5
$\begingroup$

What is the standard notation for a set of equivalence relations? Specifically, I have a pair of objects, call them $x$ and $y$ and I denote the ordered pair as $\left(x,y\right)$. I have a set of equivalence relations such as:

$\left(x,y\right) \sim \left(x^{-1},y\right) \sim \left(x,y^{-1}\right) \sim \left(y^{-1},x^{-1}\right)$

I would like to write this compactly, but I'm unsure of what the standard notation would be. Is the following appropriate?

$\left\{\left(x,y\right),\left(x^{-1},y\right),\left(x,y^{-1}\right),\left(y^{-1},x^{-1}\right)\right\}$

$\endgroup$
2
  • $\begingroup$ I think the usual way is to write just what you have written (perhaps followed by "$\forall x, y \in G$"), and say the quotient is $G/\sim$. $\endgroup$
    – Nick Matteo
    Commented Apr 23, 2013 at 17:20
  • $\begingroup$ This question is very unclear. The title asks for notation for a set of equivalence classes, the first sentence talks about a set of equivalence relations. Then it apparently says that $(x,y) \sim (x^{-1},y) \sim (x, y^{-1}) \sim (y^{-1}, x^{-1})$ is a set of equivalence relations, and I now I have no longer any idea what that might be supposed to mean. $\endgroup$
    – JiK
    Commented Dec 11, 2017 at 21:53

2 Answers 2

Reset to default
8
$\begingroup$

You can represent an equivalence class by using a representative from the class, and denoting the entire class by, say, $[(a, b)]$: this represents the set of all ordered pairs $(x, y)$ such that $(x, y) \sim (a, b)$.

So, for a given equivalence relation denoted by $\sim$: one of its equivalence classes can be denoted: $$[(a, b)] = \{(x, y)\mid (x, y) \sim (a, b)\}$$

If there are many equivalence classes determined by an equivalence relation, and you want to denote the set of equivalence classes, you can list the equivalence classes as elements of a set:

  • $\{[(a, b)], [(c,d)], [(e, f)], \cdots [(y, z)]\}$ if there is a finite set of them.
  • For example, the set of equivalence classes determined by the equivalence relation of congruence modulo $4$ on the set of integers, you could write $\{[0], [1], [2], [3]\}$,
  • or, in the case of an infinite number of equivalence classes, like those corresponding to equality/identity on the natural numbers, one can write $\{[1], [2], [3], \cdots \}$.

If you are asking how to denote a set of different equivalence relations, where the elements of the set are relations, I'm not aware of the standard notation. (That's not to say it doesn't exist.)

$\endgroup$
3
  • $\begingroup$ What I want to do is have a way to denote the entire set of equivalence relations, like $P=\left\{\left(x,y\right),\left(x^{-1},y\right),\left(x,y^{-1}\right),\left(y^{-1},x^{-1}\right)\right\}$, so that I can later define the quotient set as $G/P$, so I need a way to denote the entire set of equivlanece relations not just one of them. $\endgroup$
    – okj
    Commented Apr 22, 2013 at 19:27
  • $\begingroup$ Try using square brackets to denote that the elements of $P$ are equivalence classes:$$ P=\left\{\left[\left(x,y\right)\right],\left[\left(x^{-1},y\right)\right],\left[\left(x,y^{-1}\right)\right],\left[\left(y^{‌​-1},x^{-1}\right)\right]\right\}$$ $\endgroup$
    – amWhy
    Commented Apr 22, 2013 at 21:38
  • $\begingroup$ @amWhy: Very nice answer +1 $\endgroup$
    – Amzoti
    Commented Apr 23, 2013 at 0:17
7
$\begingroup$

If $\sim$ is an equivalence class over $A$, then in many places we write $A/{\sim}$ as the set of equivalence classes.

This notation is similar, and on purpose, to the notation from algebra when writing $V/W$ for the quotient subspace, or $G/H$ for the quotient group, or $R/I$ when taking a quotient of a ring by an ideal.

The reason is that all those quotients actually induce an equivalence relation, and we have a natural structure on the set of equivalence classes.

$\endgroup$
2
  • $\begingroup$ Is the notation trying to give the intuition that $A$ is divided by the relation $\sim$ into some equivalence classes? $\endgroup$
    – Hosein Rahnama
    Commented Aug 26, 2022 at 5:39
  • 1
    $\begingroup$ @HoseinRahnama: Yes. $\endgroup$
    – Asaf Karagila ♦
    Commented Aug 26, 2022 at 6:39

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .