3
$\begingroup$

We can define a binary relation as a set of ordered pairs. Alternatively, we can call the set of ordered pairs the "graph" of the relation, and define the relation itself as a triple $(X,Y,f)$, where $f$ is the graph, $X$ is its domain and $Y$ is its codomain. The same goes for functions; they can be defined as sets of ordered pairs, or as triples. So my question is, given that its more lightweight to define a binary relation simply as its graph, why is the ordered-triple approach much more common?

Here's a list of reasons and possible objections that I've come up with.

Reason 0: The ordered-triple approach allows us to distinguish between surjections and non-surjections. Possible Objection: This could be read another way, as suggesting that surjectivity is a contrived concept. Maybe we should only every say "$f \in \mathrm{Surj}(X,Y)$" but never simply say "$f$ is a surjection."

Reason 1: Category theory works that way. Possible Objection: Perhaps this is a "hint" that maybe there's room for improvement in the basic definitions of category theory.

Reason 2: The ordered triple approach allows us to define the complement of a relation by $f^c = X \times Y \setminus f$. Thus, the set of all relations with domain $X$ and codomain $Y$ form a Boolean algebra. Possible Objection: This is a pretty minor advantage, given that we can just write "Defining $A^c = X\times Y \setminus A$, it follows that..." whenever we need to.

So my question is, what's the payoff of the ordered triple approach?

  • 0
    Apparently, people weigh the three pro-reasons more than your three objections. I follow especially the category argument, as in other categories than **Set** you either have no "elements" of objects at all or have a great problem to speak of all morphisms from $A$ to $B$ as you must then speak of all homomorphisms from $A$ to some $C$ where $C$ happens to be a subobject (what is that?) of $B$. Then again, you are right esp. in set theory, where one merely uses predicates $\Phi(x,y)$ with $\Phi(x,y_1)\land \Phi(x,y_2)\to y_1=y_2$, as e.g. transfinite recursion might be difficult otherwise.2012-11-22
  • 1
    Good question! My undoubtedly unpopular opinion is that formal definitions usually don't matter very much, as long as we agree on what objects we are really talking about. By that criterion, a reasonably precise definition of continuous function is useful, since the nicely behaved functions of one's untutored geometric imagination are very much not "typical" continuous functions.2012-11-22
  • 0
    @AndréNicolas That's an interesting point of view. I don't think I agree, but would be interested in reading anything that you've written on this idea.2012-11-23

1 Answers 1