Talk:Axiom of pairing

Creation of the set of unordered pairs
The Cartesian product is used to create the set of ordered pairs C from the two sets A B, e.g. C:=A&times;B. What symbol is used to denote the set of unordered pairs created from those two sets ? MathMartin 14:35, 7 Jan 2005 (UTC)
 * That idea is used so seldom that I am not aware of any standard notation for it. JRSpriggs 07:22, 10 June 2006 (UTC)

Axiom of pairing is written in a correct form ?
In the formal language of the Zermelo-Frankel axioms, the axiom reads:

\forall A, \forall B, \exist C, \forall D, D \in C \Leftrightarrow (D = A \or D = B)

or in words:

Given any set A and any set B, there is a set C such that, given any set D, D is a member of C   if and only if D is equal to A or D is equal to B.

What the axiom is really saying is that, given two sets A and B, we can find a set C whose members are precisely A and B. We can use the axiom of extensionality to show that this set C is unique. We call the set C the pair of A and B, and denote it {A,B}. Thus the essence of the axiom is:

Any two sets have a pair.

the above description consider as given the fact that the reader (or a logical expressions processing program = LEPP) read and elaborate at least two times the axiom expression, that because, the first time that the LEPP will read and elaborate the axiom expression, it will produce the fact that A \in C, with the second reading the LEPP will produce the fact that B \in C, and reading and furter elaborating this two new facts the LEPP will produce the expression C = { A, B }. But how the LEPP can consider both true the expressions A \in C and B \in C when the axiom says that (D = A \or D = B), in fact, considering true the axiom as it was expressed, is possible to consider true (A \in C \or B \in C) but not (A \in C \and B \in C). why is not better expressed the axiom with the following expression ?

\exist A, \exist B, \forall C \Leftrightarrow (A \in C \and B \in C)

Maybe the axiom is well expressed and the description is wrong.

Luca Beretta, 18 march 2005, 23.55 CET.


 * I came to this talk page for a similar reason:
 * $$\forall A \, \forall B \, \exist C \, \forall D \, [ D \in C \iff (D = A \lor D = B)]$$
 * Then the article reads:
 * Given two sets, there is a set whose members are exactly the two given sets.
 * But as there is the logical disjunction or, one could also say
 * Given two sets, there is a set whose members are at least one of the two given sets. ... right? --Abdull (talk) 10:00, 5 July 2008 (UTC)


 * Luca, I don't see a problem...Doesn't the universal quantifier on A, B, and D mean that this axiom will produce all sets C that are singletons and all sets C that are pairs? As stated in the article, a singleton is just a special case of a pair. 198.70.193.2 (talk) 22:58, 6 January 2011 (UTC)

Another formula which is logically equivalent to the axiom of pairing is
 * $$\forall A \, \forall B \, \exist C \, ( A \in C \land B \in C \land \forall D \, [ ( D \neq A \land D \neq B) \Rightarrow D \notin C ] ) \,.$$

Some people might find this easier to understand. JRSpriggs (talk) 05:17, 7 January 2011 (UTC)

Does anybody know what this axiom/theorem schema is called?
-- moved this comment from the article to here. The question refers to the following paragraph:


 * Thus, one may use this as an axiom schema in the place of the axioms of empty set and pairing. Normally, however, one uses the axioms of empty set and pairing separately, and then proves this as a theorem schema. Note that adopting this as an axiom schema will not replace the axiom of union, which is still needed for other situations.


 * Oleg Alexandrov 19:40, 21 Mar 2005 (UTC)

Assessment comment
Substituted at 01:47, 5 May 2016 (UTC)