Mixing
Substructures and universal sentences.
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On page 35 in Ziegler, Tent: Model Theory it is stated:
Let $T$ be an $L$-theory. It follows from 3.1.2. that the models of
$T_forall$ are the substructures of models of $T$.
with
Lemma 3.1.2. Let $T$ be a theory, $mathfrakA$ a structure and
$Delta$ a set of formulas closed under existential quantification,
conjunction and substitution of variables. Then the following are
equivalent:
a) All sentences $varphiinDelta$ which are true in
$mathfrakA$ are consistent with $T$.
b) There is a model
$mathfrakBvDash T$ and a map $f:mathfrakAto_Delta mathfrakB$.
I don't see why it follows directly from 3.1.2.
I try to argue by this, but I am not sure if it makes much sense:
Assume that $T$ is consistent. As $T_forall=psi_foralltext universal and Tvdashpsi_forall$ we have that $T_forallsubseteqtextDed(T)$ (deductive closure) and therefore $T_forallcuptextDed(T)$ and also $T_forallcup T$ are consistent. Now, pick a model $mathfrakA_forallvDash T_forall$. By the equivalence of 3.1.2. we have that there is a model $mathfrakBvDash T$ with a map $f:mathfrakA_forallto_T_forallmathfrakB$. I see no reason why I can pick $f$ to be the inclusion map.
Conversely, I don't see why if I have $mathfrakBvDash T$ and the inclusion $iota:mathfrakAto_T_forallmathfrakB$, that then $mathfrakAvDash T_forall$.
model-theory
asked 18 hours ago
Zikrunumea
726
add a comment |Â
up vote
0
down vote
favorite
On page 35 in Ziegler, Tent: Model Theory it is stated:
Let $T$ be an $L$-theory. It follows from 3.1.2. that the models of
$T_forall$ are the substructures of models of $T$.
with
Lemma 3.1.2. Let $T$ be a theory, $mathfrakA$ a structure and
$Delta$ a set of formulas closed under existential quantification,
conjunction and substitution of variables. Then the following are
equivalent:
a) All sentences $varphiinDelta$ which are true in
$mathfrakA$ are consistent with $T$.
b) There is a model
$mathfrakBvDash T$ and a map $f:mathfrakAto_Delta mathfrakB$.
I don't see why it follows directly from 3.1.2.
I try to argue by this, but I am not sure if it makes much sense:
Assume that $T$ is consistent. As $T_forall=psi_foralltext universal and Tvdashpsi_forall$ we have that $T_forallsubseteqtextDed(T)$ (deductive closure) and therefore $T_forallcuptextDed(T)$ and also $T_forallcup T$ are consistent. Now, pick a model $mathfrakA_forallvDash T_forall$. By the equivalence of 3.1.2. we have that there is a model $mathfrakBvDash T$ with a map $f:mathfrakA_forallto_T_forallmathfrakB$. I see no reason why I can pick $f$ to be the inclusion map.
Conversely, I don't see why if I have $mathfrakBvDash T$ and the inclusion $iota:mathfrakAto_T_forallmathfrakB$, that then $mathfrakAvDash T_forall$.
model-theory
asked 18 hours ago
Zikrunumea
726
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
On page 35 in Ziegler, Tent: Model Theory it is stated:
Let $T$ be an $L$-theory. It follows from 3.1.2. that the models of
$T_forall$ are the substructures of models of $T$.
with
Lemma 3.1.2. Let $T$ be a theory, $mathfrakA$ a structure and
$Delta$ a set of formulas closed under existential quantification,
conjunction and substitution of variables. Then the following are
equivalent:
a) All sentences $varphiinDelta$ which are true in
$mathfrakA$ are consistent with $T$.
b) There is a model
$mathfrakBvDash T$ and a map $f:mathfrakAto_Delta mathfrakB$.
I don't see why it follows directly from 3.1.2.
I try to argue by this, but I am not sure if it makes much sense:
Assume that $T$ is consistent. As $T_forall=psi_foralltext universal and Tvdashpsi_forall$ we have that $T_forallsubseteqtextDed(T)$ (deductive closure) and therefore $T_forallcuptextDed(T)$ and also $T_forallcup T$ are consistent. Now, pick a model $mathfrakA_forallvDash T_forall$. By the equivalence of 3.1.2. we have that there is a model $mathfrakBvDash T$ with a map $f:mathfrakA_forallto_T_forallmathfrakB$. I see no reason why I can pick $f$ to be the inclusion map.
