At first sight there is no abstract (= structural) definition of "product" in set theory. E.g. the Cartesian product of sets $A$ and $B$ is defined as the set of all ordered pairs $(x,y)$, $x \in A$, $y \in B$, and thus depends on the definition of "ordered pair" which is notoriously arbitrary.

I wonder if the following can count as an abstract (= structural) definition of "product" in the context of set theory.

Consider a set $S$ with two equivalence relations $\sim_1$ and $\sim_2$.

Definition: $(S,\sim_1,\sim_2)$ is a product iff $$(\forall x \in S)(\forall y \in S)(\exists ! z\in S) x \sim_1 z \wedge y \sim_2 z$$ $$(\forall z\in S)(\exists ! x\in S)(\exists ! y\in S) x \sim_1 z \wedge y \sim_2 z$$

If $(S,\sim_1,\sim_2)$ is a product

  • $S$ can be understood as $S/_{\sim_1} \times S/_{\sim_2}$

  • the relations $\sim_i$ can be read as has the same $i$-th component

  • the canonical projection map $\pi_i (x) = [x]_{\sim_i}$ can be understood as the $i$-th component

Question: Isn't this definition somehow on par - concerning structuralness - with the definition of category theory? If so, why is it so rarely found, or rather: where can I find it (in which textbook, e.g.)?

Considering the product of a set with itself, i.e. $S = X \times X$, one relation $\sim$ does suffice, which does not have to be an equivalence relation, not even symmetric, but from which two equivalence relations can be defined:

$$ x \sim_1 y :\equiv (\exists z) x \sim z \wedge y \sim z $$ $$ x \sim_2 y :\equiv (\exists z) z \sim x \wedge z \sim y $$

If $(S,\sim_1,\sim_2)$ is a product the relation $x \sim y$ can be read as the first component of $x$ equals the second component of $y$.

Question: Are there conditions on a relation $\sim$ such that $\sim_1$, $\sim_2$ as defined above make $(S,\sim_1,\sim_2)$ automatically a product?

  • $\begingroup$ In your definition of a product, did you mean $\forall x \in S \forall y \in S \exists! z \in S \ldots$ instead of the unbounded quantification $\forall x \forall y \exists! z \ldots$? $\endgroup$ – Andrej Bauer Oct 9 '11 at 7:58
  • $\begingroup$ The definition of a product should involve three sets, not just one. If I tell you that $P$ is a product of two sets, you cannot always recover from $P$ the two sets (think of the case when one of the sets is empty). Your definition is unlikely going to work when one of the components is an empty set. How would we get $\emptyset \times \mathbb{N}$, for example? If you take $S = \emptyset$, as you presumably should, then you cannot recover $\mathbb{N}$. $\endgroup$ – Andrej Bauer Oct 9 '11 at 8:01
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    $\begingroup$ I would regard the `definition' of ordered pair in ZFC (or similarly) merely as a proof of existence of ordered pairs/products. Just as there are many constructions of the natural numbers, integers, rationals, real and complex numbers. Really, what matters is the relations between the objects. $\endgroup$ – George Lowther Oct 9 '11 at 9:52
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    $\begingroup$ The product of sets, together with its projection maps to the factors, could be characterized by a suitable universal mapping property. Then that construction is unique up to suitable isomorphism as all objects satisfying universal properties are. This can let you relax a little about the fact that there may be more than one way to make the construction of a product of sets (and its projections to the factors). $\endgroup$ – KConrad Oct 9 '11 at 13:08
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    $\begingroup$ @Hans Stricker: Right, so I partly agree with you there. Except, rather than adding a pair of relations, I think adding a 'Pair' keyword is neater. This is a 2-ary function symbol satisfying the axiom $$\forall x_1,x_2,y_1,y_2\;\left(({\rm Pair}(x_1,x_2)={\rm Pair}(y_1,y_2))\rightarrow(x_1=y_1 \wedge x_2=y_2)\right).$$ Or, add another pair $\pi_1,\pi_2$ of 1-ary functions satisfying $$\forall x,y\;\left(\pi_1{\rm Pair}(x,y)=x\wedge\pi_2{\rm Pair}(x,y)=y\right).$$ In any case, I don't think that ZFC style set theory on its own is really designed to be a handy framework for actually doing maths. $\endgroup$ – George Lowther Oct 9 '11 at 19:30

How do you define an equivalence relation in ZF without refering to products or ordered pairs? In case you want to define products via the usual universal property: There you have to use maps, which probably also cannot defined without ordered pairs in ZF.

Anyway, if you are interested in a category theoretic foundation of set theory and therefore of all mathematics, you might be interested in Lawvere's Elementary Theory of the Category of Sets.

  • $\begingroup$ @Martin: Is this true: do I really need ordered pairs to define an equivalence relation, don't unordered pairs suffice? $\endgroup$ – Hans-Peter Stricker Oct 10 '11 at 15:51
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    $\begingroup$ Please write down your definition of an equivalence relation which does not use ordered pairs. $\endgroup$ – Martin Brandenburg Oct 10 '11 at 18:02
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    $\begingroup$ @Martin - Because of symmetry, you should actually be able to: Define an equivalence relation on $S$ as a subset $R$ of the collection of subsets of $S$ whose cardinality is at most 2 satisfying a) For all $x \in S$, $\lbrace x \rbrace \in R$. b) For all $x, y, z \in S$, if $\lbrace x, y \rbrace \in R$ and $\lbrace y, z \rbrace \in R$, then $\lbrace x, z \rbrace \in R$. This should do the trick. $\endgroup$ – Simon Rose Oct 10 '11 at 20:12
  • $\begingroup$ That's what I had in mind. Is there a flaw? $\endgroup$ – Hans-Peter Stricker Oct 10 '11 at 22:42
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    $\begingroup$ Another natural way is to represent an equivalence relation by the partition it induces (i.e., an equivalence relation on $S$ is a set of nonempty disjoint sets whose union is $S$). $\endgroup$ – Emil Jeřábek Oct 27 '11 at 17:30

From your first part of definition from a couple $x,\ y$ of elements of $S$, there is (uniquely) associated an element $z$ of $S$ still, then essentially we have that $ S \times S \subset S $ (exist a subset of $S$ in bijection by the usual $S\times S$) and from second part this inclusion is a equality. Then your definitions dont work for finite sets for example.


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