A model of $PA^omega$ in ZFC [closed]
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It is known that the existence of a model (i.e. consistency) of $PA$ is provable in ZFC, the set $omega$ equipped with usual operation on natural numbers is indeed a model of $PA$, and we can think of it as the set of natural numbers $mathbb{N}subsetmathbb{R}$. Can the same be done with $PA^omega$? Maybe using instead of $mathbb{N}$ the set ${1}cup{1-frac{1}{n}, ninmathbb{N}}$ or something similar... I think the problem is that such a model will not be well ordered (the set of elements greater than every "natural" number contains $omega$, $omega-1$, $omega-2$, $dots$).
EDIT: $PA^omega$ is a theory constructed adding to $PA$ a new constant symbol $omega$ and new axioms $omega_n: n<omega$ for $n=1,2,dots$. Is then shown that every finite set of axioms has a model, and it's clear because in a finite set of axioms $Gamma$ we have just finite natural numbers $n$ such that $omega_ninGamma$, and so any natural number greater than all of them can play the role of $omega$. Then by Compactness Theorem we get a nonstandard model of Peano Arithmetic.
set-theory model-theory nonstandard-models
asked Dec 4 '18 at 16:58
nicolanicola
256
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closed as unclear what you're asking by Andrés E. Caicedo, Alex Kruckman, user10354138, DRF, Cesareo Dec 5 '18 at 19:31
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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show 5 more comments
$begingroup$
It is known that the existence of a model (i.e. consistency) of $PA$ is provable in ZFC, the set $omega$ equipped with usual operation on natural numbers is indeed a model of $PA$, and we can think of it as the set of natural numbers $mathbb{N}subsetmathbb{R}$. Can the same be done with $PA^omega$? Maybe using instead of $mathbb{N}$ the set ${1}cup{1-frac{1}{n}, ninmathbb{N}}$ or something similar... I think the problem is that such a model will not be well ordered (the set of elements greater than every "natural" number contains $omega$, $omega-1$, $omega-2$, $dots$).
EDIT: $PA^omega$ is a theory constructed adding to $PA$ a new constant symbol $omega$ and new axioms $omega_n: n<omega$ for $n=1,2,dots$. Is then shown that every finite set of axioms has a model, and it's clear because in a finite set of axioms $Gamma$ we have just finite natural numbers $n$ such that $omega_ninGamma$, and so any natural number greater than all of them can play the role of $omega$. Then by Compactness Theorem we get a nonstandard model of Peano Arithmetic.
set-theory model-theory nonstandard-models
asked Dec 4 '18 at 16:58
nicolanicola
256
$endgroup$
closed as unclear what you're asking by Andrés E. Caicedo, Alex Kruckman, user10354138, DRF, Cesareo Dec 5 '18 at 19:31
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
2
$begingroup$
Think about the construction of a model of ZFC by adding a new element and requiring it to be a finite ordinal but not $varnothing$ or ${varnothing}$ or so on. You still get a model of ZFC, but the model itself is ill-founded. What can you say about the object that that model think is $omega$?
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– Asaf Karagila♦
Dec 4 '18 at 17:01
3
$begingroup$
What is PA$^omega$?
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– Alex Kruckman
Dec 4 '18 at 17:18
$begingroup$
@Alex: I'm guessing that math.stackexchange.com/questions/3025809/… is related.
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– Asaf Karagila♦
Dec 4 '18 at 17:26
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Please define $PA^omega$, it is not standard notation.
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– Andrés E. Caicedo
Dec 4 '18 at 18:35
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@AsafKaragila I suppose that this new element in that new model of ZFC plays the role (is the image by interpretation function) of the constant symbol $omega$, or i’m totally wrong? That remain to me a bit vague anyway (unfortunately i’m not yet very familiar with models or forcing) and so i was looking for giving a set-theoretic “visible” structure to such a model. Thank you so much!
$endgroup$
– nicola
Dec 4 '18 at 18:38
|
show 5 more comments
$begingroup$
It is known that the existence of a model (i.e. consistency) of $PA$ is provable in ZFC, the set $omega$ equipped with usual operation on natural numbers is indeed a model of $PA$, and we can think of it as the set of natural numbers $mathbb{N}subsetmathbb{R}$. Can the same be done with $PA^omega$? Maybe using instead of $mathbb{N}$ the set ${1}cup{1-frac{1}{n}, ninmathbb{N}}$ or something similar... I think the problem is that such a model will not be well ordered (the set of elements greater than every "natural" number contains $omega$, $omega-1$, $omega-2$, $dots$).
