n$ ranges over all natural numbers. Let $a$ and $b$ be independent elements over $0$. If I amAssume that they are not mistaken, onein $acl(0)$. One has

$$acl(a)\cap acl(b)=acl(\emptyset)$$

Question 1 : Does the following equality hold? $$acl^n(a)\cap acl^n(b)=acl^n(\emptyset)$$

Am I allowed a second question? As I expect the answer to question 1 to be no, let me ask :

Question 2 : Is there a constant $k$ (depending only on $T$ and $n$) such that for all independent $a$ and $b$ over $0$, one would have $$acl^n(a)\cap acl^n(b)\subset acl^{k.n}(\emptyset)$$

$T$ is a simple first order theory as defined by Shelah, hence equiped with a "nice" notion of independence. $M$ is a model of $T$. I write $acl^n(A)$ for the set of elements in $M$ lying in a finite $A$-definable set of size at most $n$, and $acl(A)$ for the union of $acl^n(A)$ when $n$ ranges over all natural numbers. Let $a$ and $b$ be independent elements over $0$. If I am not mistaken, one has

$$acl(a)\cap acl(b)=acl(\emptyset)$$

Question 1 : Does the following equality hold? $$acl^n(a)\cap acl^n(b)=acl^n(\emptyset)$$

Am I allowed a second question? As I expect the answer to question 1 to be no, let me ask :

Question 2 : Is there a constant $k$ (depending only on $T$ and $n$) such that for all independent $a$ and $b$ over $0$, one would have $$acl^n(a)\cap acl^n(b)\subset acl^{k.n}(\emptyset)$$

$T$ is a simple first order theory as defined by Shelah, hence equiped with a "nice" notion of independence. $M$ is a model of $T$. I write $acl^n(A)$ for the set of elements in $M$ lying in a finite $A$-definable set of size at most $n$, and $acl(A)$ for the union of $acl^n(A)$ when $n$ ranges over all natural numbers. Let $a$ and $b$ be independent elements over $0$. Assume that they are not in $acl(0)$. One has

$$acl(a)\cap acl(b)=acl(\emptyset)$$

Question 1 : Does the following equality hold? $$acl^n(a)\cap acl^n(b)=acl^n(\emptyset)$$

Am I allowed a second question? As I expect the answer to question 1 to be no, let me ask :

Question 2 : Is there a constant $k$ (depending only on $T$ and $n$) such that for all independent $a$ and $b$ over $0$, one would have $$acl^n(a)\cap acl^n(b)\subset acl^{k.n}(\emptyset)$$

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