What features of a logic make possible the proof of Downward Lowenheim-Skolem?
$begingroup$
The Downward Lowenheim-Skolem Theorem asserts that if a countable first-order theory has an infinite model, then it has a countable model.
Although associated with first-order logic, the result also applies to second-order logic with Henkin semantics, and this is typically explained by the fact that a second-order logic with Henkin semantics behaves identically with a many-sorted first-order logics, and the LS theorem holds in that latter, along with Compactness and Completeness.
My general question is whether it is possible to precisely identify the boundary between logics that allow for proof of downward LST and those that do not. (I suspect that it might come down to the ability to prove the Tarski-Vaught criterion).
My specific question is whether the following theory would have a countable model (as given by a construction similar to the one for downward LST). The theory consists of the axioms of second-order ZFC except with Separation restricted to "definite" subsets. A subset is "definite" if it is defined by a "definite" property as axiomatized in Zermelo 1929a (p. 362 of Collected Works Vol I, Springer 2010) The set of definite propositions is the smallest set containing all "fundamental relations" (A "fundamental relation" is one of the form $a in b$ or $a = b$) and closed under the operations of negation, conjunction, disjunction, first-order quantification, and second-order quantification.
The axioms of this theory are Extensionality, Pairing, Second-order Separation (modulo the definiteness restriction), Powerset, Union, Foundation, and Second-order Replacement, and Choice.
logic set-theory
asked Jan 7 at 1:13
MallikMallik
1547
$endgroup$
|
show 6 more comments
$begingroup$
The Downward Lowenheim-Skolem Theorem asserts that if a countable first-order theory has an infinite model, then it has a countable model.
Although associated with first-order logic, the result also applies to second-order logic with Henkin semantics, and this is typically explained by the fact that a second-order logic with Henkin semantics behaves identically with a many-sorted first-order logics, and the LS theorem holds in that latter, along with Compactness and Completeness.
My general question is whether it is possible to precisely identify the boundary between logics that allow for proof of downward LST and those that do not. (I suspect that it might come down to the ability to prove the Tarski-Vaught criterion).
My specific question is whether the following theory would have a countable model (as given by a construction similar to the one for downward LST). The theory consists of the axioms of second-order ZFC except with Separation restricted to "definite" subsets. A subset is "definite" if it is defined by a "definite" property as axiomatized in Zermelo 1929a (p. 362 of Collected Works Vol I, Springer 2010) The set of definite propositions is the smallest set containing all "fundamental relations" (A "fundamental relation" is one of the form $a in b$ or $a = b$) and closed under the operations of negation, conjunction, disjunction, first-order quantification, and second-order quantification.
The axioms of this theory are Extensionality, Pairing, Second-order Separation (modulo the definiteness restriction), Powerset, Union, Foundation, and Second-order Replacement, and Choice.
logic set-theory
asked Jan 7 at 1:13
MallikMallik
1547
$endgroup$
$begingroup$
I don't quite understand your specific question. Downward LST for first-order logic is provable in usual ZFC. Or are you asking whether every model of your version of ZFC has a countable elementary (in what sense?) submodel? (Also you probably also want to make the same modification to Replacement as you do to Separation ...) Also, I don't see why the logic for your version of ZFC would be intermediate. If you're allowing your second-order quantifiers to quantify over all subsets of a model, then its the full semantics; the fact that the elements of the first-order part are sets is irrelevant.
$endgroup$
– Noah Schweber
Jan 7 at 16:53
$begingroup$
@NoahSchweber I'm not sure how to state the question in a way that doesn't involve too much context. Before arriving at full second-order ZFC, Zermelo tried to axiomatize a notion of "definite property" in order to restrict Separation to subsets given by "definite" properties. The idea is that properties are definite if they are given by atomic formulas, or built up from other definite properties by the operations mentioned above. So the second-order quantifiers only range over the "definite" subsets.
