12

In ZF, the pairing axiom states that for every $x,y$ there exists the set $\{x, y\}$. Wikipedia also tells us we can dispense this axiom:

This axiom is part of Z, but is redundant in ZF because it follows from the axiom schema of replacement applied to any two-member set. The existence of such a set is assured by either the axiom of infinity, or by the axiom of the power set applied twice to the empty set.

I have a few questions about it:

  1. It seems to me we only need to prove that given $x$, there exists the set $\{x\}$ and use the axiom of union for $\{x\}, \{y\}$. Am I correct?
  2. To prove $\{x\}$ exists we can apply separation on $P(x)$ - the power set of $x$ whose existence is guaranteed from the axiom of power set. Am I correct?
  3. If both the previous statements were correct, is it correct to say that the axiom of pairing can be proved without the need to assume the existence of any set (i.e. without the axiom of infinity or the axiom of the empty set).
  4. How can the axiom of infinity be stated without implicitly relying on the axiom of pairing? After all, the formulation I know of for this axiom literally says "there is an inductive set", and to formulate what "inductive set" means you have to use the expression $x\cup\{x\}$ which assumes $\{x\}$ is really an existing set...
Gadi A
  • 19,265

1 Answers1

22

You have $x,y$, let us construct the pair $\{x,y\}$.

First note that $\varnothing=\{z\in x\mid z\neq z\}$. So we have the empty set. Now by the power set axiom we have $P(\varnothing)=\{\varnothing\}$ and $P(P(\varnothing))=\{\varnothing,\{\varnothing\}\}$.

Now let us define a formula (with parameters $x,y$):

$$\varphi(u,v,x,y)\colon= (u=\varnothing\land v=x)\lor(u=\{\varnothing\}\land v=y\})$$

(Note that $\{\varnothing\}$ can be defined explicitly as the set that all its elements are the empty sets)

Using replacement now, we set the parameters $x,y$ now the axiom says that $\{u\mid\exists v\in P(P(\varnothing))\colon\varphi(v,u,x,y)\}$ exists. But this set is exactly $\{x,y\}$.

  1. You cannot use the axiom of union to prove from the existence of $\{x\}$ and $\{y\}$ the existence of the set $\{x,y\}$. The axiom of union says that if $A$ is a set then $\bigcup A$ is a set. However you want to say that $\{\{x\},\{y\}\}$ is a set therefore its union, which is $\{x,y\}$ is a set. You already assume the existence of a pair.

  2. You can indeed use separation to prove the existence of $\{x\}$ using the power set axiom as well.

  3. To use a power set, or separation argument you have already the existence of some set. Note that the power set axiom says that if $x$ is a set then there exists a set which contains all the subsets of $x$. Separation is the same, you assume the existence of a set. If you wish to use these two, adding an assumption that an empty set exists is meaningless (note that empty sets exists due to separation, so using it for $\{x\}$ is the same as using it for $\varnothing$).

Asaf Karagila
  • 393,674
  • You've already defined ${\emptyset}$--it is simply $P(\emptyset)$. – Cameron Buie May 06 '12 at 08:11
  • @CameronBuie: There is no explicit operation in the language for the power set operation. Using the power set axiom I merely prove the existence of this set. – Asaf Karagila May 06 '12 at 08:31
  • @AsafKaragila, thanks but this doesn't answer any of my questions. You simply repeat the usual proof and implicitly rely on the existence of a set (otherwise you can't prove the existence of the empty set). – Gadi A May 06 '12 at 08:33
  • @Gadi: But if ZF is inconsistent, why would it have a model, why would it be able to prove the existence of any sets in a "consistent" way? – Asaf Karagila May 06 '12 at 08:38
  • I fail to see how this is relevant to my question. Anyway, I am interested in answers to my questions 1-4, not in seeing the standard proof of the pairing axiom repeated... – Gadi A May 06 '12 at 08:50
  • @Gadi: I tried to answer the first three, I will have to answer the fourth a bit later. – Asaf Karagila May 06 '12 at 09:07
  • 2
    Your answer to 1 is exactly what I needed, thanks. – Gadi A May 06 '12 at 09:11
  • I do not understand the formula, is it a function or predicate? – LiziPizi Mar 12 '16 at 10:37
  • @LiziPizi: The formula essentially defines the function $F(\varnothing)=x$ and $F({\varnothing})=y$. – Asaf Karagila Mar 12 '16 at 10:48
  • @asafkaragila but how those functions help you? – LiziPizi Mar 12 '16 at 11:00
  • @LiziPizi: The Axiom of Replacement helps me. As explained in the answer. – Asaf Karagila Mar 12 '16 at 11:01
  • @asafkaragila correct me if I am wrong but what replacement axiom just says it is that you can create a some kind of helper function but I can't see how it is used here – LiziPizi Mar 12 '16 at 11:05
  • @LiziPizi: You're correct, this is the helper function and the power set of the power set of the empty set is how you bootstrap. It's all in the answer. – Asaf Karagila Mar 12 '16 at 11:20
  • @asafkaragila I think my main issue here is understanding the formula, I don't understand why 4 parameters and what is each one and what is the output – LiziPizi Mar 12 '16 at 11:26
  • @Lizi: This becomes too much for the comments. Ask this in your question, rather than just saying that you don't understand what's going on. Then you might receive a satisfactory explanation. – Asaf Karagila Mar 12 '16 at 11:50
  • Hey Asaf, this issue has recurred again in a recent question. It seems to me you are proving the Schema of Pairing. Can you prove the axiom with universal quantifiers from replacement ? I dont see how to do that. – Rene Schipperus Jan 31 '18 at 00:15
  • In the last line shouldn't it be {∣∃∈((∅)):(,,,)} instead? (Swapped u and v) – Mattia F. Apr 24 '19 at 19:34
  • @MattiaF.: Yes, that's correct. – Asaf Karagila Apr 24 '19 at 19:40