How it makes sense to talk about consistency of theories?

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So I was reading this post Probability axioms (Kolmogorov) and so in the answer there was said that it doesn't make sense to talk about consistency of probability axioms because it is not a formal system but just a part of measure theory, but now, can't we say in the same way that it doesn't make sense to talk about consistency of PA, Real numbers, Geometry if they all can be reduced to ZFC ? What makes probability case different from other theories ?




logic formal-systems




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  • The point, I think, is that probability is just a specific case of measure theory. It's like talking about the consistency of "the free group with one generator", rather than the consistency of group theory.
    – Asaf Karagila♦
    Aug 6 at 7:02










  • You are right about real numbers, and some parts of geometry. PA, on the other hand, is a formal system and not some constructions within ZFC. It turns out that we can prove its consistency by providing a model for it, but this model is not the whole of PA
    – Max
    Aug 6 at 7:04










  • @AsafKaragila Can't we say that Real Analysis is a specific case of ZFC ?
    – Ð®Ñ€Ñ–й Ярош
    Aug 6 at 7:12






  • 3




    I actually disagree with the answer you link to. Of course it makes sense to ask about the consistency of the axioms of probability theory. The answer is not particularly interesting, though, since their consistency follows from the many explicit examples we have of probability spaces. Of course the axioms of probability theory form a "formal system", unless one has a pretty peculiar notion of "formal system" in mind.
    – Andrés E. Caicedo
    Aug 6 at 11:18







  • 1




    In fact, I must also disagree somewhat with @Asaf's comment: there are many examples of theories $T$, models $M$ of $T$ and metatheories $S$ such that $S$ is reasonable enough to discuss consistency of theories and strong enough that $S$ proves the consistency of $T$, and yet $S$ does not prove the consistency of the theory of $M$.
    – Andrés E. Caicedo
    Aug 6 at 11:31














up vote
4
down vote

favorite












So I was reading this post Probability axioms (Kolmogorov) and so in the answer there was said that it doesn't make sense to talk about consistency of probability axioms because it is not a formal system but just a part of measure theory, but now, can't we say in the same way that it doesn't make sense to talk about consistency of PA, Real numbers, Geometry if they all can be reduced to ZFC ? What makes probability case different from other theories ?




logic formal-systems




share|cite|improve this question



















  • The point, I think, is that probability is just a specific case of measure theory. It's like talking about the consistency of "the free group with one generator", rather than the consistency of group theory.
    – Asaf Karagila♦
    Aug 6 at 7:02










  • You are right about real numbers, and some parts of geometry. PA, on the other hand, is a formal system and not some constructions within ZFC. It turns out that we can prove its consistency by providing a model for it, but this model is not the whole of PA
    – Max
    Aug 6 at 7:04










  • @AsafKaragila Can't we say that Real Analysis is a specific case of ZFC ?
    – Ð®Ñ€Ñ–й Ярош
    Aug 6 at 7:12






  • 3




    I actually disagree with the answer you link to. Of course it makes sense to ask about the consistency of the axioms of probability theory. The answer is not particularly interesting, though, since their consistency follows from the many explicit examples we have of probability spaces. Of course the axioms of probability theory form a "formal system", unless one has a pretty peculiar notion of "formal system" in mind.
    – Andrés E. Caicedo
    Aug 6 at 11:18







  • 1




    In fact, I must also disagree somewhat with @Asaf's comment: there are many examples of theories $T$, models $M$ of $T$ and metatheories $S$ such that $S$ is reasonable enough to discuss consistency of theories and strong enough that $S$ proves the consistency of $T$, and yet $S$ does not prove the consistency of the theory of $M$.
    – Andrés E. Caicedo
    Aug 6 at 11:31












up vote
4
down vote

favorite









up vote
4
down vote

favorite











So I was reading this post Probability axioms (Kolmogorov) and so in the answer there was said that it doesn't make sense to talk about consistency of probability axioms because it is not a formal system but just a part of measure theory, but now, can't we say in the same way that it doesn't make sense to talk about consistency of PA, Real numbers, Geometry if they all can be reduced to ZFC ? What makes probability case different from other theories ?




