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How can an ordered pair be a set assuming there is no order in a set?

About the notation <S,R> for ordered sets or <G,+> for groups. Is this notation absolutely rigorous or is it simply a convenient shortcut?With Choice, is any linearly ordered set well-ordered if no subset has order type $omega^*$?How many totally ordered sets can be constructed from a given finite poset?If there are order preserving injections between two countable sets, will there be an order preserving bijection?“There is no well-ordered uncountable set of real numbers”complete ordered set with least upper bound propertyPartially Ordered Set and Equivalence RelationshipHow shall I understand what the GNU utilities “comm” and “diff” do in terms of ordered sets?Can mathematical induction be applied on any total order set?Can we expand “induction principle” to a partial order $(X, leq)$?Directed set and partially/totally ordered sets.

3

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 May 23 at 8:13

5xum

94.5k498164

asked May 23 at 8:03

Eleonore Saint JamesEleonore Saint James

1,339118

$endgroup$

add a comment | 

3

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 May 23 at 8:13

5xum

94.5k498164

asked May 23 at 8:03

Eleonore Saint JamesEleonore Saint James

1,339118

$endgroup$

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 May 23 at 8:13

5xum

94.5k498164

asked May 23 at 8:03

Eleonore Saint JamesEleonore Saint James

1,339118

$endgroup$

share|cite|improve this question

edited May 23 at 8:13

5xum

94.5k498164

asked May 23 at 8:03

Eleonore Saint JamesEleonore Saint James

1,339118

share|cite|improve this question

edited May 23 at 8:13

5xum

94.5k498164

asked May 23 at 8:03

Eleonore Saint JamesEleonore Saint James

1,339118

edited May 23 at 8:13

5xum

94.5k498164

edited May 23 at 8:13

5xum

94.5k498164

asked May 23 at 8:03

Eleonore Saint JamesEleonore Saint James

1,339118

asked May 23 at 8:03

Eleonore Saint JamesEleonore Saint James

1,339118

4 Answers
4

active

oldest

votes

5

$begingroup$

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 May 23 at 15:41

David C. Ullrich

63k44298

answered May 23 at 8:07

5xum5xum

94.5k498164

$endgroup$

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 May 23 at 8:08

Yves DaoustYves Daoust

138k878237

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @YvesDaoust.I formulate (3) due to the answer I obtained here: <math.stackexchange.com/questions/3236622/…>
  $endgroup$
  – Eleonore Saint James
  May 23 at 8:10
  
  

  
  
  
  

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 May 23 at 8:16

answered May 23 at 8:11

ArthurArthur

128k7122212

$endgroup$

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 May 23 at 15:28

David C. UllrichDavid C. Ullrich

63k44298

$endgroup$

add a comment | 

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5

$begingroup$

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 May 23 at 15:41

David C. Ullrich

63k44298

answered May 23 at 8:07

5xum5xum

94.5k498164

$endgroup$

add a comment | 

5

$begingroup$

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 May 23 at 15:41

David C. Ullrich

63k44298

answered May 23 at 8:07

5xum5xum

94.5k498164

$endgroup$

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 May 23 at 15:41

David C. Ullrich

63k44298

answered May 23 at 8:07

5xum5xum

94.5k498164

$endgroup$

share|cite|improve this answer

edited May 23 at 15:41

David C. Ullrich

63k44298

answered May 23 at 8:07

5xum5xum

94.5k498164

edited May 23 at 15:41

David C. Ullrich

63k44298

edited May 23 at 15:41

David C. Ullrich

63k44298

answered May 23 at 8:07

5xum5xum

94.5k498164

answered May 23 at 8:07

5xum5xum

94.5k498164

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 May 23 at 8:08

Yves DaoustYves Daoust

138k878237

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @YvesDaoust.I formulate (3) due to the answer I obtained here: <math.stackexchange.com/questions/3236622/…>
  $endgroup$
  – Eleonore Saint James
  May 23 at 8:10
  
  

  
  
  
  

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 May 23 at 8:08

Yves DaoustYves Daoust

138k878237

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @YvesDaoust.I formulate (3) due to the answer I obtained here: <math.stackexchange.com/questions/3236622/…>
  $endgroup$
  – Eleonore Saint James
  May 23 at 8:10
  
  

  
  
  
  

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 May 23 at 8:08

Yves DaoustYves Daoust

138k878237

$endgroup$

share|cite|improve this answer

answered May 23 at 8:08

Yves DaoustYves Daoust

138k878237

answered May 23 at 8:08

Yves DaoustYves Daoust

138k878237

answered May 23 at 8:08

Yves DaoustYves Daoust

138k878237

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 May 23 at 8:16

answered May 23 at 8:11

ArthurArthur

128k7122212

$endgroup$

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 May 23 at 8:16

answered May 23 at 8:11

ArthurArthur

128k7122212

$endgroup$

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 May 23 at 8:16

answered May 23 at 8:11

ArthurArthur

128k7122212

$endgroup$

share|cite|improve this answer

edited May 23 at 8:16

answered May 23 at 8:11

ArthurArthur

128k7122212

answered May 23 at 8:11

ArthurArthur

128k7122212

answered May 23 at 8:11

ArthurArthur

128k7122212

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 May 23 at 15:28

David C. UllrichDavid C. Ullrich

63k44298

$endgroup$

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 May 23 at 15:28

David C. UllrichDavid C. Ullrich

63k44298

$endgroup$

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 May 23 at 15:28

David C. UllrichDavid C. Ullrich

63k44298

$endgroup$

share|cite|improve this answer

answered May 23 at 15:28

David C. UllrichDavid C. Ullrich

63k44298

answered May 23 at 15:28

David C. UllrichDavid C. Ullrich

63k44298

answered May 23 at 15:28

David C. UllrichDavid C. Ullrich

63k44298