EDIT: $PA^omega$ is a theory constructed adding to $PA$ a new constant symbol $omega$ and new axioms $omega_n: n<omega$ for $n=1,2,dots$. Is then shown that every finite set of axioms has a model, and it's clear because in a finite set of axioms $Gamma$ we have just finite natural numbers $n$ such that $omega_ninGamma$, and so any natural number greater than all of them can play the role of $omega$. Then by Compactness Theorem we get a nonstandard model of Peano Arithmetic.
set-theory model-theory nonstandard-models
asked Dec 4 '18 at 16:58
nicolanicola
256
$endgroup$
It is known that the existence of a model (i.e. consistency) of $PA$ is provable in ZFC, the set $omega$ equipped with usual operation on natural numbers is indeed a model of $PA$, and we can think of it as the set of natural numbers $mathbb{N}subsetmathbb{R}$. Can the same be done with $PA^omega$? Maybe using instead of $mathbb{N}$ the set ${1}cup{1-frac{1}{n}, ninmathbb{N}}$ or something similar... I think the problem is that such a model will not be well ordered (the set of elements greater than every "natural" number contains $omega$, $omega-1$, $omega-2$, $dots$).
EDIT: $PA^omega$ is a theory constructed adding to $PA$ a new constant symbol $omega$ and new axioms $omega_n: n<omega$ for $n=1,2,dots$. Is then shown that every finite set of axioms has a model, and it's clear because in a finite set of axioms $Gamma$ we have just finite natural numbers $n$ such that $omega_ninGamma$, and so any natural number greater than all of them can play the role of $omega$. Then by Compactness Theorem we get a nonstandard model of Peano Arithmetic.
set-theory model-theory nonstandard-models
set-theory model-theory nonstandard-models
asked Dec 4 '18 at 16:58
nicolanicola
256
asked Dec 4 '18 at 16:58
nicolanicola
256
edited Dec 4 '18 at 19:20
nicola
asked Dec 4 '18 at 16:58
nicolanicola
256
asked Dec 4 '18 at 16:58
nicolanicola
256
asked Dec 4 '18 at 16:58
nicolanicola
256
256
closed as unclear what you're asking by Andrés E. Caicedo, Alex Kruckman, user10354138, DRF, Cesareo Dec 5 '18 at 19:31
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Andrés E. Caicedo, Alex Kruckman, user10354138, DRF, Cesareo Dec 5 '18 at 19:31
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
2
$begingroup$
Think about the construction of a model of ZFC by adding a new element and requiring it to be a finite ordinal but not $varnothing$ or ${varnothing}$ or so on. You still get a model of ZFC, but the model itself is ill-founded. What can you say about the object that that model think is $omega$?
$endgroup$
– Asaf Karagila♦
Dec 4 '18 at 17:01
3
$begingroup$
What is PA$^omega$?
$endgroup$
– Alex Kruckman
Dec 4 '18 at 17:18
$begingroup$
@Alex: I'm guessing that math.stackexchange.com/questions/3025809/… is related.
$endgroup$
– Asaf Karagila♦
Dec 4 '18 at 17:26
$begingroup$
Please define $PA^omega$, it is not standard notation.
$endgroup$
– Andrés E. Caicedo
Dec 4 '18 at 18:35
$begingroup$
@AsafKaragila I suppose that this new element in that new model of ZFC plays the role (is the image by interpretation function) of the constant symbol $omega$, or i’m totally wrong? That remain to me a bit vague anyway (unfortunately i’m not yet very familiar with models or forcing) and so i was looking for giving a set-theoretic “visible” structure to such a model. Thank you so much!
$endgroup$
– nicola
Dec 4 '18 at 18:38
|
show 5 more comments
2
$begingroup$
Think about the construction of a model of ZFC by adding a new element and requiring it to be a finite ordinal but not $varnothing$ or ${varnothing}$ or so on. You still get a model of ZFC, but the model itself is ill-founded. What can you say about the object that that model think is $omega$?
$endgroup$
– Asaf Karagila♦
Dec 4 '18 at 17:01
3
$begingroup$
What is PA$^omega$?
$endgroup$
– Alex Kruckman
Dec 4 '18 at 17:18
$begingroup$
@Alex: I'm guessing that math.stackexchange.com/questions/3025809/… is related.