$endgroup$
– Mallik
Jan 7 at 21:24
1
$begingroup$
The general question is totally reasonable (if a bit imprecise). But the specific question doesn't make any sense. "All axioms of second-order ZFC except..." is not a logic, it is a theory in some logic. If you specify precisely what logic you want to consider this theory in, you can ask whether that logic satisfies downward LST. Or you can ask whether this particular theory has a countable model (assuming it's consistent). But as written, it's not at all clear what you mean.
$endgroup$
– Alex Kruckman
Jan 7 at 22:26
1
$begingroup$
Ah - so all the axioms are the usual first-order axioms of ZFC, except that separation applies to formulas of second-order logic, and replacement applies to all functions with no definability requirement. And you consider this theory with the full semantics. Correct?
$endgroup$
– Alex Kruckman
Jan 10 at 1:48
1
$begingroup$
If I've understood that correctly, then the second-order replacement is already a problem for having a countable model. Any model $M$ will contain a set called $omega^M$ with infinitely many elements. Now for any infinite subset $X$ of these elements, there is a function $F: omega^Mto omega^M$ with range $X$. So the replacement axiom says that $M$ contains a set whose elements are exactly $X$. And of course there are continuum-many infinite subsets of $omega$.
$endgroup$
– Alex Kruckman
Jan 10 at 1:52
|
show 6 more comments
$begingroup$
The Downward Lowenheim-Skolem Theorem asserts that if a countable first-order theory has an infinite model, then it has a countable model.
Although associated with first-order logic, the result also applies to second-order logic with Henkin semantics, and this is typically explained by the fact that a second-order logic with Henkin semantics behaves identically with a many-sorted first-order logics, and the LS theorem holds in that latter, along with Compactness and Completeness.
My general question is whether it is possible to precisely identify the boundary between logics that allow for proof of downward LST and those that do not. (I suspect that it might come down to the ability to prove the Tarski-Vaught criterion).
My specific question is whether the following theory would have a countable model (as given by a construction similar to the one for downward LST). The theory consists of the axioms of second-order ZFC except with Separation restricted to "definite" subsets. A subset is "definite" if it is defined by a "definite" property as axiomatized in Zermelo 1929a (p. 362 of Collected Works Vol I, Springer 2010) The set of definite propositions is the smallest set containing all "fundamental relations" (A "fundamental relation" is one of the form $a in b$ or $a = b$) and closed under the operations of negation, conjunction, disjunction, first-order quantification, and second-order quantification.
The axioms of this theory are Extensionality, Pairing, Second-order Separation (modulo the definiteness restriction), Powerset, Union, Foundation, and Second-order Replacement, and Choice.
logic set-theory
asked Jan 7 at 1:13
MallikMallik
1547
$endgroup$
The Downward Lowenheim-Skolem Theorem asserts that if a countable first-order theory has an infinite model, then it has a countable model.
Although associated with first-order logic, the result also applies to second-order logic with Henkin semantics, and this is typically explained by the fact that a second-order logic with Henkin semantics behaves identically with a many-sorted first-order logics, and the LS theorem holds in that latter, along with Compactness and Completeness.
My general question is whether it is possible to precisely identify the boundary between logics that allow for proof of downward LST and those that do not. (I suspect that it might come down to the ability to prove the Tarski-Vaught criterion).
My specific question is whether the following theory would have a countable model (as given by a construction similar to the one for downward LST). The theory consists of the axioms of second-order ZFC except with Separation restricted to "definite" subsets. A subset is "definite" if it is defined by a "definite" property as axiomatized in Zermelo 1929a (p. 362 of Collected Works Vol I, Springer 2010) The set of definite propositions is the smallest set containing all "fundamental relations" (A "fundamental relation" is one of the form $a in b$ or $a = b$) and closed under the operations of negation, conjunction, disjunction, first-order quantification, and second-order quantification.
