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Why can't we prove consistency of ZFC like we can for PA?
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this might be a silly question, but I was wondering: PA cannot prove its consistency by the incompleteness theorems, but we can "step outside" and exhibit a model of it, namely $mathbbN$, so we know that PA is consistent.
Why can't we do this with ZFC? I have seen things like "if $kappa$ is [some large cardinal] then $V_kappa$ models ZFC", but these stem from an "if".
Is this a case of us not having been able to do this yet, or is there a good reason why it is simply not possible?
logic set-theory model-theory incompleteness
edited 54 mins ago
Asaf Karagila♦
298k32417745
asked 1 hour ago
K. 622
31616
add a comment |Â
up vote
2
down vote
favorite
this might be a silly question, but I was wondering: PA cannot prove its consistency by the incompleteness theorems, but we can "step outside" and exhibit a model of it, namely $mathbbN$, so we know that PA is consistent.
Why can't we do this with ZFC? I have seen things like "if $kappa$ is [some large cardinal] then $V_kappa$ models ZFC", but these stem from an "if".
Is this a case of us not having been able to do this yet, or is there a good reason why it is simply not possible?
logic set-theory model-theory incompleteness
edited 54 mins ago
Asaf Karagila♦
298k32417745
asked 1 hour ago
K. 622
31616
5
Stepping outside PA is an "if". We prove PA is consistent if ZFC is.
– Malice Vidrine
1 hour ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
this might be a silly question, but I was wondering: PA cannot prove its consistency by the incompleteness theorems, but we can "step outside" and exhibit a model of it, namely $mathbbN$, so we know that PA is consistent.
Why can't we do this with ZFC? I have seen things like "if $kappa$ is [some large cardinal] then $V_kappa$ models ZFC", but these stem from an "if".
Is this a case of us not having been able to do this yet, or is there a good reason why it is simply not possible?
logic set-theory model-theory incompleteness
edited 54 mins ago
Asaf Karagila♦
298k32417745
asked 1 hour ago
K. 622
31616
this might be a silly question, but I was wondering: PA cannot prove its consistency by the incompleteness theorems, but we can "step outside" and exhibit a model of it, namely $mathbbN$, so we know that PA is consistent.
Why can't we do this with ZFC? I have seen things like "if $kappa$ is [some large cardinal] then $V_kappa$ models ZFC", but these stem from an "if".
Is this a case of us not having been able to do this yet, or is there a good reason why it is simply not possible?
logic set-theory model-theory incompleteness
logic set-theory model-theory incompleteness
edited 54 mins ago
Asaf Karagila♦
298k32417745
asked 1 hour ago
K. 622
31616
edited 54 mins ago
Asaf Karagila♦
298k32417745
asked 1 hour ago
K. 622
31616
edited 54 mins ago
Asaf Karagila♦
298k32417745
edited 54 mins ago
Asaf Karagila♦
298k32417745
edited 54 mins ago
Asaf Karagila♦
298k32417745
298k32417745
asked 1 hour ago
K. 622
31616
asked 1 hour ago
K. 622
31616
asked 1 hour ago
K. 622
31616
31616
5
Stepping outside PA is an "if". We prove PA is consistent if ZFC is.
– Malice Vidrine
1 hour ago
add a comment |Â
5
Stepping outside PA is an "if". We prove PA is consistent if ZFC is.
– Malice Vidrine
1 hour ago
5
5
Stepping outside PA is an "if". We prove PA is consistent if ZFC is.
– Malice Vidrine
1 hour ago
Stepping outside PA is an "if". We prove PA is consistent if ZFC is.
– Malice Vidrine
1 hour ago
add a comment |Â
3 Answers
3
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up vote
2
down vote
Using ZFC plus the axiom "Uncountable strong inaccessible cardinals exist" to give a model of ZFC is exactly the same kind of "stepping outside" as when you use ZF to make a model of PA.
answered 1 hour ago
Arthur
106k7101183
Ooh I see. I was essentially confused from reading "we know PA is consistent" elsewhere, when I should have interpreted that as "within the standard system we use today, PA is consistent". Thank you for clearing that up.
– K. 622
59 mins ago
add a comment |Â
up vote
2
down vote
The problem is that, unlike the case for PA, essentially all accepted mathematical reasoning can be formalized in ZFC. Any proof of the consistency of ZFC must come from a system that is stronger (at least in some ways), so we must go outside ZFC-formalizable mathematics, which is most of mathmatics. (Just like we go outside of PA-formalizable mathematics to prove the consistency of PA.)
