$exists^infty$-elimination and model companion.
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In the book Ziegler, Tent: A course in model theory it states
Exercise 5.5.7. Let $T_1$ and $T_2$ be two model complete theories in disjoint
languages $L_1$ and $L_2$. Assume that both theories eliminate $exists^infty$ . Then $T_1cup T_2$ has a model companion.
I want to prove this statement. Sine $T_1cup T_2$ also eliminates $exists^infty$ (i.e. in all models realisation sets have either a finite upper bound or are infinite) my strategy was to use the previous exercise
Exercise 5.5.6. Assume that T eliminates the quantifier $exists^infty$. Then for every formula $varphi(x_1 , . . . , x_n , overline{y})$ there is a formula $theta(overline{y})$ such that in all models $mathfrak{M}$ of
$T$ a tuple $overline{b}$ satisfies $theta(overline{y})$ if and only if $mathfrak{M}$ has an elementary extension $mathfrak{M}'$ with
elements $a_1 , . . . , a_nin M'setminus M$ such that $mathfrak{M}vDashvarphi(a_1,...,a_n)$.
I define the theory
$(T_1cup T_2)^*:={theta;|;$there exists a $L_1cup L_2$-formula $varphi(x_1,..x_n)$, that is satisfiable in all models $mathfrak{M}vDash T_1cup T_2$ such that $mathfrak{M}vDashthetaLeftrightarrow$ there exists an elementary extension $mathfrak{M}'$ with
elements $a_1 , . . . , a_nin M'setminus M$ such that $mathfrak{M}vDashvarphi(a_1,...,a_n)$$}$
To me it seemed to be right choice for the model companion of $T_1cup T_2$ until I tried to prove that $(T_1cup T_2)^*$ is model complete (without success). Does this choice make sense?
model-theory
add a comment |
up vote
2
down vote
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In the book Ziegler, Tent: A course in model theory it states
Exercise 5.5.7. Let $T_1$ and $T_2$ be two model complete theories in disjoint
languages $L_1$ and $L_2$. Assume that both theories eliminate $exists^infty$ . Then $T_1cup T_2$ has a model companion.
I want to prove this statement. Sine $T_1cup T_2$ also eliminates $exists^infty$ (i.e. in all models realisation sets have either a finite upper bound or are infinite) my strategy was to use the previous exercise
Exercise 5.5.6. Assume that T eliminates the quantifier $exists^infty$. Then for every formula $varphi(x_1 , . . . , x_n , overline{y})$ there is a formula $theta(overline{y})$ such that in all models $mathfrak{M}$ of
$T$ a tuple $overline{b}$ satisfies $theta(overline{y})$ if and only if $mathfrak{M}$ has an elementary extension $mathfrak{M}'$ with
elements $a_1 , . . . , a_nin M'setminus M$ such that $mathfrak{M}vDashvarphi(a_1,...,a_n)$.
I define the theory
$(T_1cup T_2)^*:={theta;|;$there exists a $L_1cup L_2$-formula $varphi(x_1,..x_n)$, that is satisfiable in all models $mathfrak{M}vDash T_1cup T_2$ such that $mathfrak{M}vDashthetaLeftrightarrow$ there exists an elementary extension $mathfrak{M}'$ with
elements $a_1 , . . . , a_nin M'setminus M$ such that $mathfrak{M}vDashvarphi(a_1,...,a_n)$$}$
To me it seemed to be right choice for the model companion of $T_1cup T_2$ until I tried to prove that $(T_1cup T_2)^*$ is model complete (without success). Does this choice make sense?
model-theory
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
In the book Ziegler, Tent: A course in model theory it states
Exercise 5.5.7. Let $T_1$ and $T_2$ be two model complete theories in disjoint
languages $L_1$ and $L_2$. Assume that both theories eliminate $exists^infty$ . Then $T_1cup T_2$ has a model companion.
