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Does the Axiom schema of Replacement imply the Axiom of Infinity? [duplicate]

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$begingroup$

This question already has an answer here:

• 
  Why is the Axiom of Infinity necessary?
  
  3 answers
  
  

The axiom of infinity says that the set of natural numbers exists, while the axiom of replacement says that if an object (a member of a set) exists, then all definable mappings of that object yield objects (e.g if 0 is an object, then 0+1 is also an object).
Doesn't this mean that the axiom of infinity is redundant since one can recursively prove the existence of the set of natural numbers using the successor function of the Peano axioms?

set-theory axioms natural-numbers

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asked Mar 18 at 12:27

Amr AymanAmr Ayman

1053

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marked as duplicate by Carl Mummert, Cameron Buie, Mauro ALLEGRANZA, dantopa, Xander Henderson Mar 18 at 21:44

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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  See the post why-is-the-axiom-of-infinity-necessary
  $endgroup$
  – Mauro ALLEGRANZA
  Mar 18 at 12:38
  

  
  
  
  


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  This would prove that $0,1,dots,n-1$ exists for all $n$, but not that $0,1,2,dots$ exists.
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  – Mike Earnest
  Mar 18 at 16:55
  

  
  
  
  


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  @MikeEarnest In the same manner, then the peano axioms assert the existence of $0,1,dots,n-1$ for all n but not $0,1,2,dots$. Why is this different?
  $endgroup$
  – Amr Ayman
  Mar 18 at 17:15
  
  

  
  
  
  

add a comment | 

0

$begingroup$

This question already has an answer here:

• 
  Why is the Axiom of Infinity necessary?
  
  3 answers
  
  

The axiom of infinity says that the set of natural numbers exists, while the axiom of replacement says that if an object (a member of a set) exists, then all definable mappings of that object yield objects (e.g if 0 is an object, then 0+1 is also an object).
Doesn't this mean that the axiom of infinity is redundant since one can recursively prove the existence of the set of natural numbers using the successor function of the Peano axioms?

set-theory axioms natural-numbers

share|cite|improve this question

asked Mar 18 at 12:27

Amr AymanAmr Ayman

1053

$endgroup$

marked as duplicate by Carl Mummert, Cameron Buie, Mauro ALLEGRANZA, dantopa, Xander Henderson Mar 18 at 21:44

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  See the post why-is-the-axiom-of-infinity-necessary
  $endgroup$
  – Mauro ALLEGRANZA
  Mar 18 at 12:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This would prove that $0,1,dots,n-1$ exists for all $n$, but not that $0,1,2,dots$ exists.
  $endgroup$
  – Mike Earnest
  Mar 18 at 16:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MikeEarnest In the same manner, then the peano axioms assert the existence of $0,1,dots,n-1$ for all n but not $0,1,2,dots$. Why is this different?
  $endgroup$
  – Amr Ayman
  Mar 18 at 17:15
  
  

  
  
  
  

add a comment | 

$begingroup$

This question already has an answer here:

• 
  Why is the Axiom of Infinity necessary?
  
  3 answers
  
  

The axiom of infinity says that the set of natural numbers exists, while the axiom of replacement says that if an object (a member of a set) exists, then all definable mappings of that object yield objects (e.g if 0 is an object, then 0+1 is also an object).
Doesn't this mean that the axiom of infinity is redundant since one can recursively prove the existence of the set of natural numbers using the successor function of the Peano axioms?

set-theory axioms natural-numbers

share|cite|improve this question

asked Mar 18 at 12:27

Amr AymanAmr Ayman

1053

$endgroup$

share|cite|improve this question

asked Mar 18 at 12:27

Amr AymanAmr Ayman

1053

share|cite|improve this question

asked Mar 18 at 12:27

Amr AymanAmr Ayman

1053

asked Mar 18 at 12:27

Amr AymanAmr Ayman

1053

asked Mar 18 at 12:27

Amr AymanAmr Ayman

1053

1 Answer
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2

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No, replacement doesn't imply infinity. What you've found is that if $S$ is a set so is $xcupx$, hence so is $xcupx$. Each element of $omega$ can be proven to be a set by this method, but we can't use this on its own to prove $omega$ is a set, even though it's just the union of its elements.

share|cite|improve this answer

answered Mar 18 at 12:57

J.G.J.G.