logic formal-systems




share|cite|improve this question











So I was reading this post Probability axioms (Kolmogorov) and so in the answer there was said that it doesn't make sense to talk about consistency of probability axioms because it is not a formal system but just a part of measure theory, but now, can't we say in the same way that it doesn't make sense to talk about consistency of PA, Real numbers, Geometry if they all can be reduced to ZFC ? What makes probability case different from other theories ?





logic formal-systems





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Aug 6 at 6:59









Юрій Ярош

995513




995513











  • The point, I think, is that probability is just a specific case of measure theory. It's like talking about the consistency of "the free group with one generator", rather than the consistency of group theory.
    – Asaf Karagila♦
    Aug 6 at 7:02










  • You are right about real numbers, and some parts of geometry. PA, on the other hand, is a formal system and not some constructions within ZFC. It turns out that we can prove its consistency by providing a model for it, but this model is not the whole of PA
    – Max
    Aug 6 at 7:04










  • @AsafKaragila Can't we say that Real Analysis is a specific case of ZFC ?
    – Ð®Ñ€Ñ–й Ярош
    Aug 6 at 7:12






  • 3




    I actually disagree with the answer you link to. Of course it makes sense to ask about the consistency of the axioms of probability theory. The answer is not particularly interesting, though, since their consistency follows from the many explicit examples we have of probability spaces. Of course the axioms of probability theory form a "formal system", unless one has a pretty peculiar notion of "formal system" in mind.
    – Andrés E. Caicedo
    Aug 6 at 11:18







  • 1




    In fact, I must also disagree somewhat with @Asaf's comment: there are many examples of theories $T$, models $M$ of $T$ and metatheories $S$ such that $S$ is reasonable enough to discuss consistency of theories and strong enough that $S$ proves the consistency of $T$, and yet $S$ does not prove the consistency of the theory of $M$.
    – Andrés E. Caicedo
    Aug 6 at 11:31
















  • The point, I think, is that probability is just a specific case of measure theory. It's like talking about the consistency of "the free group with one generator", rather than the consistency of group theory.
    – Asaf Karagila♦
    Aug 6 at 7:02










  • You are right about real numbers, and some parts of geometry. PA, on the other hand, is a formal system and not some constructions within ZFC. It turns out that we can prove its consistency by providing a model for it, but this model is not the whole of PA
    – Max
    Aug 6 at 7:04










  • @AsafKaragila Can't we say that Real Analysis is a specific case of ZFC ?
    – Ð®Ñ€Ñ–й Ярош
    Aug 6 at 7:12






  • 3




    I actually disagree with the answer you link to. Of course it makes sense to ask about the consistency of the axioms of probability theory. The answer is not particularly interesting, though, since their consistency follows from the many explicit examples we have of probability spaces. Of course the axioms of probability theory form a "formal system", unless one has a pretty peculiar notion of "formal system" in mind.
    – Andrés E. Caicedo
    Aug 6 at 11:18







  • 1




    In fact, I must also disagree somewhat with @Asaf's comment: there are many examples of theories $T$, models $M$ of $T$ and metatheories $S$ such that $S$ is reasonable enough to discuss consistency of theories and strong enough that $S$ proves the consistency of $T$, and yet $S$ does not prove the consistency of the theory of $M$.
    – Andrés E. Caicedo
    Aug 6 at 11:31















The point, I think, is that probability is just a specific case of measure theory. It's like talking about the consistency of "the free group with one generator", rather than the consistency of group theory.
– Asaf Karagila♦
Aug 6 at 7:02




The point, I think, is that probability is just a specific case of measure theory. It's like talking about the consistency of "the free group with one generator", rather than the consistency of group theory.
– Asaf Karagila♦
Aug 6 at 7:02












You are right about real numbers, and some parts of geometry. PA, on the other hand, is a formal system and not some constructions within ZFC. It turns out that we can prove its consistency by providing a model for it, but this model is not the whole of PA
– Max
Aug 6 at 7:04




