Talk:Unordered Pairs Exist

What I'm puzzled about is that we have an axiom that states that ordered pairs exist, and now we have a theorem that states that ordered pairs exist which just states what the axiom says. Why do we need that second theorem?

If a) the axiom states their existence in the realm of sets, and a class is something which contains sets, then it should not necessarily mean that we need a separate theorem to state their existence in the realm of classes. What obvious fact am I missing?

Oh, and see whether you can learn to use the "Eqn" template, then that will be less work for me to do. :-) --prime mover 00:28, 15 September 2011 (CDT)


 * The fact that the Kuratowski Ordered Pair exists is simply a special case of the fact that Unordered Pairs exist (here). Really, this is a restatement of the Axiom of Pairing in a more usable format.  About "Eqn" I'll see what I can do... -Andrew Salmon 00:31, 15 September 2011 (CDT)
 * I'm not talking about the Kuratowski ordered pair, I was specifically referring to the Axiom of Pairing. Why is it less usable than this entry? If you specifically need to refer in another proof to the existence of an unordered pair, then surely you can just link back to the Axiom of Pairing? Again, what am I missing here? --prime mover 00:36, 15 September 2011 (CDT)
 * Axioms of ZFC only use primitive terms. This one uses defined terms, and thus is more usable in that it can immediately be applied via direct substitution to situations that use defined terms. Andrew Salmon 00:43, 15 September 2011 (CDT)
 * Well okay, then ... but can you make sure define the terms you use in all of these proofs? What are $A$ and $B$, are they classes or sets, and what's the context? Also make sure you put a definition for $U$. --prime mover 01:23, 15 September 2011 (CDT)