The axioms of this theory are Extensionality, Pairing, Second-order Separation (modulo the definiteness restriction), Powerset, Union, Foundation, and Second-order Replacement, and Choice.
logic set-theory
logic set-theory
asked Jan 7 at 1:13
MallikMallik
1547
asked Jan 7 at 1:13
MallikMallik
1547
edited Jan 10 at 1:32
Mallik
asked Jan 7 at 1:13
MallikMallik
1547
asked Jan 7 at 1:13
MallikMallik
1547
asked Jan 7 at 1:13
MallikMallik
1547
1547
$begingroup$
I don't quite understand your specific question. Downward LST for first-order logic is provable in usual ZFC. Or are you asking whether every model of your version of ZFC has a countable elementary (in what sense?) submodel? (Also you probably also want to make the same modification to Replacement as you do to Separation ...) Also, I don't see why the logic for your version of ZFC would be intermediate. If you're allowing your second-order quantifiers to quantify over all subsets of a model, then its the full semantics; the fact that the elements of the first-order part are sets is irrelevant.
$endgroup$
– Noah Schweber
Jan 7 at 16:53
$begingroup$
@NoahSchweber I'm not sure how to state the question in a way that doesn't involve too much context. Before arriving at full second-order ZFC, Zermelo tried to axiomatize a notion of "definite property" in order to restrict Separation to subsets given by "definite" properties. The idea is that properties are definite if they are given by atomic formulas, or built up from other definite properties by the operations mentioned above. So the second-order quantifiers only range over the "definite" subsets.
$endgroup$
– Mallik
Jan 7 at 21:24
1
$begingroup$
The general question is totally reasonable (if a bit imprecise). But the specific question doesn't make any sense. "All axioms of second-order ZFC except..." is not a logic, it is a theory in some logic. If you specify precisely what logic you want to consider this theory in, you can ask whether that logic satisfies downward LST. Or you can ask whether this particular theory has a countable model (assuming it's consistent). But as written, it's not at all clear what you mean.
$endgroup$
– Alex Kruckman
Jan 7 at 22:26
1
$begingroup$
Ah - so all the axioms are the usual first-order axioms of ZFC, except that separation applies to formulas of second-order logic, and replacement applies to all functions with no definability requirement. And you consider this theory with the full semantics. Correct?
$endgroup$
– Alex Kruckman
Jan 10 at 1:48
1
$begingroup$
If I've understood that correctly, then the second-order replacement is already a problem for having a countable model. Any model $M$ will contain a set called $omega^M$ with infinitely many elements. Now for any infinite subset $X$ of these elements, there is a function $F: omega^Mto omega^M$ with range $X$. So the replacement axiom says that $M$ contains a set whose elements are exactly $X$. And of course there are continuum-many infinite subsets of $omega$.
$endgroup$
– Alex Kruckman
Jan 10 at 1:52
|
show 6 more comments
$begingroup$
I don't quite understand your specific question. Downward LST for first-order logic is provable in usual ZFC. Or are you asking whether every model of your version of ZFC has a countable elementary (in what sense?) submodel? (Also you probably also want to make the same modification to Replacement as you do to Separation ...) Also, I don't see why the logic for your version of ZFC would be intermediate. If you're allowing your second-order quantifiers to quantify over all subsets of a model, then its the full semantics; the fact that the elements of the first-order part are sets is irrelevant.
$endgroup$
– Noah Schweber
Jan 7 at 16:53
$begingroup$
@NoahSchweber I'm not sure how to state the question in a way that doesn't involve too much context. Before arriving at full second-order ZFC, Zermelo tried to axiomatize a notion of "definite property" in order to restrict Separation to subsets given by "definite" properties. The idea is that properties are definite if they are given by atomic formulas, or built up from other definite properties by the operations mentioned above. So the second-order quantifiers only range over the "definite" subsets.
$endgroup$
– Mallik
Jan 7 at 21:24
1
$begingroup$
The general question is totally reasonable (if a bit imprecise). But the specific question doesn't make any sense. "All axioms of second-order ZFC except..." is not a logic, it is a theory in some logic. If you specify precisely what logic you want to consider this theory in, you can ask whether that logic satisfies downward LST. Or you can ask whether this particular theory has a countable model (assuming it's consistent). But as written, it's not at all clear what you mean.