You are right about real numbers, and some parts of geometry. PA, on the other hand, is a formal system and not some constructions within ZFC. It turns out that we can prove its consistency by providing a model for it, but this model is not the whole of PA
– Max
Aug 6 at 7:04












@AsafKaragila Can't we say that Real Analysis is a specific case of ZFC ?
– Ð®Ñ€Ñ–й Ярош
Aug 6 at 7:12




@AsafKaragila Can't we say that Real Analysis is a specific case of ZFC ?
– Ð®Ñ€Ñ–й Ярош
Aug 6 at 7:12




3




3




I actually disagree with the answer you link to. Of course it makes sense to ask about the consistency of the axioms of probability theory. The answer is not particularly interesting, though, since their consistency follows from the many explicit examples we have of probability spaces. Of course the axioms of probability theory form a "formal system", unless one has a pretty peculiar notion of "formal system" in mind.
– Andrés E. Caicedo
Aug 6 at 11:18





I actually disagree with the answer you link to. Of course it makes sense to ask about the consistency of the axioms of probability theory. The answer is not particularly interesting, though, since their consistency follows from the many explicit examples we have of probability spaces. Of course the axioms of probability theory form a "formal system", unless one has a pretty peculiar notion of "formal system" in mind.
– Andrés E. Caicedo
Aug 6 at 11:18





1




1




In fact, I must also disagree somewhat with @Asaf's comment: there are many examples of theories $T$, models $M$ of $T$ and metatheories $S$ such that $S$ is reasonable enough to discuss consistency of theories and strong enough that $S$ proves the consistency of $T$, and yet $S$ does not prove the consistency of the theory of $M$.
– Andrés E. Caicedo
Aug 6 at 11:31




In fact, I must also disagree somewhat with @Asaf's comment: there are many examples of theories $T$, models $M$ of $T$ and metatheories $S$ such that $S$ is reasonable enough to discuss consistency of theories and strong enough that $S$ proves the consistency of $T$, and yet $S$ does not prove the consistency of the theory of $M$.
– Andrés E. Caicedo
Aug 6 at 11:31










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I'm with Andres on this one: I disagree with the linked answer and Asaf's comment.



Let me respond to Asaf's comment first. While "the free group on two generators is consistent" doesn't make any sense, there is an obvious set of axioms for "the free group on two generators" - namely, the axioms of group theory together with the negation of every nontrivial identity involving two new constant symbols - and we can ask whether that set of axioms is consistent; the most natural proof that it is, of course, consists of first proving the soundness theorem and then constructing the free group on two generators and showing it satisfies the above-mentioned theory. This argument requires some "metamathematical overhead" - we need to be able to talk about groups, axioms, and satisfaction - but for example is easily performable in ZFC.



Now on to the probability space example. Unlike the above situation, the axioms for probability spaces are not first-order (e.g. even saying that something is a $sigma$-algebra takes us well outside of FOL). That's not inherently a problem, but it does mean that we have to be working in a metatheory which can make sense of the relevant logic (second-order logic will do the job just fine). ZFC, for example, will more than suffice: ZFC proves that the axioms for probability spaces are satisfiable.



Note that I said "satisfiable" rather than "consistent." Generalized logics don't necessarily come with sound and complete proof systems; for example, in a very strong sense there is no good proof system for second-order logic. When we work with first-order logic, consistency is a perfectly meaningful notion, but when handling more complicated logics the semantic side of things is much better than the syntactic side of things. However, one immediate corollary of the above-mentioned fact is that ZFC proves "For every proof system which is sound with respect to second-order logic, the axioms of probability spaces are consistent with respect to that proof system." Note that this really is trivial (since soundness exactly tells us that no new entailments can be produced), so despite the above concerns there is a very good sense in which ZFC proves "the probability space axioms are consistent."






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  • So is it impossible to formulate the definition of σ-algebra with ZFC and FOL ? Or did I misunderstand you ?
    – Ð®Ñ€Ñ–й Ярош
    Aug 8 at 8:02










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I'm with Andres on this one: I disagree with the linked answer and Asaf's comment.



