I read that the definition of an ordered pair $(a,b)$ is $(a,b)=\{\{a\},\{a,b\}\}$. This definition is based on the concept of unordered pair, which, in turn, is based on the Axiom of pair:

For every two sets $A$ and $B$ there exists a set $C$ such that $x\in C$ iff $x=A$ or $x=B$.

As we see, this axiom declares the existence of a set $\{A,B\}$ where $A$ and $B$ are both sets. This is what we call an unordered pair. Then considering $\{A\}:=\{A,A\}$ we define the ordered pair as $(A,B):=\{\{A\},\{A,B\}\}$. Thus, we have defined an ordered pair of sets. But, how can we define or work with ordered pairs of objets that are not necessarily sets? In the books I am reading this is not specified, and I don´t figure it out yet. Any help is welcome.

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    $\begingroup$ In Set Theory, there are no “objets that are not necessarily sets”. $\endgroup$ – José Carlos Santos Oct 17 at 18:30
  • $\begingroup$ @JoséCarlosSantos That is what I was thinking, but in the books when the topic of Relations is treated, after the topic of ordered pairs, the examples don't use ordered pais of sets, but ordered pairs of number, for example. That is what confuses me. $\endgroup$ – Joshua Oct 17 at 18:47
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    $\begingroup$ In set theory, it's routine to consider numbers as themselves sets. You start by building up the natural numbers and defining $+$ on the natural numbers. Then use equivalence classes of ordered pairs of natural numbers to represent the integers {$(a, b) \sim (x, y) \iff a+y=b+x$, so $(a, b)$ represents $a-b$). Define the usual ordering relation. Then use equivalence classes of ordered pairs of integers to represent the rationals. Define the usual ordering relation on the rational numbers. Then use Dedekind cuts to define the real numbers. Each of these steps gives you a set. $\endgroup$ – Robert Shore Oct 17 at 19:05

In principle, if we assume a set theoretic foundation, then all objects are sets. The number $\pi$ can be interpreted as a particular set (which one depends on how you encode it), or so on. But things exist if and only they are sets.

One can weaken this foundation and allow atoms, or urelements, which are objects that are not sets. We can then decide that $\Bbb R$ and all the "other atomic ideas" are just these atoms. But that doesn't matter for the definition of an ordered pair, since any object will still be an element of a set (yes, we need to slightly modify our axioms of set theory, but not by much).

And if any object can be an element of a set, then $\{\{A\},\{A,B\}\}$ makes sense even when $A$ and $B$ are not themselves sets.

Finally, some set theories include also proper classes. This is sometimes used naively, since it allows us to think about "the collection of all sets" as a bona fide mathematical object. But then sets (or atoms) have the distinguished property that they are exactly the objects which are elements of some class. So a proper class is not an element of anything, so $\{A\}$ does not exist when $A$ is a proper class. And indeed, $(A,B)$ doesn't formally exist in that case. We can use some tricks, sometimes, to still define ordered pairs with classes, but not by the Kuratowski encoding.

Luckily for us, in the absolute majority of cases, even if we consider proper classes, we don't really need those as elements of ordered pairs.

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  • $\begingroup$ So when we define binary relations as sets of ordered pairs, we are still considering sets of ordered pairs of sets? After that, the same with funtions and so on? $\endgroup$ – Joshua Oct 17 at 19:04
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    $\begingroup$ Yes. But the point is that once we have a way to interpret ordered pairs as sets (and indeed we have many ways to do it), none of it matters anymore. $A\times B$ is just the collection of objects satisfying the property of being an ordered pair, $(a,b)$ with $a\in A$ and $b\in B$. Think of it as a template. $\endgroup$ – Asaf Karagila Oct 17 at 19:05

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