For instance, as you mention, we could go to a stronger system where there is an axiom giving the existence of inaccessible cardinals. Working in Morse-Kelley is another possibility (though this is in a lot of ways pretty similar to positing inaccessible cardinals). In any case, you are in the same position with the stronger system as you were with ZFC: you cannot use it to prove its own consistency.
answered 1 hour ago
spaceisdarkgreen
30.6k21551
Thank you your answer, that makes perfect sense. As I mentioned on the other answer, I think the confusion came from reading "we know PA is consistent" in the wrong light.
– K. 622
57 mins ago
@K. 622 That’s right. In a very crude (and possibly contentious) approximation “we know†$approx$ “ZFC provesâ€Â
– spaceisdarkgreen
49 mins ago
add a comment |Â
up vote
1
down vote
Of course we can. And we do just that.
To "step outside of $sf PA$" means that you assume the consistency of a far stronger theory, e.g. $sf ZFC$, which lets you construct $Bbb N$ as an object and prove it satisfies $sf PA$.
When you say "assume that $kappa$ is a cardinal such that $V_kappa$ is a model of $sf ZFC$" you effectively saying "We are stepping outside of $sf ZFC$ into a theory "$sf ZFC+varphi$ for a suitable axiom $varphi$, and there we can can find a model of $sf ZFC$.
In fact, when you assume something like $V_kappa$ is a model of $sf ZFC$, you can even find a fairly canonical model of $sf ZFC$: $L_alpha$ for the least $alpha$ satisfying $sf ZFC$, where $L_alpha$ is the $alpha$th step in the constructible hierarchy. As to why it exists, that's another question. But it is there.
answered 54 mins ago
Asaf Karagila♦
298k32417745
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Using ZFC plus the axiom "Uncountable strong inaccessible cardinals exist" to give a model of ZFC is exactly the same kind of "stepping outside" as when you use ZF to make a model of PA.
answered 1 hour ago
Arthur
106k7101183
Ooh I see. I was essentially confused from reading "we know PA is consistent" elsewhere, when I should have interpreted that as "within the standard system we use today, PA is consistent". Thank you for clearing that up.
– K. 622
59 mins ago
add a comment |Â
up vote
2
down vote
Using ZFC plus the axiom "Uncountable strong inaccessible cardinals exist" to give a model of ZFC is exactly the same kind of "stepping outside" as when you use ZF to make a model of PA.
answered 1 hour ago
Arthur
106k7101183
Ooh I see. I was essentially confused from reading "we know PA is consistent" elsewhere, when I should have interpreted that as "within the standard system we use today, PA is consistent". Thank you for clearing that up.
– K. 622
59 mins ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Using ZFC plus the axiom "Uncountable strong inaccessible cardinals exist" to give a model of ZFC is exactly the same kind of "stepping outside" as when you use ZF to make a model of PA.
answered 1 hour ago
Arthur
106k7101183
Using ZFC plus the axiom "Uncountable strong inaccessible cardinals exist" to give a model of ZFC is exactly the same kind of "stepping outside" as when you use ZF to make a model of PA.
answered 1 hour ago
Arthur
106k7101183
answered 1 hour ago
Arthur
106k7101183
answered 1 hour ago
Arthur
106k7101183
answered 1 hour ago
Arthur
106k7101183
106k7101183
Ooh I see. I was essentially confused from reading "we know PA is consistent" elsewhere, when I should have interpreted that as "within the standard system we use today, PA is consistent". Thank you for clearing that up.
– K. 622
59 mins ago
add a comment |Â
Ooh I see. I was essentially confused from reading "we know PA is consistent" elsewhere, when I should have interpreted that as "within the standard system we use today, PA is consistent". Thank you for clearing that up.
– K. 622
59 mins ago
Ooh I see. I was essentially confused from reading "we know PA is consistent" elsewhere, when I should have interpreted that as "within the standard system we use today, PA is consistent". Thank you for clearing that up.
– K. 622
59 mins ago
Ooh I see. I was essentially confused from reading "we know PA is consistent" elsewhere, when I should have interpreted that as "within the standard system we use today, PA is consistent". Thank you for clearing that up.
– K. 622
59 mins ago
add a comment |Â
up vote
2
down vote
The problem is that, unlike the case for PA, essentially all accepted mathematical reasoning can be formalized in ZFC. Any proof of the consistency of ZFC must come from a system that is stronger (at least in some ways), so we must go outside ZFC-formalizable mathematics, which is most of mathmatics. (Just like we go outside of PA-formalizable mathematics to prove the consistency of PA.)