Let me respond to Asaf's comment first. While "the free group on two generators is consistent" doesn't make any sense, there is an obvious set of axioms for "the free group on two generators" - namely, the axioms of group theory together with the negation of every nontrivial identity involving two new constant symbols - and we can ask whether that set of axioms is consistent; the most natural proof that it is, of course, consists of first proving the soundness theorem and then constructing the free group on two generators and showing it satisfies the above-mentioned theory. This argument requires some "metamathematical overhead" - we need to be able to talk about groups, axioms, and satisfaction - but for example is easily performable in ZFC.



Now on to the probability space example. Unlike the above situation, the axioms for probability spaces are not first-order (e.g. even saying that something is a $sigma$-algebra takes us well outside of FOL). That's not inherently a problem, but it does mean that we have to be working in a metatheory which can make sense of the relevant logic (second-order logic will do the job just fine). ZFC, for example, will more than suffice: ZFC proves that the axioms for probability spaces are satisfiable.



Note that I said "satisfiable" rather than "consistent." Generalized logics don't necessarily come with sound and complete proof systems; for example, in a very strong sense there is no good proof system for second-order logic. When we work with first-order logic, consistency is a perfectly meaningful notion, but when handling more complicated logics the semantic side of things is much better than the syntactic side of things. However, one immediate corollary of the above-mentioned fact is that ZFC proves "For every proof system which is sound with respect to second-order logic, the axioms of probability spaces are consistent with respect to that proof system." Note that this really is trivial (since soundness exactly tells us that no new entailments can be produced), so despite the above concerns there is a very good sense in which ZFC proves "the probability space axioms are consistent."






share|cite|improve this answer





















  • So is it impossible to formulate the definition of σ-algebra with ZFC and FOL ? Or did I misunderstand you ?
    – Ð®Ñ€Ñ–й Ярош
    Aug 8 at 8:02














up vote
4
down vote



accepted










I'm with Andres on this one: I disagree with the linked answer and Asaf's comment.



Let me respond to Asaf's comment first. While "the free group on two generators is consistent" doesn't make any sense, there is an obvious set of axioms for "the free group on two generators" - namely, the axioms of group theory together with the negation of every nontrivial identity involving two new constant symbols - and we can ask whether that set of axioms is consistent; the most natural proof that it is, of course, consists of first proving the soundness theorem and then constructing the free group on two generators and showing it satisfies the above-mentioned theory. This argument requires some "metamathematical overhead" - we need to be able to talk about groups, axioms, and satisfaction - but for example is easily performable in ZFC.



Now on to the probability space example. Unlike the above situation, the axioms for probability spaces are not first-order (e.g. even saying that something is a $sigma$-algebra takes us well outside of FOL). That's not inherently a problem, but it does mean that we have to be working in a metatheory which can make sense of the relevant logic (second-order logic will do the job just fine). ZFC, for example, will more than suffice: ZFC proves that the axioms for probability spaces are satisfiable.



Note that I said "satisfiable" rather than "consistent." Generalized logics don't necessarily come with sound and complete proof systems; for example, in a very strong sense there is no good proof system for second-order logic. When we work with first-order logic, consistency is a perfectly meaningful notion, but when handling more complicated logics the semantic side of things is much better than the syntactic side of things. However, one immediate corollary of the above-mentioned fact is that ZFC proves "For every proof system which is sound with respect to second-order logic, the axioms of probability spaces are consistent with respect to that proof system." Note that this really is trivial (since soundness exactly tells us that no new entailments can be produced), so despite the above concerns there is a very good sense in which ZFC proves "the probability space axioms are consistent."






share|cite|improve this answer





















  • So is it impossible to formulate the definition of σ-algebra with ZFC and FOL ? Or did I misunderstand you ?
    – Ð®Ñ€Ñ–й Ярош
    Aug 8 at 8:02












up vote
4
down vote



accepted







up vote
4
down vote



accepted






I'm with Andres on this one: I disagree with the linked answer and Asaf's comment.



