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2

0

$begingroup$

The following propositions, I think, are generally considered as true ( though they may not all have the same level of rigor).

(1) In a set there is no order ( due to the extensionality axion) : $a,b = b,a$

(2) In an ordered pair there is an order : $(a, b)$ is not equal to $(b,a)$.

(3) An ordered pair is a set: $(a,b) = a , a,b $.

Does the problem lie in proposition (2) : should one say, instead of " in an ordered pair there is an order" that " an ordered pair is an order"?

How to formulate rigorously these propositions in order to make them compatible?

elementary-set-theory soft-question definition order-theory

share|cite|improve this question

edited 9 hours ago

5xum

94.2k498164

asked 10 hours ago

Eleonore Saint JamesEleonore Saint James

1,086118

$endgroup$

add a comment | 

2

0

$begingroup$

The following propositions, I think, are generally considered as true ( though they may not all have the same level of rigor).

(1) In a set there is no order ( due to the extensionality axion) : $a,b = b,a$

(2) In an ordered pair there is an order : $(a, b)$ is not equal to $(b,a)$.

(3) An ordered pair is a set: $(a,b) = a , a,b $.

Does the problem lie in proposition (2) : should one say, instead of " in an ordered pair there is an order" that " an ordered pair is an order"?

How to formulate rigorously these propositions in order to make them compatible?

elementary-set-theory soft-question definition order-theory

share|cite|improve this question

edited 9 hours ago

5xum

94.2k498164

asked 10 hours ago

Eleonore Saint JamesEleonore Saint James

1,086118

$endgroup$

add a comment | 

$begingroup$

The following propositions, I think, are generally considered as true ( though they may not all have the same level of rigor).

(1) In a set there is no order ( due to the extensionality axion) : $a,b = b,a$

(2) In an ordered pair there is an order : $(a, b)$ is not equal to $(b,a)$.

(3) An ordered pair is a set: $(a,b) = a , a,b $.

Does the problem lie in proposition (2) : should one say, instead of " in an ordered pair there is an order" that " an ordered pair is an order"?

How to formulate rigorously these propositions in order to make them compatible?

elementary-set-theory soft-question definition order-theory

share|cite|improve this question

edited 9 hours ago

5xum

94.2k498164

asked 10 hours ago

Eleonore Saint JamesEleonore Saint James

1,086118

$endgroup$

share|cite|improve this question

edited 9 hours ago

5xum

94.2k498164

asked 10 hours ago

Eleonore Saint JamesEleonore Saint James

1,086118

share|cite|improve this question

edited 9 hours ago

5xum

94.2k498164

asked 10 hours ago

Eleonore Saint JamesEleonore Saint James

1,086118

edited 9 hours ago

5xum

94.2k498164

edited 9 hours ago

5xum

94.2k498164

asked 10 hours ago

Eleonore Saint JamesEleonore Saint James

1,086118

asked 10 hours ago

Eleonore Saint JamesEleonore Saint James

1,086118

4 Answers
4

active

oldest

votes

4

$begingroup$

You formally define $(a,b)$ to be equal to $a, a,b$.

That's the only formal definition.

The rest is our interpretation. We notice that using the definition above, $(a,b)$ is not equal to $(b,a)$ if $aneq b$ (*), and therefore, in the newly introduced symbol $(.,.)$, order matters. That's why we call this newly introduced symbol an "ordered pair".

In other words, (2) is not a proposition, it is a consequence of (3).

---

(*) Note that the only thing you need to prove $(a,b)neq (b,a)$ is that $aneq b$. This is because, if $aneq b$, then the set $a$ is an element of $(a,b)$ (because $(a,b)=a, a,b$, but it is not an element of $(b,a)$ (because $aneq b$ and $aneqb,a$ and there are no other elements in $(b,a)$.

share|cite|improve this answer

edited 2 hours ago

David C. Ullrich

62.5k44197

answered 9 hours ago

5xum5xum

94.2k498164

$endgroup$

add a comment | 

2

$begingroup$

An order is a relation between elements, not the elements themselves. Propositions 1 and 2 are fine as they stand.

share|cite|improve this answer

answered 9 hours ago

Yves DaoustYves Daoust

137k877237

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @YvesDaoust.I formulate (3) due to the answer I obtained here: <math.stackexchange.com/questions/3236622/…>
  $endgroup$
  – Eleonore Saint James
  9 hours ago
  
