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Re: Testing for the equivalence relation

From: Dan <guntermann_at_verizon.net>
Date: 1 Jul 2005 12:42:56 -0700
Message-ID: <1120246976.568742.298870@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.

Received on Fri Jul 01 2005 - 14:42:56 CDT

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