I want to prove this statement. Sine $T_1cup T_2$ also eliminates $exists^infty$ (i.e. in all models realisation sets have either a finite upper bound or are infinite) my strategy was to use the previous exercise
Exercise 5.5.6. Assume that T eliminates the quantifier $exists^infty$. Then for every formula $varphi(x_1 , . . . , x_n , overline{y})$ there is a formula $theta(overline{y})$ such that in all models $mathfrak{M}$ of
$T$ a tuple $overline{b}$ satisfies $theta(overline{y})$ if and only if $mathfrak{M}$ has an elementary extension $mathfrak{M}'$ with
elements $a_1 , . . . , a_nin M'setminus M$ such that $mathfrak{M}vDashvarphi(a_1,...,a_n)$.
I define the theory
$(T_1cup T_2)^*:={theta;|;$there exists a $L_1cup L_2$-formula $varphi(x_1,..x_n)$, that is satisfiable in all models $mathfrak{M}vDash T_1cup T_2$ such that $mathfrak{M}vDashthetaLeftrightarrow$ there exists an elementary extension $mathfrak{M}'$ with
elements $a_1 , . . . , a_nin M'setminus M$ such that $mathfrak{M}vDashvarphi(a_1,...,a_n)$$}$
To me it seemed to be right choice for the model companion of $T_1cup T_2$ until I tried to prove that $(T_1cup T_2)^*$ is model complete (without success). Does this choice make sense?
model-theory
In the book Ziegler, Tent: A course in model theory it states
Exercise 5.5.7. Let $T_1$ and $T_2$ be two model complete theories in disjoint
languages $L_1$ and $L_2$. Assume that both theories eliminate $exists^infty$ . Then $T_1cup T_2$ has a model companion.
I want to prove this statement. Sine $T_1cup T_2$ also eliminates $exists^infty$ (i.e. in all models realisation sets have either a finite upper bound or are infinite) my strategy was to use the previous exercise
Exercise 5.5.6. Assume that T eliminates the quantifier $exists^infty$. Then for every formula $varphi(x_1 , . . . , x_n , overline{y})$ there is a formula $theta(overline{y})$ such that in all models $mathfrak{M}$ of
$T$ a tuple $overline{b}$ satisfies $theta(overline{y})$ if and only if $mathfrak{M}$ has an elementary extension $mathfrak{M}'$ with
elements $a_1 , . . . , a_nin M'setminus M$ such that $mathfrak{M}vDashvarphi(a_1,...,a_n)$.
I define the theory
$(T_1cup T_2)^*:={theta;|;$there exists a $L_1cup L_2$-formula $varphi(x_1,..x_n)$, that is satisfiable in all models $mathfrak{M}vDash T_1cup T_2$ such that $mathfrak{M}vDashthetaLeftrightarrow$ there exists an elementary extension $mathfrak{M}'$ with
elements $a_1 , . . . , a_nin M'setminus M$ such that $mathfrak{M}vDashvarphi(a_1,...,a_n)$$}$
To me it seemed to be right choice for the model companion of $T_1cup T_2$ until I tried to prove that $(T_1cup T_2)^*$ is model complete (without success). Does this choice make sense?
model-theory
model-theory
asked Nov 14 at 6:10
Zikrunumea
438
438
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Since $T_1$ and $T_2$ are model complete, they are $forallexists$-axiomatizable, and $T_1cup T_2$ is also $forallexists$-axiomatizable. So to show the model companion exists, you need to axiomatize the existentially closed models of $T_1cup T_2$.
So suppose $Msubseteq M'$, both models of $T_1cup T_2$. We want to write down a sufficient condition for $M$ to be existentially closed in $M'$. Let $psi(x)$ be a quantifier-free formula with parameters from $M$ such that $M'models exists x, psi(x)$. You want to observe the following things:
It suffices to assume that $psi(x)$ is a conjunction of atomic and negated atomic $(L_1cup L_2)$-formulas.