Conversely, I don't see why if I have $mathfrakBvDash T$ and the inclusion $iota:mathfrakAto_T_forallmathfrakB$, that then $mathfrakAvDash T_forall$.
model-theory
asked 18 hours ago
Zikrunumea
726
On page 35 in Ziegler, Tent: Model Theory it is stated:
Let $T$ be an $L$-theory. It follows from 3.1.2. that the models of
$T_forall$ are the substructures of models of $T$.
with
Lemma 3.1.2. Let $T$ be a theory, $mathfrakA$ a structure and
$Delta$ a set of formulas closed under existential quantification,
conjunction and substitution of variables. Then the following are
equivalent:
a) All sentences $varphiinDelta$ which are true in
$mathfrakA$ are consistent with $T$.
b) There is a model
$mathfrakBvDash T$ and a map $f:mathfrakAto_Delta mathfrakB$.
I don't see why it follows directly from 3.1.2.
I try to argue by this, but I am not sure if it makes much sense:
Assume that $T$ is consistent. As $T_forall=psi_foralltext universal and Tvdashpsi_forall$ we have that $T_forallsubseteqtextDed(T)$ (deductive closure) and therefore $T_forallcuptextDed(T)$ and also $T_forallcup T$ are consistent. Now, pick a model $mathfrakA_forallvDash T_forall$. By the equivalence of 3.1.2. we have that there is a model $mathfrakBvDash T$ with a map $f:mathfrakA_forallto_T_forallmathfrakB$. I see no reason why I can pick $f$ to be the inclusion map.
Conversely, I don't see why if I have $mathfrakBvDash T$ and the inclusion $iota:mathfrakAto_T_forallmathfrakB$, that then $mathfrakAvDash T_forall$.
model-theory
asked 18 hours ago
Zikrunumea
726
asked 18 hours ago
Zikrunumea
726
asked 18 hours ago
Zikrunumea
726
asked 18 hours ago
Zikrunumea
726
726
add a comment |Â
add a comment |Â
1 Answer
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0
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I think the authors' intention is to apply 3.1.2 with $Delta$ being the set of existential formulas. Then the proof of the alleged consequence would go like this. Suppose $mathfrak Amodels T_forall$. Then I claim clause (a) of 3.1.2 is true. Indeed, if $phi$ is an existential sentence true in $mathfrak A$, then $negphi$ is (up to logical equivalence) a universal sentence that is false in $mathfrak A$ and therefore is not in $T_forall$. So $phi$ is consistent with $T$. By 3.1.2, we infer that clause (b) is also true, so we have $mathfrak Bmodels T$ and $f:mathfrak Ato_Deltamathfrak B$.
Now we need another (standard but easy to get wrong) maneuver to get $f$ to be an inclusion. As it stands, $f$ is an isomorphism from $mathfrak A$ to a substructure $mathfrak A'$ of $mathfrak B$ (because all atomic formulas and their negations are in $Delta$). The idea is to modify $mathfrak B$ by replacing this copy $mathfrak A'$ of $mathfrak A$ with $mathfrak A$ itself. That way you get an isomorphic copy of $mathfrak B$ (hence still a model of $T$) that contains literally $mathfrak A$ rather than a copy $mathfrak A'$ of it. The problem (the reason it's easy to get wrong) is that, if we're unlucky, $mathfrak B$ might already contain some elements of $mathfrak A$, and not in the place where we want them; then if we just replace $mathfrak A'$ with $mathfrak A$ we'll end up with some elements of $mathfrak A$ trying to be in two different places in $mathfrak B$. Fortunately, the cure for this problem is easy: First replace $mathfrak B$ by an isomorphic copy that is disjoint from $mathfrak A$, and then proceed as above.
Finally, to show the converse, that all substructures of models of $T$ are models of $T_forall$, prove, by induction on universal formulas $theta$, that, if $mathfrak A$ is a substructure of $mathfrak B$ and if some assignment of values for variables in $mathfrak A$ satisfies $theta$ in $mathfrak B$, then it also satisfies $theta$ in $mathfrak A$. The only nontrivial step in the induction is the universal quantifier step, and there you just use that all elements of $mathfrak A$ (that might serve as counterexamples) are also in $mathfrak B$ (and therefore cannot be counterexamples).
answered 17 hours ago
Andreas Blass
47.3k348103
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
I think the authors' intention is to apply 3.1.2 with $Delta$ being the set of existential formulas. Then the proof of the alleged consequence would go like this. Suppose $mathfrak Amodels T_forall$. Then I claim clause (a) of 3.1.2 is true. Indeed, if $phi$ is an existential sentence true in $mathfrak A$, then $negphi$ is (up to logical equivalence) a universal sentence that is false in $mathfrak A$ and therefore is not in $T_forall$. So $phi$ is consistent with $T$. By 3.1.2, we infer that clause (b) is also true, so we have $mathfrak Bmodels T$ and $f:mathfrak Ato_Deltamathfrak B$.
