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:
Distinct equivalence classes
[a] = {a, b}
[b] = {a, b}
[c] = {c, d}
[d] = {c, d}
{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