So if $L_1$ and $L_2$ are relational, $psi(x)$ is actually a conjunction $varphi_1(x)land varphi_2(x)$, where $varphi_1$ is an $L_1$-formula and $varphi_2$ is an $L_2$-formula. If the languages aren't relational, there's an issue that $psi$ could mention terms formed using function symbols from both $L_1$ and $L_2$. Then you have to think about replacing $psi$ with another formula obtained by "unnesting" all terms. It might be better to work out the details in the relational case first and then come back to this complication.
It suffices to assume that there is a witness $M'models psi(a')$ such that each element of $a'$ is in $M'setminus M$ and the elements of $a'$ are all distinct.
This leads us to the following axiomatization: Let $varphi_1(x,y)$ be a quantifier-free $L_1$-formula, and let $varphi_2(x,z)$ be a quantifier-free $L_2$-formula (here $x$, $y$, $z$ are tuples of variables). Let $theta_1(y)$ and $theta_2(z)$ be the formulas provided by Exercise 5.5.6. for $varphi_1$ and $varphi_2$, respectively. Let $theta'_1(y)$ be the conjunction of $theta_1(y)$ and inequations $y_ineq y_j$ for $ineq j$, and similarly for $theta'_2(z)$. Then look at the following sentence: $$forall y, forall z, ((theta'_1(y)land theta'_2(z))rightarrow exists x, (varphi_1(x,y)land varphi_2(x,z))).$$
The model companion of $T_1cup T_2$ is axiomatized by $T_1cup T_2$ together with all sentences of the above form. The discussion above can be viewed as an extended hint that every model of this theory is an existentially closed model of $T_1cup T_2$. You also need to show the converse: that every existentially closed model of $T_1cup T_2$ satisfies these extra axioms. Explicitly, given $varphi_1(x,y)$ and $varphi_2(x,z)$, if $Mmodels theta_1'(b)land theta_2'(c)$, then you can embed $M$ in a model $M'$ of $T_1cup T_2$ such that $M'models varphi_1(a',b)land varphi_2(a',c)$ for some $a'in M'$, and use the fact that $M$ is existentially closed to find a witness in $M$.
Aside: The statement of the exercise is a result from Peter Winkler's 1975 PhD thesis. It's a nice coincidence that you posted this problem now. Minh Tran, Erik Walsberg, and I are working on a project that we call "Interpolative Fusions" which at its heart is a generalization of this result of Winkler's. We just posted the first paper from this project to the arXiv this week: https://arxiv.org/abs/1811.06108
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Since $T_1$ and $T_2$ are model complete, they are $forallexists$-axiomatizable, and $T_1cup T_2$ is also $forallexists$-axiomatizable. So to show the model companion exists, you need to axiomatize the existentially closed models of $T_1cup T_2$.
So suppose $Msubseteq M'$, both models of $T_1cup T_2$. We want to write down a sufficient condition for $M$ to be existentially closed in $M'$. Let $psi(x)$ be a quantifier-free formula with parameters from $M$ such that $M'models exists x, psi(x)$. You want to observe the following things:
It suffices to assume that $psi(x)$ is a conjunction of atomic and negated atomic $(L_1cup L_2)$-formulas.
So if $L_1$ and $L_2$ are relational, $psi(x)$ is actually a conjunction $varphi_1(x)land varphi_2(x)$, where $varphi_1$ is an $L_1$-formula and $varphi_2$ is an $L_2$-formula. If the languages aren't relational, there's an issue that $psi$ could mention terms formed using function symbols from both $L_1$ and $L_2$. Then you have to think about replacing $psi$ with another formula obtained by "unnesting" all terms. It might be better to work out the details in the relational case first and then come back to this complication.
It suffices to assume that there is a witness $M'models psi(a')$ such that each element of $a'$ is in $M'setminus M$ and the elements of $a'$ are all distinct.