  

  
  
  
  

add a comment | 

2

$begingroup$

(2) is a definition on "reader-level", and defines a piece of notation we would like to make use of and how to think about its use intuitively. (3) is a formal, "lower-level" definition of that same notaiton that makes sure we haven't really made any new set theory in the process, but rather that we are still (under the hood) using only the regular (unordered) sets we already had.

share|cite|improve this answer

edited 9 hours ago

answered 9 hours ago

ArthurArthur

127k7122212

$endgroup$

add a comment | 

1

$begingroup$

There's no incompatibility to be fixed. In fact $$(a,b)= a , a,b
= a,b ,a = a , b,a
= b,a ,a .$$

share|cite|improve this answer

answered 2 hours ago

David C. UllrichDavid C. Ullrich

62.5k44197

$endgroup$

add a comment | 

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4

$begingroup$

You formally define $(a,b)$ to be equal to $a, a,b$.

That's the only formal definition.

The rest is our interpretation. We notice that using the definition above, $(a,b)$ is not equal to $(b,a)$ if $aneq b$ (*), and therefore, in the newly introduced symbol $(.,.)$, order matters. That's why we call this newly introduced symbol an "ordered pair".

In other words, (2) is not a proposition, it is a consequence of (3).

---

(*) Note that the only thing you need to prove $(a,b)neq (b,a)$ is that $aneq b$. This is because, if $aneq b$, then the set $a$ is an element of $(a,b)$ (because $(a,b)=a, a,b$, but it is not an element of $(b,a)$ (because $aneq b$ and $aneqb,a$ and there are no other elements in $(b,a)$.

share|cite|improve this answer

edited 2 hours ago

David C. Ullrich

62.5k44197

answered 9 hours ago

5xum5xum

94.2k498164

$endgroup$

add a comment | 

4

$begingroup$

You formally define $(a,b)$ to be equal to $a, a,b$.

That's the only formal definition.

The rest is our interpretation. We notice that using the definition above, $(a,b)$ is not equal to $(b,a)$ if $aneq b$ (*), and therefore, in the newly introduced symbol $(.,.)$, order matters. That's why we call this newly introduced symbol an "ordered pair".

In other words, (2) is not a proposition, it is a consequence of (3).

---

(*) Note that the only thing you need to prove $(a,b)neq (b,a)$ is that $aneq b$. This is because, if $aneq b$, then the set $a$ is an element of $(a,b)$ (because $(a,b)=a, a,b$, but it is not an element of $(b,a)$ (because $aneq b$ and $aneqb,a$ and there are no other elements in $(b,a)$.

share|cite|improve this answer

edited 2 hours ago

David C. Ullrich

62.5k44197

answered 9 hours ago

5xum5xum

94.2k498164

$endgroup$

add a comment | 

$begingroup$

You formally define $(a,b)$ to be equal to $a, a,b$.

That's the only formal definition.

The rest is our interpretation. We notice that using the definition above, $(a,b)$ is not equal to $(b,a)$ if $aneq b$ (*), and therefore, in the newly introduced symbol $(.,.)$, order matters. That's why we call this newly introduced symbol an "ordered pair".

In other words, (2) is not a proposition, it is a consequence of (3).

---

(*) Note that the only thing you need to prove $(a,b)neq (b,a)$ is that $aneq b$. This is because, if $aneq b$, then the set $a$ is an element of $(a,b)$ (because $(a,b)=a, a,b$, but it is not an element of $(b,a)$ (because $aneq b$ and $aneqb,a$ and there are no other elements in $(b,a)$.

share|cite|improve this answer

edited 2 hours ago

David C. Ullrich

62.5k44197

answered 9 hours ago

5xum5xum

94.2k498164

$endgroup$

share|cite|improve this answer

edited 2 hours ago

David C. Ullrich

62.5k44197

answered 9 hours ago

5xum5xum

94.2k498164

edited 2 hours ago

David C. Ullrich

62.5k44197

edited 2 hours ago

David C. Ullrich

62.5k44197

answered 9 hours ago

5xum5xum

94.2k498164

answered 9 hours ago

5xum5xum

94.2k498164

2

$begingroup$

An order is a relation between elements, not the elements themselves. Propositions 1 and 2 are fine as they stand.