$endgroup$
– Alex Kruckman
Jan 7 at 22:26
1
$begingroup$
Ah - so all the axioms are the usual first-order axioms of ZFC, except that separation applies to formulas of second-order logic, and replacement applies to all functions with no definability requirement. And you consider this theory with the full semantics. Correct?
$endgroup$
– Alex Kruckman
Jan 10 at 1:48
1
$begingroup$
If I've understood that correctly, then the second-order replacement is already a problem for having a countable model. Any model $M$ will contain a set called $omega^M$ with infinitely many elements. Now for any infinite subset $X$ of these elements, there is a function $F: omega^Mto omega^M$ with range $X$. So the replacement axiom says that $M$ contains a set whose elements are exactly $X$. And of course there are continuum-many infinite subsets of $omega$.
$endgroup$
– Alex Kruckman
Jan 10 at 1:52
$begingroup$
I don't quite understand your specific question. Downward LST for first-order logic is provable in usual ZFC. Or are you asking whether every model of your version of ZFC has a countable elementary (in what sense?) submodel? (Also you probably also want to make the same modification to Replacement as you do to Separation ...) Also, I don't see why the logic for your version of ZFC would be intermediate. If you're allowing your second-order quantifiers to quantify over all subsets of a model, then its the full semantics; the fact that the elements of the first-order part are sets is irrelevant.
$endgroup$
– Noah Schweber
Jan 7 at 16:53
$begingroup$
I don't quite understand your specific question. Downward LST for first-order logic is provable in usual ZFC. Or are you asking whether every model of your version of ZFC has a countable elementary (in what sense?) submodel? (Also you probably also want to make the same modification to Replacement as you do to Separation ...) Also, I don't see why the logic for your version of ZFC would be intermediate. If you're allowing your second-order quantifiers to quantify over all subsets of a model, then its the full semantics; the fact that the elements of the first-order part are sets is irrelevant.
$endgroup$
– Noah Schweber
Jan 7 at 16:53
$begingroup$
@NoahSchweber I'm not sure how to state the question in a way that doesn't involve too much context. Before arriving at full second-order ZFC, Zermelo tried to axiomatize a notion of "definite property" in order to restrict Separation to subsets given by "definite" properties. The idea is that properties are definite if they are given by atomic formulas, or built up from other definite properties by the operations mentioned above. So the second-order quantifiers only range over the "definite" subsets.
$endgroup$
– Mallik
Jan 7 at 21:24
$begingroup$
@NoahSchweber I'm not sure how to state the question in a way that doesn't involve too much context. Before arriving at full second-order ZFC, Zermelo tried to axiomatize a notion of "definite property" in order to restrict Separation to subsets given by "definite" properties. The idea is that properties are definite if they are given by atomic formulas, or built up from other definite properties by the operations mentioned above. So the second-order quantifiers only range over the "definite" subsets.
$endgroup$
– Mallik
Jan 7 at 21:24
1
1
$begingroup$
The general question is totally reasonable (if a bit imprecise). But the specific question doesn't make any sense. "All axioms of second-order ZFC except..." is not a logic, it is a theory in some logic. If you specify precisely what logic you want to consider this theory in, you can ask whether that logic satisfies downward LST. Or you can ask whether this particular theory has a countable model (assuming it's consistent). But as written, it's not at all clear what you mean.
$endgroup$
– Alex Kruckman
Jan 7 at 22:26
$begingroup$
The general question is totally reasonable (if a bit imprecise). But the specific question doesn't make any sense. "All axioms of second-order ZFC except..." is not a logic, it is a theory in some logic. If you specify precisely what logic you want to consider this theory in, you can ask whether that logic satisfies downward LST. Or you can ask whether this particular theory has a countable model (assuming it's consistent). But as written, it's not at all clear what you mean.