This leads us to the following axiomatization: Let $varphi_1(x,y)$ be a quantifier-free $L_1$-formula, and let $varphi_2(x,z)$ be a quantifier-free $L_2$-formula (here $x$, $y$, $z$ are tuples of variables). Let $theta_1(y)$ and $theta_2(z)$ be the formulas provided by Exercise 5.5.6. for $varphi_1$ and $varphi_2$, respectively. Let $theta'_1(y)$ be the conjunction of $theta_1(y)$ and inequations $y_ineq y_j$ for $ineq j$, and similarly for $theta'_2(z)$. Then look at the following sentence: $$forall y, forall z, ((theta'_1(y)land theta'_2(z))rightarrow exists x, (varphi_1(x,y)land varphi_2(x,z))).$$
The model companion of $T_1cup T_2$ is axiomatized by $T_1cup T_2$ together with all sentences of the above form. The discussion above can be viewed as an extended hint that every model of this theory is an existentially closed model of $T_1cup T_2$. You also need to show the converse: that every existentially closed model of $T_1cup T_2$ satisfies these extra axioms. Explicitly, given $varphi_1(x,y)$ and $varphi_2(x,z)$, if $Mmodels theta_1'(b)land theta_2'(c)$, then you can embed $M$ in a model $M'$ of $T_1cup T_2$ such that $M'models varphi_1(a',b)land varphi_2(a',c)$ for some $a'in M'$, and use the fact that $M$ is existentially closed to find a witness in $M$.
Aside: The statement of the exercise is a result from Peter Winkler's 1975 PhD thesis. It's a nice coincidence that you posted this problem now. Minh Tran, Erik Walsberg, and I are working on a project that we call "Interpolative Fusions" which at its heart is a generalization of this result of Winkler's. We just posted the first paper from this project to the arXiv this week: https://arxiv.org/abs/1811.06108
add a comment |
up vote
2
down vote
accepted
Since $T_1$ and $T_2$ are model complete, they are $forallexists$-axiomatizable, and $T_1cup T_2$ is also $forallexists$-axiomatizable. So to show the model companion exists, you need to axiomatize the existentially closed models of $T_1cup T_2$.
So suppose $Msubseteq M'$, both models of $T_1cup T_2$. We want to write down a sufficient condition for $M$ to be existentially closed in $M'$. Let $psi(x)$ be a quantifier-free formula with parameters from $M$ such that $M'models exists x, psi(x)$. You want to observe the following things:
It suffices to assume that $psi(x)$ is a conjunction of atomic and negated atomic $(L_1cup L_2)$-formulas.
So if $L_1$ and $L_2$ are relational, $psi(x)$ is actually a conjunction $varphi_1(x)land varphi_2(x)$, where $varphi_1$ is an $L_1$-formula and $varphi_2$ is an $L_2$-formula. If the languages aren't relational, there's an issue that $psi$ could mention terms formed using function symbols from both $L_1$ and $L_2$. Then you have to think about replacing $psi$ with another formula obtained by "unnesting" all terms. It might be better to work out the details in the relational case first and then come back to this complication.
It suffices to assume that there is a witness $M'models psi(a')$ such that each element of $a'$ is in $M'setminus M$ and the elements of $a'$ are all distinct.