For instance, as you mention, we could go to a stronger system where there is an axiom giving the existence of inaccessible cardinals. Working in Morse-Kelley is another possibility (though this is in a lot of ways pretty similar to positing inaccessible cardinals). In any case, you are in the same position with the stronger system as you were with ZFC: you cannot use it to prove its own consistency.
answered 1 hour ago
spaceisdarkgreen
30.6k21551
Thank you your answer, that makes perfect sense. As I mentioned on the other answer, I think the confusion came from reading "we know PA is consistent" in the wrong light.
– K. 622
57 mins ago
@K. 622 That’s right. In a very crude (and possibly contentious) approximation “we know†$approx$ “ZFC provesâ€Â
– spaceisdarkgreen
49 mins ago
add a comment |Â
up vote
2
down vote
The problem is that, unlike the case for PA, essentially all accepted mathematical reasoning can be formalized in ZFC. Any proof of the consistency of ZFC must come from a system that is stronger (at least in some ways), so we must go outside ZFC-formalizable mathematics, which is most of mathmatics. (Just like we go outside of PA-formalizable mathematics to prove the consistency of PA.)
For instance, as you mention, we could go to a stronger system where there is an axiom giving the existence of inaccessible cardinals. Working in Morse-Kelley is another possibility (though this is in a lot of ways pretty similar to positing inaccessible cardinals). In any case, you are in the same position with the stronger system as you were with ZFC: you cannot use it to prove its own consistency.
answered 1 hour ago
spaceisdarkgreen
30.6k21551
Thank you your answer, that makes perfect sense. As I mentioned on the other answer, I think the confusion came from reading "we know PA is consistent" in the wrong light.
– K. 622
57 mins ago
@K. 622 That’s right. In a very crude (and possibly contentious) approximation “we know†$approx$ “ZFC provesâ€Â
– spaceisdarkgreen
49 mins ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
The problem is that, unlike the case for PA, essentially all accepted mathematical reasoning can be formalized in ZFC. Any proof of the consistency of ZFC must come from a system that is stronger (at least in some ways), so we must go outside ZFC-formalizable mathematics, which is most of mathmatics. (Just like we go outside of PA-formalizable mathematics to prove the consistency of PA.)
For instance, as you mention, we could go to a stronger system where there is an axiom giving the existence of inaccessible cardinals. Working in Morse-Kelley is another possibility (though this is in a lot of ways pretty similar to positing inaccessible cardinals). In any case, you are in the same position with the stronger system as you were with ZFC: you cannot use it to prove its own consistency.
answered 1 hour ago
spaceisdarkgreen
30.6k21551
The problem is that, unlike the case for PA, essentially all accepted mathematical reasoning can be formalized in ZFC. Any proof of the consistency of ZFC must come from a system that is stronger (at least in some ways), so we must go outside ZFC-formalizable mathematics, which is most of mathmatics. (Just like we go outside of PA-formalizable mathematics to prove the consistency of PA.)
For instance, as you mention, we could go to a stronger system where there is an axiom giving the existence of inaccessible cardinals. Working in Morse-Kelley is another possibility (though this is in a lot of ways pretty similar to positing inaccessible cardinals). In any case, you are in the same position with the stronger system as you were with ZFC: you cannot use it to prove its own consistency.
answered 1 hour ago
spaceisdarkgreen
30.6k21551
edited 53 mins ago
answered 1 hour ago
spaceisdarkgreen
30.6k21551
answered 1 hour ago
spaceisdarkgreen
30.6k21551
answered 1 hour ago
spaceisdarkgreen
30.6k21551
30.6k21551
Thank you your answer, that makes perfect sense. As I mentioned on the other answer, I think the confusion came from reading "we know PA is consistent" in the wrong light.
– K. 622
57 mins ago
@K. 622 That’s right. In a very crude (and possibly contentious) approximation “we know†$approx$ “ZFC provesâ€Â
– spaceisdarkgreen
49 mins ago
add a comment |Â
Thank you your answer, that makes perfect sense. As I mentioned on the other answer, I think the confusion came from reading "we know PA is consistent" in the wrong light.
– K. 622
57 mins ago
@K. 622 That’s right. In a very crude (and possibly contentious) approximation “we know†$approx$ “ZFC provesâ€Â
– spaceisdarkgreen
49 mins ago
Thank you your answer, that makes perfect sense. As I mentioned on the other answer, I think the confusion came from reading "we know PA is consistent" in the wrong light.
– K. 622
57 mins ago
Thank you your answer, that makes perfect sense. As I mentioned on the other answer, I think the confusion came from reading "we know PA is consistent" in the wrong light.