32.3k23250

$endgroup$

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  So the problem lies in "infinitely applying the pairwise union axiom"? I don't see how that's different from the axiom of induction of the natural numbers..
  $endgroup$
  – Amr Ayman
  Mar 18 at 13:00
  

  
  
  
  


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  @AmrAyman No, the problem is you don't have a way of proving these individual items all belong to the same set $omega$. If you actually try to write down a proof that such a set exists, you'll find none of the other axioms let you do so.
  $endgroup$
  – J.G.
  Mar 18 at 13:01
  

  
  
  
  


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  Ok. But why can't $omega$ be proven to exist, and simply be defined as the union of its elements?
  $endgroup$
  – Amr Ayman
  Mar 18 at 13:29
  

  
  
  
  


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  @AmrAyman Because you can't glue together arbitrary sets, e.g. as $phi(x)$ or the union thereof for a unary predicate $phi$.
  $endgroup$
  – J.G.
  Mar 18 at 13:50
  

  
  
  
  


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  Oh, that's only possible under unrestricted comprehension. My bad.
  $endgroup$
  – Amr Ayman
  Mar 18 at 17:59
  

  
  
  
  

add a comment | 

2

$begingroup$

No, replacement doesn't imply infinity. What you've found is that if $S$ is a set so is $xcupx$, hence so is $xcupx$. Each element of $omega$ can be proven to be a set by this method, but we can't use this on its own to prove $omega$ is a set, even though it's just the union of its elements.

share|cite|improve this answer

answered Mar 18 at 12:57

J.G.J.G.

32.3k23250

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So the problem lies in "infinitely applying the pairwise union axiom"? I don't see how that's different from the axiom of induction of the natural numbers..
  $endgroup$
  – Amr Ayman
  Mar 18 at 13:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @AmrAyman No, the problem is you don't have a way of proving these individual items all belong to the same set $omega$. If you actually try to write down a proof that such a set exists, you'll find none of the other axioms let you do so.
  $endgroup$
  – J.G.
  Mar 18 at 13:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok. But why can't $omega$ be proven to exist, and simply be defined as the union of its elements?
  $endgroup$
  – Amr Ayman
  Mar 18 at 13:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @AmrAyman Because you can't glue together arbitrary sets, e.g. as $phi(x)$ or the union thereof for a unary predicate $phi$.
  $endgroup$
  – J.G.
  Mar 18 at 13:50
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Oh, that's only possible under unrestricted comprehension. My bad.
  $endgroup$
  – Amr Ayman
  Mar 18 at 17:59
  

  
  
  
  

add a comment | 

2

$begingroup$

No, replacement doesn't imply infinity. What you've found is that if $S$ is a set so is $xcupx$, hence so is $xcupx$. Each element of $omega$ can be proven to be a set by this method, but we can't use this on its own to prove $omega$ is a set, even though it's just the union of its elements.

share|cite|improve this answer

answered Mar 18 at 12:57

J.G.J.G.

32.3k23250

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So the problem lies in "infinitely applying the pairwise union axiom"? I don't see how that's different from the axiom of induction of the natural numbers..
  $endgroup$
  – Amr Ayman
  Mar 18 at 13:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @AmrAyman No, the problem is you don't have a way of proving these individual items all belong to the same set $omega$. If you actually try to write down a proof that such a set exists, you'll find none of the other axioms let you do so.
  $endgroup$
  – J.G.
  Mar 18 at 13:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok. But why can't $omega$ be proven to exist, and simply be defined as the union of its elements?
  $endgroup$
  – Amr Ayman
  Mar 18 at 13:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @AmrAyman Because you can't glue together arbitrary sets, e.g. as $phi(x)$ or the union thereof for a unary predicate $phi$.
  $endgroup$
  – J.G.
  Mar 18 at 13:50
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Oh, that's only possible under unrestricted comprehension. My bad.
  $endgroup$
  – Amr Ayman
  Mar 18 at 17:59
  

  
  
  
  

add a comment | 

$begingroup$

No, replacement doesn't imply infinity. What you've found is that if $S$ is a set so is $xcupx$, hence so is $xcupx$. Each element of $omega$ can be proven to be a set by this method, but we can't use this on its own to prove $omega$ is a set, even though it's just the union of its elements.

share|cite|improve this answer

answered Mar 18 at 12:57

J.G.J.G.

32.3k23250

$endgroup$

share|cite|improve this answer

answered Mar 18 at 12:57

J.G.J.G.

32.3k23250

answered Mar 18 at 12:57

J.G.J.G.

32.3k23250

answered Mar 18 at 12:57

J.G.J.G.

32.3k23250