Axiom of Pairing states that if $a,b$ are sets, $\exists$ a set $A$ such that $A=\{{a,b\}}$. My question is that why we can't use Axiom of specification to define $A=\{x|x=a \vee x=b\}$?


Its called the axiom of separation and it only allows you to define a subset of another set. An axiom such as $\{x|\varphi(x)\}$ exists leads to contradictions.

  • $\begingroup$ I see. But why we can't revise the axiom without the restriction of subset $\endgroup$ – 89085731 Jun 23 '14 at 12:24
  • $\begingroup$ Without the subset you could have the set of all sets and other stuff that causes trouble. $\endgroup$ – Rene Schipperus Jun 23 '14 at 12:27
  • $\begingroup$ can u give me some example $\endgroup$ – 89085731 Jun 23 '14 at 12:41
  • 5
    $\begingroup$ The Russell paradox $y=\{x|x\not \in x\}$. Is $y \in y$ ? $\endgroup$ – Rene Schipperus Jun 23 '14 at 12:44

The axiom of pairing is not a consequence of the axiom of specification, as Rene Schipperus pointed out.

However, it is a consequence of the axioms of replacement, power set and empty set. Because now you can prove there exists a set of two elements $\{\varnothing,\{\varnothing\}\}$, and using replacement you can now generate any other pair:

$$\varphi(x,y,p_1,p_2)= (x=\varnothing\land y=p_1)\lor(x=\{\varnothing\}\land y=p_2)$$

(And I have used here some shortcuts in the form of $\varnothing$ and $\{\varnothing\}$, otherwise this would be very very long.)

  • $\begingroup$ Ingenious. Or we could use Infinity plus Replacement. $\endgroup$ – hmakholm left over Monica Jun 23 '14 at 14:08
  • $\begingroup$ Yes, anything which proves that there are two sets would suffice. This is how I learned about pairing (I had a teacher who did his best to use as little axioms as possible, because the course was mainly about constructing models of the theory, so we would have less to verify). $\endgroup$ – Asaf Karagila Jun 23 '14 at 14:10
  • $\begingroup$ I thought in reality the axiom of empty set is not needed in ZFC to prove an empty set exists? Also, it turns out that the axiom of pairing is not needed also? The axiom schema of replacement seems to settle a lot of things... $\endgroup$ – user132181 Jun 23 '14 at 15:24
  • $\begingroup$ @user132181: Yes, you can prove the existence of the empty set from other axioms. But if you are trying to prove a particular theorem, it might be provable directly from "less axioms" like that. Of course if you know that the empty set exists, then you don't need the axiom for it, instead you can use whatever it was that you used to prove its existence. $\endgroup$ – Asaf Karagila Jun 23 '14 at 16:02

User wrote:

My question is that why we can't use Axiom of specification to define $A=\{x|x=a \vee x=b\}$?

To apply Specification to construct $A=\{a,b\}$, $a$ and $b$ would have to have been assumed or proven to be elements of some other set $B$, i.e. $a\in B$ and $b\in B$. Then $A=\{x|x\in B \land [x=a \vee x=b]\}$. Pairing has no such requirement.


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