Proving the pairing axiom from the rest of ZF

Solution 1:

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 $\varnothing$).