– K. 622
57 mins ago
@K. 622 That’s right. In a very crude (and possibly contentious) approximation “we know†$approx$ “ZFC provesâ€Â
– spaceisdarkgreen
49 mins ago
@K. 622 That’s right. In a very crude (and possibly contentious) approximation “we know†$approx$ “ZFC provesâ€Â
– spaceisdarkgreen
49 mins ago
add a comment |Â
up vote
1
down vote
Of course we can. And we do just that.
To "step outside of $sf PA$" means that you assume the consistency of a far stronger theory, e.g. $sf ZFC$, which lets you construct $Bbb N$ as an object and prove it satisfies $sf PA$.
When you say "assume that $kappa$ is a cardinal such that $V_kappa$ is a model of $sf ZFC$" you effectively saying "We are stepping outside of $sf ZFC$ into a theory "$sf ZFC+varphi$ for a suitable axiom $varphi$, and there we can can find a model of $sf ZFC$.
In fact, when you assume something like $V_kappa$ is a model of $sf ZFC$, you can even find a fairly canonical model of $sf ZFC$: $L_alpha$ for the least $alpha$ satisfying $sf ZFC$, where $L_alpha$ is the $alpha$th step in the constructible hierarchy. As to why it exists, that's another question. But it is there.
answered 54 mins ago
Asaf Karagila♦
298k32417745
add a comment |Â
up vote
1
down vote
Of course we can. And we do just that.
To "step outside of $sf PA$" means that you assume the consistency of a far stronger theory, e.g. $sf ZFC$, which lets you construct $Bbb N$ as an object and prove it satisfies $sf PA$.
When you say "assume that $kappa$ is a cardinal such that $V_kappa$ is a model of $sf ZFC$" you effectively saying "We are stepping outside of $sf ZFC$ into a theory "$sf ZFC+varphi$ for a suitable axiom $varphi$, and there we can can find a model of $sf ZFC$.
In fact, when you assume something like $V_kappa$ is a model of $sf ZFC$, you can even find a fairly canonical model of $sf ZFC$: $L_alpha$ for the least $alpha$ satisfying $sf ZFC$, where $L_alpha$ is the $alpha$th step in the constructible hierarchy. As to why it exists, that's another question. But it is there.
answered 54 mins ago
Asaf Karagila♦
298k32417745
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Of course we can. And we do just that.
To "step outside of $sf PA$" means that you assume the consistency of a far stronger theory, e.g. $sf ZFC$, which lets you construct $Bbb N$ as an object and prove it satisfies $sf PA$.
When you say "assume that $kappa$ is a cardinal such that $V_kappa$ is a model of $sf ZFC$" you effectively saying "We are stepping outside of $sf ZFC$ into a theory "$sf ZFC+varphi$ for a suitable axiom $varphi$, and there we can can find a model of $sf ZFC$.
In fact, when you assume something like $V_kappa$ is a model of $sf ZFC$, you can even find a fairly canonical model of $sf ZFC$: $L_alpha$ for the least $alpha$ satisfying $sf ZFC$, where $L_alpha$ is the $alpha$th step in the constructible hierarchy. As to why it exists, that's another question. But it is there.
answered 54 mins ago
Asaf Karagila♦
298k32417745
Of course we can. And we do just that.
To "step outside of $sf PA$" means that you assume the consistency of a far stronger theory, e.g. $sf ZFC$, which lets you construct $Bbb N$ as an object and prove it satisfies $sf PA$.
When you say "assume that $kappa$ is a cardinal such that $V_kappa$ is a model of $sf ZFC$" you effectively saying "We are stepping outside of $sf ZFC$ into a theory "$sf ZFC+varphi$ for a suitable axiom $varphi$, and there we can can find a model of $sf ZFC$.
In fact, when you assume something like $V_kappa$ is a model of $sf ZFC$, you can even find a fairly canonical model of $sf ZFC$: $L_alpha$ for the least $alpha$ satisfying $sf ZFC$, where $L_alpha$ is the $alpha$th step in the constructible hierarchy. As to why it exists, that's another question. But it is there.
answered 54 mins ago
Asaf Karagila♦
298k32417745
answered 54 mins ago
Asaf Karagila♦
298k32417745
answered 54 mins ago
Asaf Karagila♦
298k32417745
answered 54 mins ago
Asaf Karagila♦
298k32417745
298k32417745
add a comment |Â
add a comment |Â
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5
Stepping outside PA is an "if". We prove PA is consistent if ZFC is.
– Malice Vidrine
1 hour ago