Re: is this a set?
From: shedar (nobody_at_nonesuch.com)
Date: 09/21/04
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Date: Tue, 21 Sep 2004 16:42:39 GMT
"Peter Fairbrother" <zenadsl6186@zen.co.uk> wrote in message
news:BD756229.62BC7%zenadsl6186@zen.co.uk...
> nsgi_2004 wrote:
>
> > Ok thanks for still trying to help. I must admit my set theory is not
very
> > good, I just began reading about it the last few days to get back into
math
> > as I start my jr. year courses next week.
> >
> > Q={w in Z | there exist x,y in Z such that (x > y) and w={x,y}}
>
> If w is a pair, ordered or not, then w is not in Z.
That was a typographical error of Shedar which has already been corrected in
an earlier post.
> Try:
> Q = { (x,y) | x,y in Z; x > y }
>
> The two conditions are x,y in Z and x > y.
>
> You should probably make clear in the text whether the pair is ordered - I
> don't know of a standard notation that distinguishes between ordered and
> non-ordered pairs.
That is incorrect. There IS a notation for unordered pairs. Indeed, the
notation {x,y} IS the notation for the UNORDERED pair x and y. Observe that
for x,y in Z, {x,y} = { u in Z | (u=x) or (u=y) }.
>
> However, if a non-ordered pair is intended, then the second condition
would
> probably be written as x != y.
>
Please read all the previous messages in the thread before jumping in. The
original poster clearly wants the members of his set "Q" to be UNORDERED
pairs {x,y}. Had he wanted the ORDERED pair (x,y), he would have pointed
this out a long time ago.
> That it isn't would imply that the pair is ordered - math notation is a
> language, and it allows such implications. The symbol (x,y) is also used
for
> a greatest common divisor for instance, and only context implies that here
> it is used to mean a pair.
Again, please read previous messages in the thread. This question is about
{x,y}, not about (x,y). Nor is it about the GCD of x and y. The original
poster's context is clear.
> Which brings us to the question, what is a set? My definition is the
> collection of items to which "yes" results are associated when applying to
> all items some function that always gives a yes/no answer for each and
every
> item.
>
> The function here is the list of conditions for membership. Note that the
> first part of the list, (x,y) is also in some sense a condition - any item
> that is not a pair of numbers is excluded.
>
> Under that definition - assuming that the ambiguity about whether the pair
> is ordered is settled - then we have a set, as the function will give an
> answer for all items*.
>
> The empty set is a set such that no item has a yes answer. To show that
this
> set is not empty you would have to show an item with a yes answer, ie a
pair
> consisting of two integers, the first being greater than the second.
>
>
> * it will give an answer for any item that is a pair, and for any item
that
> is not a pair. Whether all items are either [pairs] or [not pairs] is not
> discussed here. Defining "item" is hard.
>
As for the rest of your comments, it betrays a lack the proper training in
axiomatic set theory.
The question of "is this a set or not a set" is to be settled ONLY using the
axioms of the underlying set theory. In most applications of mathematics
(except in set theory, model theory, logic where different lists of axioms
are being investigated or employed), the accepted underlying axioms are
usually those known as "the axioms of ZFC" (Zermelo-Frankel set theory with
"axiom of choice"). As it has already been pointed in much earlier posts by
Dennis May, Dave Seaman and Dave Ullrich, the original poster's set "Q" is a
set according to the axioms of ZF.
Shedar
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