I first heard of the ZF Aziom of Pairing watching this. I don't get why it is necessary to have an axiom which states that a set exists. Doesn't a set exist simply by virtue of the fact that it has a definition?

Given that $x$ and $y$ exist, why can I not claim that $S=\{x,y\}$ exists without this axiom?

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    $\begingroup$ Nope. $\endgroup$
    – J.G.
    Mar 5, 2022 at 19:13
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    $\begingroup$ The question is reversed: in mathematics, the burden of proof is on the mathematician, it's up to you to prove that something is possible, things are not "possible until proven otherwise". $\endgroup$ Mar 5, 2022 at 19:16
  • $\begingroup$ Are you looking for a model of ZFC - pairing such that $\{x,y\}$ is not a set? $\endgroup$
    – Couchy
    Mar 5, 2022 at 19:54
  • $\begingroup$ See also Proving the pairing axiom from the rest of ZF. $\endgroup$
    – Couchy
    Mar 5, 2022 at 19:59
  • $\begingroup$ What's the definition of a set? I don't know one. $\endgroup$
    – user
    Mar 6, 2022 at 9:10

3 Answers 3


I think that your view of set theory is not formal enough. Anyone will happily introduce set theory to a newcomer by describing sets as "collections of stuff". Then one might say that the formula $x\in A$ means that "the object $x$ is a member of the collection $A$", and that $\{a,b,c\}$ is a collection that contains exactly the objects $a$, $b$ and $c$. But this is just psychological.

Formally speaking, set theory deals with completely opaque objects, which we choose to call "sets", and which can be related by some binary relation $\in$, which we choose to call "membership". But just as Hilbert famously suggested to rename points, lines and planes in geometry as chairs, tables and beer glasses, we might rename "sets" as "giraffes" and "membership" as "bumbleness" (so a giraffe $y$ might bumble another giraffe $x$, which we will denote as $x\in y$).

Then Giraffe Theory explains how the bumbleness relation behaves and relates giraffes. We might ask: given two giraffes $x$ and $y$, is there a giraffe $A$ such that the complete list of giraffes which $A$ bumbles consists of exactly $x$ and $y$? Well, a priori there is no reason why that would be the case. I don't even really know what "bumble" means!

And then someone comes along and says: "Well of course there is one, I denote it by $\{x, y\}$! It exists because I can write that symbol." I would ask "But what does $\{x,y\}$ mean?" And that person would answer "It is precisely the giraffe which bumbles $x$ and $y$, and nothing else. By definition." And the obvious reaction would be "The fact that you can write this symbol and declare that it represents such a giraffe does not mean that this giraffe actually exists."

The point of this little charade is to make you realize that your mistake is to think that the set $\{x,y\}$ "obviously" exists, because you know what a set is supposed to be and how sets are supposed to behave. But the point of set theory is precisely to formalize that, so we may not assume anything. The reason why the pairing axiom is necessary is that otherwise we can't prove that there is such a pair set, using the other axioms. You may view the pairing axiom as the formal justification for which you are allowed to write the symbol $\{x,y\}$ (and more generally $\{x_1,\dots,x_n\}$ for a finite number of objects) and declare that this is a well-defined set which contains exactly $x$ and $y$.