This leads us to the following axiomatization: Let $varphi_1(x,y)$ be a quantifier-free $L_1$-formula, and let $varphi_2(x,z)$ be a quantifier-free $L_2$-formula (here $x$, $y$, $z$ are tuples of variables). Let $theta_1(y)$ and $theta_2(z)$ be the formulas provided by Exercise 5.5.6. for $varphi_1$ and $varphi_2$, respectively. Let $theta'_1(y)$ be the conjunction of $theta_1(y)$ and inequations $y_ineq y_j$ for $ineq j$, and similarly for $theta'_2(z)$. Then look at the following sentence: $$forall y, forall z, ((theta'_1(y)land theta'_2(z))rightarrow exists x, (varphi_1(x,y)land varphi_2(x,z))).$$
The model companion of $T_1cup T_2$ is axiomatized by $T_1cup T_2$ together with all sentences of the above form. The discussion above can be viewed as an extended hint that every model of this theory is an existentially closed model of $T_1cup T_2$. You also need to show the converse: that every existentially closed model of $T_1cup T_2$ satisfies these extra axioms. Explicitly, given $varphi_1(x,y)$ and $varphi_2(x,z)$, if $Mmodels theta_1'(b)land theta_2'(c)$, then you can embed $M$ in a model $M'$ of $T_1cup T_2$ such that $M'models varphi_1(a',b)land varphi_2(a',c)$ for some $a'in M'$, and use the fact that $M$ is existentially closed to find a witness in $M$.
Aside: The statement of the exercise is a result from Peter Winkler's 1975 PhD thesis. It's a nice coincidence that you posted this problem now. Minh Tran, Erik Walsberg, and I are working on a project that we call "Interpolative Fusions" which at its heart is a generalization of this result of Winkler's. We just posted the first paper from this project to the arXiv this week: https://arxiv.org/abs/1811.06108
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Since $T_1$ and $T_2$ are model complete, they are $forallexists$-axiomatizable, and $T_1cup T_2$ is also $forallexists$-axiomatizable. So to show the model companion exists, you need to axiomatize the existentially closed models of $T_1cup T_2$.
So suppose $Msubseteq M'$, both models of $T_1cup T_2$. We want to write down a sufficient condition for $M$ to be existentially closed in $M'$. Let $psi(x)$ be a quantifier-free formula with parameters from $M$ such that $M'models exists x, psi(x)$. You want to observe the following things:
It suffices to assume that $psi(x)$ is a conjunction of atomic and negated atomic $(L_1cup L_2)$-formulas.
So if $L_1$ and $L_2$ are relational, $psi(x)$ is actually a conjunction $varphi_1(x)land varphi_2(x)$, where $varphi_1$ is an $L_1$-formula and $varphi_2$ is an $L_2$-formula. If the languages aren't relational, there's an issue that $psi$ could mention terms formed using function symbols from both $L_1$ and $L_2$. Then you have to think about replacing $psi$ with another formula obtained by "unnesting" all terms. It might be better to work out the details in the relational case first and then come back to this complication.
It suffices to assume that there is a witness $M'models psi(a')$ such that each element of $a'$ is in $M'setminus M$ and the elements of $a'$ are all distinct.
This leads us to the following axiomatization: Let $varphi_1(x,y)$ be a quantifier-free $L_1$-formula, and let $varphi_2(x,z)$ be a quantifier-free $L_2$-formula (here $x$, $y$, $z$ are tuples of variables). Let $theta_1(y)$ and $theta_2(z)$ be the formulas provided by Exercise 5.5.6. for $varphi_1$ and $varphi_2$, respectively. Let $theta'_1(y)$ be the conjunction of $theta_1(y)$ and inequations $y_ineq y_j$ for $ineq j$, and similarly for $theta'_2(z)$. Then look at the following sentence: $$forall y, forall z, ((theta'_1(y)land theta'_2(z))rightarrow exists x, (varphi_1(x,y)land varphi_2(x,z))).$$
The model companion of $T_1cup T_2$ is axiomatized by $T_1cup T_2$ together with all sentences of the above form. The discussion above can be viewed as an extended hint that every model of this theory is an existentially closed model of $T_1cup T_2$. You also need to show the converse: that every existentially closed model of $T_1cup T_2$ satisfies these extra axioms. Explicitly, given $varphi_1(x,y)$ and $varphi_2(x,z)$, if $Mmodels theta_1'(b)land theta_2'(c)$, then you can embed $M$ in a model $M'$ of $T_1cup T_2$ such that $M'models varphi_1(a',b)land varphi_2(a',c)$ for some $a'in M'$, and use the fact that $M$ is existentially closed to find a witness in $M$.
Aside: The statement of the exercise is a result from Peter Winkler's 1975 PhD thesis. It's a nice coincidence that you posted this problem now. Minh Tran, Erik Walsberg, and I are working on a project that we call "Interpolative Fusions" which at its heart is a generalization of this result of Winkler's. We just posted the first paper from this project to the arXiv this week: https://arxiv.org/abs/1811.06108
Since $T_1$ and $T_2$ are model complete, they are $forallexists$-axiomatizable, and $T_1cup T_2$ is also $forallexists$-axiomatizable. So to show the model companion exists, you need to axiomatize the existentially closed models of $T_1cup T_2$.
So suppose $Msubseteq M'$, both models of $T_1cup T_2$. We want to write down a sufficient condition for $M$ to be existentially closed in $M'$. Let $psi(x)$ be a quantifier-free formula with parameters from $M$ such that $M'models exists x, psi(x)$. You want to observe the following things:
It suffices to assume that $psi(x)$ is a conjunction of atomic and negated atomic $(L_1cup L_2)$-formulas.
So if $L_1$ and $L_2$ are relational, $psi(x)$ is actually a conjunction $varphi_1(x)land varphi_2(x)$, where $varphi_1$ is an $L_1$-formula and $varphi_2$ is an $L_2$-formula. If the languages aren't relational, there's an issue that $psi$ could mention terms formed using function symbols from both $L_1$ and $L_2$. Then you have to think about replacing $psi$ with another formula obtained by "unnesting" all terms. It might be better to work out the details in the relational case first and then come back to this complication.
It suffices to assume that there is a witness $M'models psi(a')$ such that each element of $a'$ is in $M'setminus M$ and the elements of $a'$ are all distinct.
This leads us to the following axiomatization: Let $varphi_1(x,y)$ be a quantifier-free $L_1$-formula, and let $varphi_2(x,z)$ be a quantifier-free $L_2$-formula (here $x$, $y$, $z$ are tuples of variables). Let $theta_1(y)$ and $theta_2(z)$ be the formulas provided by Exercise 5.5.6. for $varphi_1$ and $varphi_2$, respectively. Let $theta'_1(y)$ be the conjunction of $theta_1(y)$ and inequations $y_ineq y_j$ for $ineq j$, and similarly for $theta'_2(z)$. Then look at the following sentence: $$forall y, forall z, ((theta'_1(y)land theta'_2(z))rightarrow exists x, (varphi_1(x,y)land varphi_2(x,z))).$$
The model companion of $T_1cup T_2$ is axiomatized by $T_1cup T_2$ together with all sentences of the above form. The discussion above can be viewed as an extended hint that every model of this theory is an existentially closed model of $T_1cup T_2$. You also need to show the converse: that every existentially closed model of $T_1cup T_2$ satisfies these extra axioms. Explicitly, given $varphi_1(x,y)$ and $varphi_2(x,z)$, if $Mmodels theta_1'(b)land theta_2'(c)$, then you can embed $M$ in a model $M'$ of $T_1cup T_2$ such that $M'models varphi_1(a',b)land varphi_2(a',c)$ for some $a'in M'$, and use the fact that $M$ is existentially closed to find a witness in $M$.
Aside: The statement of the exercise is a result from Peter Winkler's 1975 PhD thesis. It's a nice coincidence that you posted this problem now. Minh Tran, Erik Walsberg, and I are working on a project that we call "Interpolative Fusions" which at its heart is a generalization of this result of Winkler's. We just posted the first paper from this project to the arXiv this week: https://arxiv.org/abs/1811.06108
edited Nov 18 at 11:15
answered Nov 18 at 9:29
Alex Kruckman
25.7k22455
25.7k22455
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