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Can the successor function be applied to N itself? ( With ' N' denoting the set of natural numbers)

0

$begingroup$

I don't think my question leads to anything, but does it have an answer?

My question is: I am allowed to form the set

 N U N , 

that is the set :

 0, 1, 2 ,3 ............................... 0,1,2,3 ....... , 

applying the successor function S to N itself, where S(x) = x U x ?

Remark: I'm not asking whether N is itself a natural number, that would be forbidden, I think, by the rule according to which no set can be a member of itself.

elementary-set-theory

share|cite|improve this question

edited Mar 31 at 11:59

Andrés E. Caicedo

65.9k8160252

asked Mar 30 at 18:01

Ray LittleRockRay LittleRock

9610

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  In set theory yes. See Von Neumann definition of ordinal.
  $endgroup$
  – Mauro ALLEGRANZA
  Mar 30 at 18:06
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks. May I ask you whether this set is of any interest? Could it qualify as a number of some sort?
  $endgroup$
  – Ray LittleRock
  Mar 30 at 18:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ Mauro Allegranza. Thanks for the link.
  $endgroup$
  – Ray LittleRock
  Mar 30 at 18:12
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, it can qualify as a number... it's the ordinal number $omega + 1$.
  $endgroup$
  – mjqxxxx
  Mar 30 at 18:14
  

  
  
  
  

add a comment | 

0

$begingroup$

I don't think my question leads to anything, but does it have an answer?

My question is: I am allowed to form the set

 N U N , 

that is the set :

 0, 1, 2 ,3 ............................... 0,1,2,3 ....... , 

applying the successor function S to N itself, where S(x) = x U x ?

Remark: I'm not asking whether N is itself a natural number, that would be forbidden, I think, by the rule according to which no set can be a member of itself.

elementary-set-theory

share|cite|improve this question

edited Mar 31 at 11:59

Andrés E. Caicedo

65.9k8160252

asked Mar 30 at 18:01

Ray LittleRockRay LittleRock

9610

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  In set theory yes. See Von Neumann definition of ordinal.
  $endgroup$
  – Mauro ALLEGRANZA
  Mar 30 at 18:06
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks. May I ask you whether this set is of any interest? Could it qualify as a number of some sort?
  $endgroup$
  – Ray LittleRock
  Mar 30 at 18:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ Mauro Allegranza. Thanks for the link.
  $endgroup$
  – Ray LittleRock
  Mar 30 at 18:12
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, it can qualify as a number... it's the ordinal number $omega + 1$.
  $endgroup$
  – mjqxxxx
  Mar 30 at 18:14
  

  
  
  
  

add a comment | 

$begingroup$

I don't think my question leads to anything, but does it have an answer?

My question is: I am allowed to form the set

 N U N , 

that is the set :

 0, 1, 2 ,3 ............................... 0,1,2,3 ....... , 

applying the successor function S to N itself, where S(x) = x U x ?

Remark: I'm not asking whether N is itself a natural number, that would be forbidden, I think, by the rule according to which no set can be a member of itself.

elementary-set-theory

share|cite|improve this question

edited Mar 31 at 11:59

Andrés E. Caicedo

65.9k8160252

asked Mar 30 at 18:01

Ray LittleRockRay LittleRock

9610

$endgroup$

 N U N , 

 0, 1, 2 ,3 ............................... 0,1,2,3 ....... , 

share|cite|improve this question

edited Mar 31 at 11:59

Andrés E. Caicedo

65.9k8160252

asked Mar 30 at 18:01

Ray LittleRockRay LittleRock

9610

share|cite|improve this question

edited Mar 31 at 11:59

Andrés E. Caicedo

65.9k8160252

asked Mar 30 at 18:01

Ray LittleRockRay LittleRock

9610

edited Mar 31 at 11:59

Andrés E. Caicedo

65.9k8160252

edited Mar 31 at 11:59

Andrés E. Caicedo

65.9k8160252

asked Mar 30 at 18:01

Ray LittleRockRay LittleRock

9610

asked Mar 30 at 18:01

Ray LittleRockRay LittleRock

9610

1 Answer
1

active

oldest

votes

1

$begingroup$

For any set $X,$ the axiom of pairing implies $X$ is a set, and then applied again, says $X,X$ is a set, and then the axiom of union says $cupX,X=XcupX$ is a set. (This is only one way to construct it... there are lots of ways.) So the successor operation is defined for any set (although it's really only called the successor operation when it's used on ordinals).

