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  1. An operation is a kind of function.

  2. A function is a kind of relation.

  3. A relation is a subset of a Cartesian product.

  4. A Cartesian product is an operation.

  5. Back to 1.

It seems to me that there's something wrong. Can we explain $X$ in terms of $Y$, while $Y$ needs $X$ in order to be explained?

Daniel W. Farlow
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    A Cartesian product is an operation? Not exactly. One can think of it as a sort of "class operation" on the class of all sets, but typically, one doesn't. – Cameron Buie Apr 05 '15 at 16:03
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    It would be better to think of a relation as a set of ordered pairs. One can then define a Cartesian product as a special kind of relation. – Cameron Buie Apr 05 '15 at 16:07
  • @CameronBuie: How can you talk about ordered pairs without knowing what a Cartesian product is already? And of course, a Cartesian product is a special kind of relation. – Bey Apr 05 '15 at 16:21
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    @Bey: Because we know, as human beings who shared some experience with others, what are the properties of ordered pairs, then we give a definition in the universe of set theory of sets which have a certain property called Blorgordibupks. Then we prove that these Blorgordibupks have all the properties we wanted from ordered pairs, so we say "Why not... we can just define ordered pairs to be Blorgordibupks!" and we do it, then we have a concept of ordered pairs, internalized to the universe of set theory, and with that we can do the rest. – Asaf Karagila Apr 05 '15 at 16:24
  • Perhaps I have missed your point, but I think ultimately my question is this: how would we rigorously define "ordered pair" without recreating the concept of "Cartesian product"?

    And I still am not sure I follow @CameronBuie. For example, consider the set of ordered pairs $(n,m)$, where $n,m\in \mathbb{N}$. When you define the Cartesian product, how do you know I'm not thinking of these ordered pairs as a relation in $\mathbb{R}\times\mathbb{R}$?

     – Bey Apr 05 '15 at 16:42
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    Ah! It seems, Bey, that you've only seen ordered pairs defined as the elements of a Cartesian product. As @Asaf points out, we need only define some object with the properties desired in ordered pairs, using our axioms, then call them ordered pairs. Among the most standard ways is to define $$(x,y):=\bigl{{x},{x,y}\bigr},$$ but there are a number of ways to go about it. Ultimately, the formal definition is less important than proving that something exists meeting our standards for ordered pairs, so ordered pairs exist, in some form. – Cameron Buie Apr 06 '15 at 03:44

2 Answers2

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The Cartesian product between two sets $A,B$, noted $A \times B$ is defined as the set $$A \times B = \left \{ (x,y) : x \in A \wedge y \in B \right \}$$

A relation $R$ is a subset of a cartesian product: $$R \subseteq A \times B$$

A function $f$ is a triplet $f=(F,A,B)$, where $A,B$ are sets ($A$ is called the domain of $f$, $B$ the codomain) and $F$ is a relation $F \subseteq A \times B$ with the additional properties:

$$(x,y)\in F \wedge (x,z) \in F \Rightarrow y=z$$

$$\forall x \in A \exists y \in B \ \text{such that} \ (x,y)\in F$$

The first is the usual property of functions and the second means, in layman's terms, that "$f$ is defined for every element of $A$".

We note this by saying that $f: A \to B$.

Finally, given a non-empty set $A$, a binary operation $\ast$ on $A$ is a function $$\ast: A \times A \to A$$

By convention, the image $\ast(x,y)$ is usually denoted by $x \ast y$

Aleksei Shchekotikhin
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Reveillark
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    You have defined a binary operation on $A$. – Bey Apr 05 '15 at 16:20
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    In many places a function is just identified with its graph, what you denoted by $F$. – Asaf Karagila Apr 05 '15 at 16:21
  • Also, the formatting in your second property of a function could use some work. In particular, there should at least be something between $B$ and $(x,y)$; perhaps "with" or "such that." – Bey Apr 05 '15 at 16:23
  • @AsafKaragila I was taught this definition on the premise that changing the domain ($A$) or codomain ($B$) gives a different function, and that say, $f(x)=x^2$ was not an injection nor a surjection from $\mathbb{R} \to \mathbb{R}$, but if was from $(0,\infty) \to (0,\infty)$, but this information would also be contained in $F$. Is there an advange, besides simplicity, to either definition? – Reveillark Apr 05 '15 at 16:32
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    @Bey Whoops, you're right on both cases. I tried to edit accordingly. – Reveillark Apr 05 '15 at 16:32
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    There are set theoretic advantages to this approach, for example when you want to talk about arbitrary functions, without having to worry about specifying a common domain, or codomain. And sometimes you can't do that at all, in the sense that you have a proper class of functions and no set which is a valid codomain for all of them. – Asaf Karagila Apr 05 '15 at 16:34
  • Regarding the definition of function as "$f = (F, A, B)$" with "$F$ is a relation $F \subset A x B$", I noticed you used "subset" instead of "strict subset". Is there ever a case where $F = A x B$? I can only think of very limited cases where $F = A x B$ while still satisfying the the "additional properties", namely, $A = \emptyset$ or $B$ is a singleton set. Are there any other cases? – joseville Nov 12 '21 at 01:35
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In a certain sense, that's all right. But as Rota said, be careful not to confuse the medicine with the food!

Aside from relation vs. function (and Cartrsian Product is kind of the black sheep here), most of the differences in terminology are to provide context clues.

Yes, operations are a kind of function, but typically they're meant to portray a very fundamental kind of function. For example, there are uncountable many functions of two variables from $\Bbb R \times \Bbb R \to \Bbb R$, but we typically reserve operation to denote those special functions like addition and multiplication. Generally, operations are those special functions with which an object is "naturally" equipped (eg, the operations of groups and rings).

Yes, one can think of the Cartesian product as a kind of function (or operation, if you prefer) that produces a set from two input sets, or you can just think of it as a bunch of ordered pairs, and any subset of these ordered pairs as relations (and certain subsets of those relations as functions).

My take-away is that a "function" is ubiquitously useful, and anything that produced a unique output from a given input, is a function. You can be careful to avoid all use of the word (as in the example of functions as certain kinds of subsets of Cartesian products), but you'll just turn around and say "Hey, that felt like I was working with functions of a different sort, all along!"

pjs36
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