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    $\begingroup$ This is by far the best answer and the only one that really addresses the actual question here. No idea why it hasn't received any other upvotes... $\endgroup$ Mar 6, 2022 at 4:53
  • $\begingroup$ @CaptainLama why it is axiom of pairing? It seems a bit too specific. Why couldn't there be a generic axiom of grouping, which simply, given n giraffes, states that a giraffe A exists, and list of giraffes that A bumbles contains exactly those initial n giraffes. $\endgroup$ Aug 1, 2022 at 11:40
  • $\begingroup$ @CaptainLama, or the other way, if it was intended to be specific, why not an axiom of singleton, which along with axiom of union, can imply axiom of pairing! $\endgroup$ Aug 1, 2022 at 11:43
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    $\begingroup$ @SouravKannanthaB For your idea of grouping, the thing is that you would need a separate axiom for each $n$, because you can't quantify on the number of variables in a statement. So that would require an infinite scheme of axioms, and the axiom of pairing would be the case $n=2$. But this axiom of pairing is already enough to prove all the other $n$, so they would just be unnecessary. $\endgroup$ Aug 1, 2022 at 16:17
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    $\begingroup$ @SouravKannanthaB For your idea of singleton, I think you are mistaken about the axiom of union. The axiom of union + an axiom of singleton would not be enough to prove the axiom of pairing. That is because the axiom of union does not say "given $A$ and $B$ there exists $A\cup B$". It rather says "given $\{A,B\}$ there exists $\bigcup \{A,B\}$" (which is usually denoted as $A\cup B$). But to do that you need to form the set $\{A,B\}$, therefore you need pairing. $\endgroup$ Aug 1, 2022 at 16:20

Without being able to prove in ZFC that the unordered pair $\{x,y\}$ exists whenever $x, y$ are sets, all you have is a notation, $\{x,y\}$, for that unique set which, if it exists, contains just $x$ and $y$. The notion that "a set exist[s] simply by virtue of the fact that it has a definition" is codified by the (false, inconsistent) "naive Comprehension" principle, which claims that for any formula $\psi(x)$, the collection $\{x\,|\,\psi(x)\}$ is a set. Taking $\psi(x)$ to be $x\notin x$ gives the Russell paradox.

Pairing is redundant in the presence of other axioms. Given the Axiom of Infinity (in its usual formulation) and Separation, you can show that the set $2 = \{\emptyset, \{\emptyset\}\} = \{0, 1\}$ exists. Similarly, If you have existence of $\emptyset$ and the Powerset axiom, again $2 = \mathcal{P}(\mathcal{P}(\emptyset))$ is provably a set.

Now, given any sets $a, b$, the class function $F$ given by $0 \mapsto a, 1 \mapsto b$ is definable — for example, by this formula with parameters $a, b$:

$$\varphi(x, y) =_{def} (x=0\land y=a)\lor(x=1\land y=b).$$

Because $2$ as above is a set, by Replacement its image under $F$ is a set. That image is $F[2] = \{y\,|\, (\exists x\in 2)\, \varphi(x, y)\} = \{a, b\}$.

Pairing exists as a separate axiom because we want to be able to study the systems that result from dropping other ZFC axioms such as Infinity, Replacement, or Powerset. The remaining core of axioms should characterize basic aspects of "sethood" that don't involve those notions. Pairing is a basic requirement we have for universes of sets — whatever else a model of set theory does or doesn't contain, surely it ought to be closed under pairing.

Note that Separation can be proved from Replacement and the other ZFC axioms, but it's retained in most presentations as a separate axiom schema, for pedagogical reasons — it provides an opportunity to discuss naive Comprehension and the resulting inconsistencies — and because ZC, ZFC minus Replacement, came first and remains a thing.

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    $\begingroup$ This is a good answer to a question that looks similar to the one asked here but is actually totally different. OP's question is why just writing down the notation "$\{x,y\}$" is not sufficient to prove the set exists. $\endgroup$ Mar 6, 2022 at 4:56
  • $\begingroup$ @EricWofsey I added a bit to address OP's question directtly. $\endgroup$
    – BrianO
    Mar 6, 2022 at 5:15

"Doesn’t a set exist by virtue of it’s definition"
No, essentially you are saying that if you have a property $P$, then there should exist a set $X$ whose elements are exactly those which satisfy $P$.

Now this may seem like a very natural axiom, however it’s inconsistent(using it you can prove absurd statements like $0=1$!).

One can see it’s inconsistency by letting $P(x)=\neg(x\in x)$ and forming the “set of all sets which do not contain themselves” which leads to the famous Russel Paradox.
See https://en.wikipedia.org/wiki/Russell%27s_paradox


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