# Re: Testing for the equivalence relation

Date: 1 Jul 2005 12:42:56 -0700

Message-ID: <1120246976.568742.298870_at_g49g2000cwa.googlegroups.com>

"No, there are two equivalence classes. Period. Saying that there are 4 is sloppy terminology."

Equivalance classes, denoted with brackets:

[a] = {a, b} [b] = {a, b} [c] = {c, d} [d] = {c, d}

Distinct equivalence classes

{a,b}

{c,d}

I make my comments based on the following as a defintion of equivalence class (from Discrete Mathematics with Applications by Susan Epp). If you think it is sloppy, I suppose you'll have to take it up with the author: "Suppose A is a set and R is an equivalence relation on A. For each element a in A, the equivalence class of a, denoted [a] and called the class of a for short, is the set of all elements in A such that x is related to a by R.

[a] = {x is an element of S | x R a}

- OR -

for all x in S, x is an element of [a] if, and only if, x is related to
a.

Jan: "Er, actually, you cannot. The notion of 'reflexivity' is defined
wrt.

(1) a domain and (2) a binary relation over that domain. The notion is
not really defined if you only have (2). Of course you could introduce
a

related notion that only needs (2), but that would be a different
notion."

So why wouldn't the set S suffice as being the basis of (1)?. And yes, I am curious whether another notion applies here.

> For me, one of the benfits of answering this question is that it goes a

*> long way ascertaining a distinction between type and domain.
*

Hm, you think there is a difference?

Actually, in one sense, yes and in another sense, no.

- Dan