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:

  1. Extensionality
  2. Foundation
  3. Unordered Pair
  4. Empty Set
  5. Infinity
  6. Union
  7. Power Set
  8. Separation
  • $\begingroup$ Which axioms do you have in your Z? This is one of the reasons people will often include redundant axioms in ZF, since it's easier to talk about subtheories this way. $\endgroup$
    – Asaf Karagila
    Oct 24, 2022 at 15:09
  • $\begingroup$ The axioms of $\mathsf{Z}$ are the same as defined in your notes: (0) Extensionality • (1) Foundation • (2) Unordered Pair • (3) Empty Set • (4) Infinity • (5) Union • (6) Power Set • (7) Separation $\endgroup$
    – Roy Sht
    Oct 24, 2022 at 16:31
  • $\begingroup$ I mean, sure, but my notes don't include 2 and 3. $\endgroup$
    – Asaf Karagila
    Oct 24, 2022 at 19:54

1 Answer 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.

clearly $\overline a,\{\overline a,\overline b\}\notin N_{\overline a}$ and $\overline b,\{\overline a,\overline b\}\notin N_{\overline b}$.

Assume $\overline b\notin N_{\overline a}$, in particular $b∈\cl{\overline a}$, because $|b|>2$ this implies $b=a$ (which we assumed it is not), or $b\in \operatorname{cl}(a)$ (which is impossible as they have the same rank), so $\overline b\in N_{\overline a}$, symmetric argument will show that $\overline a\in N_{\overline b}$, so $\overline a,\overline b\in N_{\overline a}∪N_{\overline b}∌\{\overline a,\overline b\}$

  • $\begingroup$ Thanks! Do you know where to find any references for this (or similar) construction in the literature? $\endgroup$
    – Roy Sht
    Nov 5, 2022 at 22:11
  • 1
    $\begingroup$ Here is a paper from Andreas about similar points (although Andreas doesn't use this construction to show that Z-pairing+¬pairing is consistent). Note that in this mention Andreas defines the models $N_X$ as proper classes, inwhere I took some "ground model" $V_{ω+ω}$, the advantage of my construction is that you get a model, and not a class model. The advantage of Andreas construction is that you don't need the full power of ZF in the universe (which I used when I assumed $V_{ω+ω}$ exists), only Z $\endgroup$
    – ℋolo
    Nov 5, 2022 at 22:55

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .