Re: Testing for the equivalence relation
Date: Sun, 3 Jul 2005 08:20:04 -0400
Message-ID: <9-edncJMMJJ0SlrfRVn-iw_at_comcast.com>
"Dan Guntermann" <guntermann_at_verizon.net> wrote in message
news:f3Oxe.6649$Fy4.5701_at_trnddc04...
>
> "VC" <boston103_at_hotmail.com> wrote in message
[...]
>> There are just two equivalence classes. What's a 'distinct' equivalence
>> class ? Is it some kind of 'computer'-math speak ?
>>
> Erratum
>
> Change the words "4 equivalence classes" to "four elements with
> equivalence classes, denoted [a], [b], [c], [d]."
The words "four elements with equivalence classes, denoted [a], [b], [c], [d]." just does not make sense because, say, [a] is not 'an element with an equivalence class. It's just an equivalence class, that's all. You might say let's introduce a pair (a, [a]) but I fail to see what the use of this construction would be.
>>>> {a,b} subset is different from the {a, b} subset ?
>> Saw your later comment. Since an equivalence class is a subset of the
>> original set over which the equivalence relation is defined, how the
>>
> Yes, it is a partition and the sets are the same. Perhaps if one applies
> a name to both sets, we now have a basis for distinguishing the first from
> the second, even if they were to be proven equal?
See above.
>[...] when we ask whether set A is equal to set B. They might be exactly
>the same, but we still distinguish them because they might not be the same.
??? How is it possible to be exactly the same and not the same at once ?
vc Received on Sun Jul 03 2005 - 14:20:04 CEST