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I understand that a function may be considered as a set of ordered pairs which relate the elements between two sets. I understand that a function is a subset of the cartesian product between the two sets and it can be defined by an equation like $y=x+1$ or $f(x)=x+1$, on a specified domain for $x$.

I am struggling to understand relations though. i know a function is a relation (set of ordered pairs) and a relation between to sets is nothing more than a subset of the cartesian product and that you could use $y=x+1$ or $y^2 + x^2 =1$ to define a relation.

What i dont get is why "equals" is referred to as a relation and "less than". Is the equals relation the set of ordered pairs $(x,y)$ defined by $y=x$, on some domain for x? and similarly for $y<x$?

Carlos Bacca
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2 Answers2

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Short answer: yes (in both cases).

For instance, if you are working on $\mathbb N$, then the binary relation “$=$” is the set$$\{(n,n)\,|\,n\in\mathbb{N}\},$$whereas the binary relation “$\leqslant$” is the set$$\{(m,n)\,|\,m,n\in\mathbb{N}\wedge m\leqslant n\}.$$

If $A$ is a set, a binary relation on $A$ is subset $R$ of $A^2$. It's, in fact, the set of all pairs of elements $(a,b)$ of $A^2$ for which the relation holds. (This is not a definition, of course. I am just trying to convey the idea.)

José Carlos Santos
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  • Thanks, would it make sense to talk of the relation $y^2 + x^2 = 1$ on $R$ as being the set of ordered pairs for which this equation holds. Also even though i referred to $y^2 + x^2 = 1$ as the relation would it be ok to think of this a property involving 2 objects which characterises the relation, just as with "equals" or "greater than", which would make sense as a relation is a set of ordered pairs or is this not ok. Thanks – Carlos Bacca Oct 17 '18 at 18:22
  • Yes, ${(x,y)\in\mathbb{R}^2,|,x^2+y^2=1}$ is a binary relation on $\mathbb R$. It is symmetric, but not reflexive. – José Carlos Santos Oct 17 '18 at 20:18
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A function is a "special" relation where there are no two pairs $(x,y_1)$ and $(x,y_2)$.

Example : the relation "father-to-child" is not a function, because a father may have more than one children.

The relation "child-to-father" is a function because a child has exactly one father.

"Less than" is a relation but not a function : $1 < 2$ and $1 < 3$ and ...

Mauro ALLEGRANZA
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