So if $mathbb N$ is a set, then we can form the successor set. Almost always, we define $mathbb N$ to be $omega,$ the first infinite ordinal, which is obtained as the closure of the empty set under the successor operation (this closure exists by the axiom of infinity). The successor of $omega,$ $omegacupomega,$ is the next ordinal, denoted $omega+1.$ Applying it again we get $omega+2,$ $omega + 3$ and so on.

share|cite|improve this answer

answered Mar 30 at 20:23

spaceisdarkgreenspaceisdarkgreen

34k21754

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is $omega-1$ a well defined element of $mathbbN$?
  $endgroup$
  – Count Iblis
  Mar 30 at 20:30
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @CountIblis no, subtraction for infinite ordinals is usually left undefined
  $endgroup$
  – Holo
  Mar 30 at 20:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @CountIblis This certainly isn't a natural or an ordinal. There are some exotic systems that make sense of it though, such as the surreal numbers.
  $endgroup$
  – spaceisdarkgreen
  Mar 30 at 20:59
  
  

  
  
  
  

add a comment | 

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1

$begingroup$

For any set $X,$ the axiom of pairing implies $X$ is a set, and then applied again, says $X,X$ is a set, and then the axiom of union says $cupX,X=XcupX$ is a set. (This is only one way to construct it... there are lots of ways.) So the successor operation is defined for any set (although it's really only called the successor operation when it's used on ordinals).

So if $mathbb N$ is a set, then we can form the successor set. Almost always, we define $mathbb N$ to be $omega,$ the first infinite ordinal, which is obtained as the closure of the empty set under the successor operation (this closure exists by the axiom of infinity). The successor of $omega,$ $omegacupomega,$ is the next ordinal, denoted $omega+1.$ Applying it again we get $omega+2,$ $omega + 3$ and so on.

share|cite|improve this answer

answered Mar 30 at 20:23

spaceisdarkgreenspaceisdarkgreen

34k21754

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is $omega-1$ a well defined element of $mathbbN$?
  $endgroup$
  – Count Iblis
  Mar 30 at 20:30
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @CountIblis no, subtraction for infinite ordinals is usually left undefined
  $endgroup$
  – Holo
  Mar 30 at 20:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @CountIblis This certainly isn't a natural or an ordinal. There are some exotic systems that make sense of it though, such as the surreal numbers.
  $endgroup$
  – spaceisdarkgreen
  Mar 30 at 20:59
  
  

  
  
  
  

add a comment | 

1

$begingroup$

For any set $X,$ the axiom of pairing implies $X$ is a set, and then applied again, says $X,X$ is a set, and then the axiom of union says $cupX,X=XcupX$ is a set. (This is only one way to construct it... there are lots of ways.) So the successor operation is defined for any set (although it's really only called the successor operation when it's used on ordinals).

So if $mathbb N$ is a set, then we can form the successor set. Almost always, we define $mathbb N$ to be $omega,$ the first infinite ordinal, which is obtained as the closure of the empty set under the successor operation (this closure exists by the axiom of infinity). The successor of $omega,$ $omegacupomega,$ is the next ordinal, denoted $omega+1.$ Applying it again we get $omega+2,$ $omega + 3$ and so on.

share|cite|improve this answer

answered Mar 30 at 20:23

spaceisdarkgreenspaceisdarkgreen

34k21754

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is $omega-1$ a well defined element of $mathbbN$?
  $endgroup$
  – Count Iblis
  Mar 30 at 20:30
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @CountIblis no, subtraction for infinite ordinals is usually left undefined
  $endgroup$
  – Holo
  Mar 30 at 20:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @CountIblis This certainly isn't a natural or an ordinal. There are some exotic systems that make sense of it though, such as the surreal numbers.
  $endgroup$
  – spaceisdarkgreen
  Mar 30 at 20:59
  
  

  
  
  
  

add a comment | 

$begingroup$

For any set $X,$ the axiom of pairing implies $X$ is a set, and then applied again, says $X,X$ is a set, and then the axiom of union says $cupX,X=XcupX$ is a set. (This is only one way to construct it... there are lots of ways.) So the successor operation is defined for any set (although it's really only called the successor operation when it's used on ordinals).

So if $mathbb N$ is a set, then we can form the successor set. Almost always, we define $mathbb N$ to be $omega,$ the first infinite ordinal, which is obtained as the closure of the empty set under the successor operation (this closure exists by the axiom of infinity). The successor of $omega,$ $omegacupomega,$ is the next ordinal, denoted $omega+1.$ Applying it again we get $omega+2,$ $omega + 3$ and so on.

share|cite|improve this answer

answered Mar 30 at 20:23

spaceisdarkgreenspaceisdarkgreen

34k21754

$endgroup$

share|cite|improve this answer

answered Mar 30 at 20:23

spaceisdarkgreenspaceisdarkgreen

34k21754

answered Mar 30 at 20:23

spaceisdarkgreenspaceisdarkgreen

34k21754

answered Mar 30 at 20:23

spaceisdarkgreenspaceisdarkgreen

34k21754