Now we need another (standard but easy to get wrong) maneuver to get $f$ to be an inclusion. As it stands, $f$ is an isomorphism from $mathfrak A$ to a substructure $mathfrak A'$ of $mathfrak B$ (because all atomic formulas and their negations are in $Delta$). The idea is to modify $mathfrak B$ by replacing this copy $mathfrak A'$ of $mathfrak A$ with $mathfrak A$ itself. That way you get an isomorphic copy of $mathfrak B$ (hence still a model of $T$) that contains literally $mathfrak A$ rather than a copy $mathfrak A'$ of it. The problem (the reason it's easy to get wrong) is that, if we're unlucky, $mathfrak B$ might already contain some elements of $mathfrak A$, and not in the place where we want them; then if we just replace $mathfrak A'$ with $mathfrak A$ we'll end up with some elements of $mathfrak A$ trying to be in two different places in $mathfrak B$. Fortunately, the cure for this problem is easy: First replace $mathfrak B$ by an isomorphic copy that is disjoint from $mathfrak A$, and then proceed as above.
Finally, to show the converse, that all substructures of models of $T$ are models of $T_forall$, prove, by induction on universal formulas $theta$, that, if $mathfrak A$ is a substructure of $mathfrak B$ and if some assignment of values for variables in $mathfrak A$ satisfies $theta$ in $mathfrak B$, then it also satisfies $theta$ in $mathfrak A$. The only nontrivial step in the induction is the universal quantifier step, and there you just use that all elements of $mathfrak A$ (that might serve as counterexamples) are also in $mathfrak B$ (and therefore cannot be counterexamples).
answered 17 hours ago
Andreas Blass
47.3k348103
add a comment |Â
up vote
0
down vote
I think the authors' intention is to apply 3.1.2 with $Delta$ being the set of existential formulas. Then the proof of the alleged consequence would go like this. Suppose $mathfrak Amodels T_forall$. Then I claim clause (a) of 3.1.2 is true. Indeed, if $phi$ is an existential sentence true in $mathfrak A$, then $negphi$ is (up to logical equivalence) a universal sentence that is false in $mathfrak A$ and therefore is not in $T_forall$. So $phi$ is consistent with $T$. By 3.1.2, we infer that clause (b) is also true, so we have $mathfrak Bmodels T$ and $f:mathfrak Ato_Deltamathfrak B$.
Now we need another (standard but easy to get wrong) maneuver to get $f$ to be an inclusion. As it stands, $f$ is an isomorphism from $mathfrak A$ to a substructure $mathfrak A'$ of $mathfrak B$ (because all atomic formulas and their negations are in $Delta$). The idea is to modify $mathfrak B$ by replacing this copy $mathfrak A'$ of $mathfrak A$ with $mathfrak A$ itself. That way you get an isomorphic copy of $mathfrak B$ (hence still a model of $T$) that contains literally $mathfrak A$ rather than a copy $mathfrak A'$ of it. The problem (the reason it's easy to get wrong) is that, if we're unlucky, $mathfrak B$ might already contain some elements of $mathfrak A$, and not in the place where we want them; then if we just replace $mathfrak A'$ with $mathfrak A$ we'll end up with some elements of $mathfrak A$ trying to be in two different places in $mathfrak B$. Fortunately, the cure for this problem is easy: First replace $mathfrak B$ by an isomorphic copy that is disjoint from $mathfrak A$, and then proceed as above.
Finally, to show the converse, that all substructures of models of $T$ are models of $T_forall$, prove, by induction on universal formulas $theta$, that, if $mathfrak A$ is a substructure of $mathfrak B$ and if some assignment of values for variables in $mathfrak A$ satisfies $theta$ in $mathfrak B$, then it also satisfies $theta$ in $mathfrak A$. The only nontrivial step in the induction is the universal quantifier step, and there you just use that all elements of $mathfrak A$ (that might serve as counterexamples) are also in $mathfrak B$ (and therefore cannot be counterexamples).
answered 17 hours ago
Andreas Blass
47.3k348103
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I think the authors' intention is to apply 3.1.2 with $Delta$ being the set of existential formulas. Then the proof of the alleged consequence would go like this. Suppose $mathfrak Amodels T_forall$. Then I claim clause (a) of 3.1.2 is true. Indeed, if $phi$ is an existential sentence true in $mathfrak A$, then $negphi$ is (up to logical equivalence) a universal sentence that is false in $mathfrak A$ and therefore is not in $T_forall$. So $phi$ is consistent with $T$. By 3.1.2, we infer that clause (b) is also true, so we have $mathfrak Bmodels T$ and $f:mathfrak Ato_Deltamathfrak B$.
