Equivalence relations and bijections without ordered pairs

Question

Is it correct that — opposed to general relations and functions — equivalence relations and bijective functions can be defined without reference to ordered pairs? Especially, do the following definitions capture the usual notions of equivalence and bijection?

Definition: $X$ is an equivalence relation on set $Y$ if

1. $X \subset \mathcal{P}(Y)$

2. $(\forall x \in X)\ |x| = 1 \vee |x| = 2$

3. $(\forall y \in Y)\ \lbrace y \rbrace \in X$

4. $(\forall x,y,z \in Y)\ \lbrace x,y \rbrace \in X \wedge \lbrace y,z \rbrace \in X \rightarrow \lbrace x,z \rbrace \in X$

Definition: $X$ is a bijection between sets $Y$ and $Z$ if

1. $(\forall x \in X)(\exists y \in Y)(\exists z \in Z)\ \lbrace y,z\rbrace = x$

2. $(\forall y \in Y)(\forall z \in Z)(\exists x \in X)\ \lbrace y,z\rbrace = x$

Or is there a mistake in one of these definitions?

Answer 1

The "equivalence" definition looks okay (and the same technique can encode every symmetric relation), but the "bijection" one has trouble.

It can only begin to work if $Y$ and $Z$ are disjoint. And even so, the second condition must be replaced with two:

• $(\forall y \in Y)(\exists_1 z \in Z)\ \lbrace y,z\rbrace \in X$
• $(\forall z \in Z)(\exists_1 y \in Y)\ \lbrace y,z\rbrace \in X$

where $\exists_1$ is the "there exists exactly one" quantifier.

Answer 2

I would say that a bijection is, in particular, a function, and therefore must have a specified domain and codomain. How would you define the inverse of a bijection in your definition?

In other words, your bijections are not in bijection with my bijections, because for any such set $X$ in your definition, there are two bijections in my definition, namely, $f_X:Y\to Z$ and $f_X^{-1}:Z\to Y$.

Answer 3

Every equivalence relation on a set $X$ defines a (unique) partition on said set, so you can conversely define an equivalence relation using not ordered pairs, but partitions.