share|cite|improve this answer

answered 9 hours ago

Yves DaoustYves Daoust

137k877237

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @YvesDaoust.I formulate (3) due to the answer I obtained here: <math.stackexchange.com/questions/3236622/…>
  $endgroup$
  – Eleonore Saint James
  9 hours ago
  
  

  
  
  
  

add a comment | 

2

$begingroup$

An order is a relation between elements, not the elements themselves. Propositions 1 and 2 are fine as they stand.

share|cite|improve this answer

answered 9 hours ago

Yves DaoustYves Daoust

137k877237

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @YvesDaoust.I formulate (3) due to the answer I obtained here: <math.stackexchange.com/questions/3236622/…>
  $endgroup$
  – Eleonore Saint James
  9 hours ago
  
  

  
  
  
  

add a comment | 

$begingroup$

An order is a relation between elements, not the elements themselves. Propositions 1 and 2 are fine as they stand.

share|cite|improve this answer

answered 9 hours ago

Yves DaoustYves Daoust

137k877237

$endgroup$

share|cite|improve this answer

answered 9 hours ago

Yves DaoustYves Daoust

137k877237

answered 9 hours ago

Yves DaoustYves Daoust

137k877237

answered 9 hours ago

Yves DaoustYves Daoust

137k877237

2

$begingroup$

(2) is a definition on "reader-level", and defines a piece of notation we would like to make use of and how to think about its use intuitively. (3) is a formal, "lower-level" definition of that same notaiton that makes sure we haven't really made any new set theory in the process, but rather that we are still (under the hood) using only the regular (unordered) sets we already had.

share|cite|improve this answer

edited 9 hours ago

answered 9 hours ago

ArthurArthur

127k7122212

$endgroup$

add a comment | 

2

$begingroup$

(2) is a definition on "reader-level", and defines a piece of notation we would like to make use of and how to think about its use intuitively. (3) is a formal, "lower-level" definition of that same notaiton that makes sure we haven't really made any new set theory in the process, but rather that we are still (under the hood) using only the regular (unordered) sets we already had.

share|cite|improve this answer

edited 9 hours ago

answered 9 hours ago

ArthurArthur

127k7122212

$endgroup$

add a comment | 

$begingroup$

(2) is a definition on "reader-level", and defines a piece of notation we would like to make use of and how to think about its use intuitively. (3) is a formal, "lower-level" definition of that same notaiton that makes sure we haven't really made any new set theory in the process, but rather that we are still (under the hood) using only the regular (unordered) sets we already had.

share|cite|improve this answer

edited 9 hours ago

answered 9 hours ago

ArthurArthur

127k7122212

$endgroup$

share|cite|improve this answer

edited 9 hours ago

answered 9 hours ago

ArthurArthur

127k7122212

answered 9 hours ago

ArthurArthur

127k7122212

answered 9 hours ago

ArthurArthur

127k7122212

1

$begingroup$

There's no incompatibility to be fixed. In fact $$(a,b)= a , a,b
= a,b ,a = a , b,a
= b,a ,a .$$

share|cite|improve this answer

answered 2 hours ago

David C. UllrichDavid C. Ullrich

62.5k44197

$endgroup$

add a comment | 

1

$begingroup$

There's no incompatibility to be fixed. In fact $$(a,b)= a , a,b
= a,b ,a = a , b,a
= b,a ,a .$$

share|cite|improve this answer

answered 2 hours ago

David C. UllrichDavid C. Ullrich

62.5k44197

$endgroup$

add a comment | 

$begingroup$

There's no incompatibility to be fixed. In fact $$(a,b)= a , a,b
= a,b ,a = a , b,a
= b,a ,a .$$

share|cite|improve this answer

answered 2 hours ago

David C. UllrichDavid C. Ullrich

62.5k44197

$endgroup$

share|cite|improve this answer

answered 2 hours ago

David C. UllrichDavid C. Ullrich

62.5k44197

answered 2 hours ago

David C. UllrichDavid C. Ullrich

62.5k44197

answered 2 hours ago

David C. UllrichDavid C. Ullrich

62.5k44197