Now we need another (standard but easy to get wrong) maneuver to get $f$ to be an inclusion. As it stands, $f$ is an isomorphism from $mathfrak A$ to a substructure $mathfrak A'$ of $mathfrak B$ (because all atomic formulas and their negations are in $Delta$). The idea is to modify $mathfrak B$ by replacing this copy $mathfrak A'$ of $mathfrak A$ with $mathfrak A$ itself. That way you get an isomorphic copy of $mathfrak B$ (hence still a model of $T$) that contains literally $mathfrak A$ rather than a copy $mathfrak A'$ of it. The problem (the reason it's easy to get wrong) is that, if we're unlucky, $mathfrak B$ might already contain some elements of $mathfrak A$, and not in the place where we want them; then if we just replace $mathfrak A'$ with $mathfrak A$ we'll end up with some elements of $mathfrak A$ trying to be in two different places in $mathfrak B$. Fortunately, the cure for this problem is easy: First replace $mathfrak B$ by an isomorphic copy that is disjoint from $mathfrak A$, and then proceed as above.
Finally, to show the converse, that all substructures of models of $T$ are models of $T_forall$, prove, by induction on universal formulas $theta$, that, if $mathfrak A$ is a substructure of $mathfrak B$ and if some assignment of values for variables in $mathfrak A$ satisfies $theta$ in $mathfrak B$, then it also satisfies $theta$ in $mathfrak A$. The only nontrivial step in the induction is the universal quantifier step, and there you just use that all elements of $mathfrak A$ (that might serve as counterexamples) are also in $mathfrak B$ (and therefore cannot be counterexamples).
answered 17 hours ago
Andreas Blass
47.3k348103
I think the authors' intention is to apply 3.1.2 with $Delta$ being the set of existential formulas. Then the proof of the alleged consequence would go like this. Suppose $mathfrak Amodels T_forall$. Then I claim clause (a) of 3.1.2 is true. Indeed, if $phi$ is an existential sentence true in $mathfrak A$, then $negphi$ is (up to logical equivalence) a universal sentence that is false in $mathfrak A$ and therefore is not in $T_forall$. So $phi$ is consistent with $T$. By 3.1.2, we infer that clause (b) is also true, so we have $mathfrak Bmodels T$ and $f:mathfrak Ato_Deltamathfrak B$.
Now we need another (standard but easy to get wrong) maneuver to get $f$ to be an inclusion. As it stands, $f$ is an isomorphism from $mathfrak A$ to a substructure $mathfrak A'$ of $mathfrak B$ (because all atomic formulas and their negations are in $Delta$). The idea is to modify $mathfrak B$ by replacing this copy $mathfrak A'$ of $mathfrak A$ with $mathfrak A$ itself. That way you get an isomorphic copy of $mathfrak B$ (hence still a model of $T$) that contains literally $mathfrak A$ rather than a copy $mathfrak A'$ of it. The problem (the reason it's easy to get wrong) is that, if we're unlucky, $mathfrak B$ might already contain some elements of $mathfrak A$, and not in the place where we want them; then if we just replace $mathfrak A'$ with $mathfrak A$ we'll end up with some elements of $mathfrak A$ trying to be in two different places in $mathfrak B$. Fortunately, the cure for this problem is easy: First replace $mathfrak B$ by an isomorphic copy that is disjoint from $mathfrak A$, and then proceed as above.
Finally, to show the converse, that all substructures of models of $T$ are models of $T_forall$, prove, by induction on universal formulas $theta$, that, if $mathfrak A$ is a substructure of $mathfrak B$ and if some assignment of values for variables in $mathfrak A$ satisfies $theta$ in $mathfrak B$, then it also satisfies $theta$ in $mathfrak A$. The only nontrivial step in the induction is the universal quantifier step, and there you just use that all elements of $mathfrak A$ (that might serve as counterexamples) are also in $mathfrak B$ (and therefore cannot be counterexamples).
answered 17 hours ago
Andreas Blass
47.3k348103
answered 17 hours ago
Andreas Blass
47.3k348103
answered 17 hours ago
Andreas Blass
47.3k348103
answered 17 hours ago
Andreas Blass
47.3k348103
47.3k348103
add a comment |Â
add a comment |Â
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