$endgroup$
– Alex Kruckman
Jan 7 at 22:26
1
1
$begingroup$
Ah - so all the axioms are the usual first-order axioms of ZFC, except that separation applies to formulas of second-order logic, and replacement applies to all functions with no definability requirement. And you consider this theory with the full semantics. Correct?
$endgroup$
– Alex Kruckman
Jan 10 at 1:48
$begingroup$
Ah - so all the axioms are the usual first-order axioms of ZFC, except that separation applies to formulas of second-order logic, and replacement applies to all functions with no definability requirement. And you consider this theory with the full semantics. Correct?
$endgroup$
– Alex Kruckman
Jan 10 at 1:48
1
1
$begingroup$
If I've understood that correctly, then the second-order replacement is already a problem for having a countable model. Any model $M$ will contain a set called $omega^M$ with infinitely many elements. Now for any infinite subset $X$ of these elements, there is a function $F: omega^Mto omega^M$ with range $X$. So the replacement axiom says that $M$ contains a set whose elements are exactly $X$. And of course there are continuum-many infinite subsets of $omega$.
$endgroup$
– Alex Kruckman
Jan 10 at 1:52
$begingroup$
If I've understood that correctly, then the second-order replacement is already a problem for having a countable model. Any model $M$ will contain a set called $omega^M$ with infinitely many elements. Now for any infinite subset $X$ of these elements, there is a function $F: omega^Mto omega^M$ with range $X$. So the replacement axiom says that $M$ contains a set whose elements are exactly $X$. And of course there are continuum-many infinite subsets of $omega$.
$endgroup$
– Alex Kruckman
Jan 10 at 1:52
|
show 6 more comments
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$begingroup$
I don't quite understand your specific question. Downward LST for first-order logic is provable in usual ZFC. Or are you asking whether every model of your version of ZFC has a countable elementary (in what sense?) submodel? (Also you probably also want to make the same modification to Replacement as you do to Separation ...) Also, I don't see why the logic for your version of ZFC would be intermediate. If you're allowing your second-order quantifiers to quantify over all subsets of a model, then its the full semantics; the fact that the elements of the first-order part are sets is irrelevant.
$endgroup$
– Noah Schweber
Jan 7 at 16:53
$begingroup$
@NoahSchweber I'm not sure how to state the question in a way that doesn't involve too much context. Before arriving at full second-order ZFC, Zermelo tried to axiomatize a notion of "definite property" in order to restrict Separation to subsets given by "definite" properties. The idea is that properties are definite if they are given by atomic formulas, or built up from other definite properties by the operations mentioned above. So the second-order quantifiers only range over the "definite" subsets.
$endgroup$
– Mallik
Jan 7 at 21:24
1
$begingroup$
The general question is totally reasonable (if a bit imprecise). But the specific question doesn't make any sense. "All axioms of second-order ZFC except..." is not a logic, it is a theory in some logic. If you specify precisely what logic you want to consider this theory in, you can ask whether that logic satisfies downward LST. Or you can ask whether this particular theory has a countable model (assuming it's consistent). But as written, it's not at all clear what you mean.
$endgroup$
– Alex Kruckman
Jan 7 at 22:26
1
$begingroup$
Ah - so all the axioms are the usual first-order axioms of ZFC, except that separation applies to formulas of second-order logic, and replacement applies to all functions with no definability requirement. And you consider this theory with the full semantics. Correct?
$endgroup$
– Alex Kruckman
Jan 10 at 1:48
1
$begingroup$
If I've understood that correctly, then the second-order replacement is already a problem for having a countable model. Any model $M$ will contain a set called $omega^M$ with infinitely many elements. Now for any infinite subset $X$ of these elements, there is a function $F: omega^Mto omega^M$ with range $X$. So the replacement axiom says that $M$ contains a set whose elements are exactly $X$. And of course there are continuum-many infinite subsets of $omega$.
$endgroup$
– Alex Kruckman
Jan 10 at 1:52