$endgroup$
– Asaf Karagila♦
Dec 4 '18 at 17:26
$begingroup$
Please define $PA^omega$, it is not standard notation.
$endgroup$
– Andrés E. Caicedo
Dec 4 '18 at 18:35
$begingroup$
@AsafKaragila I suppose that this new element in that new model of ZFC plays the role (is the image by interpretation function) of the constant symbol $omega$, or i’m totally wrong? That remain to me a bit vague anyway (unfortunately i’m not yet very familiar with models or forcing) and so i was looking for giving a set-theoretic “visible” structure to such a model. Thank you so much!
$endgroup$
– nicola
Dec 4 '18 at 18:38
2
2
$begingroup$
Think about the construction of a model of ZFC by adding a new element and requiring it to be a finite ordinal but not $varnothing$ or ${varnothing}$ or so on. You still get a model of ZFC, but the model itself is ill-founded. What can you say about the object that that model think is $omega$?
$endgroup$
– Asaf Karagila♦
Dec 4 '18 at 17:01
$begingroup$
Think about the construction of a model of ZFC by adding a new element and requiring it to be a finite ordinal but not $varnothing$ or ${varnothing}$ or so on. You still get a model of ZFC, but the model itself is ill-founded. What can you say about the object that that model think is $omega$?
$endgroup$
– Asaf Karagila♦
Dec 4 '18 at 17:01
3
3
$begingroup$
What is PA$^omega$?
$endgroup$
– Alex Kruckman
Dec 4 '18 at 17:18
$begingroup$
What is PA$^omega$?
$endgroup$
– Alex Kruckman
Dec 4 '18 at 17:18
$begingroup$
@Alex: I'm guessing that math.stackexchange.com/questions/3025809/… is related.
$endgroup$
– Asaf Karagila♦
Dec 4 '18 at 17:26
$begingroup$
@Alex: I'm guessing that math.stackexchange.com/questions/3025809/… is related.
$endgroup$
– Asaf Karagila♦
Dec 4 '18 at 17:26
$begingroup$
Please define $PA^omega$, it is not standard notation.
$endgroup$
– Andrés E. Caicedo
Dec 4 '18 at 18:35
$begingroup$
Please define $PA^omega$, it is not standard notation.
$endgroup$
– Andrés E. Caicedo
Dec 4 '18 at 18:35
$begingroup$
@AsafKaragila I suppose that this new element in that new model of ZFC plays the role (is the image by interpretation function) of the constant symbol $omega$, or i’m totally wrong? That remain to me a bit vague anyway (unfortunately i’m not yet very familiar with models or forcing) and so i was looking for giving a set-theoretic “visible” structure to such a model. Thank you so much!
$endgroup$
– nicola
Dec 4 '18 at 18:38
$begingroup$
@AsafKaragila I suppose that this new element in that new model of ZFC plays the role (is the image by interpretation function) of the constant symbol $omega$, or i’m totally wrong? That remain to me a bit vague anyway (unfortunately i’m not yet very familiar with models or forcing) and so i was looking for giving a set-theoretic “visible” structure to such a model. Thank you so much!
$endgroup$
– nicola
Dec 4 '18 at 18:38
|
show 5 more comments
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$begingroup$
Think about the construction of a model of ZFC by adding a new element and requiring it to be a finite ordinal but not $varnothing$ or ${varnothing}$ or so on. You still get a model of ZFC, but the model itself is ill-founded. What can you say about the object that that model think is $omega$?
$endgroup$
– Asaf Karagila♦
Dec 4 '18 at 17:01
3
$begingroup$
What is PA$^omega$?
$endgroup$
– Alex Kruckman
Dec 4 '18 at 17:18
$begingroup$
@Alex: I'm guessing that math.stackexchange.com/questions/3025809/… is related.
$endgroup$
– Asaf Karagila♦
Dec 4 '18 at 17:26
$begingroup$
Please define $PA^omega$, it is not standard notation.
$endgroup$
– Andrés E. Caicedo
Dec 4 '18 at 18:35
$begingroup$
@AsafKaragila I suppose that this new element in that new model of ZFC plays the role (is the image by interpretation function) of the constant symbol $omega$, or i’m totally wrong? That remain to me a bit vague anyway (unfortunately i’m not yet very familiar with models or forcing) and so i was looking for giving a set-theoretic “visible” structure to such a model. Thank you so much!
$endgroup$
– nicola
Dec 4 '18 at 18:38