Is Zermelo's axiom system, $\mathsf{Z}$ (which is, $\mathsf{ZF}-\text{replacement}$), consistent with $\neg \text{pairing}$?
I know that $(\mathsf{ZF}-\text{pairing})\vdash \text{pairing}$, by replacing $\mathcal{P(P(\emptyset))}=\{\emptyset,\{\emptyset\}\}$ with any given two sets $a,b$, as described here. But is it easy to find a model of $(\mathsf{Z}-\text{pairing})+\neg\text{pairing}$, assuming $\mathsf{Z}$ is consistant?
Edit: For clarification, I assume $\mathsf{Z}$ includes the following axioms:
- Extensionality
- Foundation
- Unordered Pair
- Empty Set
- Infinity
- Union
- Power Set
- Separation
Best Answer
$\newcommand{\cl}[1]{\operatorname{cl}(#1)}$
One possible construction (I believe it is by Andreas):
A set $X$ is called "avoidable" if it satisfy the weak pairing property: $∀x,y\in X\exists z\in X\;(\{x,y\}⊆z)$
Start with $N=V_{ω+ω}$, and an avoidable $X\in N$,
Define $N_X=\{a\in N\mid X\nsubseteq \cl{a}\}$, you can verify that $N_X$ is a model of $Z$ minus infinity (the only tricky part is pairing), moreover, if $X\nsubseteq ω$ then it is a model of $Z$.
Now take two different elements, $a,b∈N$ such that $|a|,|b|>2$ and $\operatorname{rank}(a)=\operatorname{rank}(b)>ω$.
Now we need to make $a,b$ into avoidable sets, define $\overline a$ to be the closure of $\{a\}$ under pairing, and similarly for $\{b\}$.
From that we can get the structure $N_{\overline a}∪N_{\overline b}$ of Z-pairing+¬pairing
Again the only tricky part is to show it doesn't satisfy pairing.