Let me respond to Asaf's comment first. While "the free group on two generators is consistent" doesn't make any sense, there is an obvious set of axioms for "the free group on two generators" - namely, the axioms of group theory together with the negation of every nontrivial identity involving two new constant symbols - and we can ask whether that set of axioms is consistent; the most natural proof that it is, of course, consists of first proving the soundness theorem and then constructing the free group on two generators and showing it satisfies the above-mentioned theory. This argument requires some "metamathematical overhead" - we need to be able to talk about groups, axioms, and satisfaction - but for example is easily performable in ZFC.



Now on to the probability space example. Unlike the above situation, the axioms for probability spaces are not first-order (e.g. even saying that something is a $sigma$-algebra takes us well outside of FOL). That's not inherently a problem, but it does mean that we have to be working in a metatheory which can make sense of the relevant logic (second-order logic will do the job just fine). ZFC, for example, will more than suffice: ZFC proves that the axioms for probability spaces are satisfiable.



Note that I said "satisfiable" rather than "consistent." Generalized logics don't necessarily come with sound and complete proof systems; for example, in a very strong sense there is no good proof system for second-order logic. When we work with first-order logic, consistency is a perfectly meaningful notion, but when handling more complicated logics the semantic side of things is much better than the syntactic side of things. However, one immediate corollary of the above-mentioned fact is that ZFC proves "For every proof system which is sound with respect to second-order logic, the axioms of probability spaces are consistent with respect to that proof system." Note that this really is trivial (since soundness exactly tells us that no new entailments can be produced), so despite the above concerns there is a very good sense in which ZFC proves "the probability space axioms are consistent."






share|cite|improve this answer













I'm with Andres on this one: I disagree with the linked answer and Asaf's comment.



Let me respond to Asaf's comment first. While "the free group on two generators is consistent" doesn't make any sense, there is an obvious set of axioms for "the free group on two generators" - namely, the axioms of group theory together with the negation of every nontrivial identity involving two new constant symbols - and we can ask whether that set of axioms is consistent; the most natural proof that it is, of course, consists of first proving the soundness theorem and then constructing the free group on two generators and showing it satisfies the above-mentioned theory. This argument requires some "metamathematical overhead" - we need to be able to talk about groups, axioms, and satisfaction - but for example is easily performable in ZFC.



Now on to the probability space example. Unlike the above situation, the axioms for probability spaces are not first-order (e.g. even saying that something is a $sigma$-algebra takes us well outside of FOL). That's not inherently a problem, but it does mean that we have to be working in a metatheory which can make sense of the relevant logic (second-order logic will do the job just fine). ZFC, for example, will more than suffice: ZFC proves that the axioms for probability spaces are satisfiable.



Note that I said "satisfiable" rather than "consistent." Generalized logics don't necessarily come with sound and complete proof systems; for example, in a very strong sense there is no good proof system for second-order logic. When we work with first-order logic, consistency is a perfectly meaningful notion, but when handling more complicated logics the semantic side of things is much better than the syntactic side of things. However, one immediate corollary of the above-mentioned fact is that ZFC proves "For every proof system which is sound with respect to second-order logic, the axioms of probability spaces are consistent with respect to that proof system." Note that this really is trivial (since soundness exactly tells us that no new entailments can be produced), so despite the above concerns there is a very good sense in which ZFC proves "the probability space axioms are consistent."







share|cite|improve this answer













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answered Aug 6 at 18:02









Noah Schweber

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  • So is it impossible to formulate the definition of σ-algebra with ZFC and FOL ? Or did I misunderstand you ?
    – Ð®Ñ€Ñ–й Ярош
    Aug 8 at 8:02
















  • So is it impossible to formulate the definition of σ-algebra with ZFC and FOL ? Or did I misunderstand you ?
    – Ð®Ñ€Ñ–й Ярош
    Aug 8 at 8:02















So is it impossible to formulate the definition of σ-algebra with ZFC and FOL ? Or did I misunderstand you ?
– Ð®Ñ€Ñ–й Ярош
Aug 8 at 8:02




So is it impossible to formulate the definition of σ-algebra with ZFC and FOL ? Or did I misunderstand you ?
– Ð®Ñ€Ñ–й Ярош
Aug 8 at 8:02












 

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