Re: counting the elements of finite sets

Peter Webb wrote:
"Calvin" <crice5@xxxxxxxxxxxxxx> wrote in message
news:1162004714.997017.6260@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
The sets {x} and {x,x} supposedly are equivalent
because each element of one is also in the other. But
that doesn't seem right because, eg., suppose we have
the two sets, {2} and {sol(2x=4),sol(3x=6)} where 'sol'
means 'solution of the equation'. It seems clear
that the second set consists of two solutions, not
one element, no matter that the two solutions happen to
be equal. (The second set is purposely not described
as {x:2x=4,x:3x=6} because that would tend to obscure
the point.)

Set theory defines {x} and {x,x} as equal in the axiom of regularity. If you
want to create something different where {x,x} <> {x} you are welcome to;
but don't use any results of set theory (which depend on this axiom) or call
your structures "sets".

So it follows that there can be no same-size set of solutions to a
set of distinct equations if the solutions are not themselves
distinct? Or, there couldn't be a set of twenty grades for a class
of twenty students if two of the grades were the same? In the latter
case you don't necessarily have to order the grades to correspond
to the students. Maybe all you want to do is sum the elements of
the set (if it was allowed to exist) and divide by 20 to find the
average.

For another example, consider the set of all numbers
of the form p/q where p and q are positive integers
smaller than 10. Clearly there are 9*9 such numbers,
but their distinct values number fewer than 81 since,
for example, 2/4=4/8. But, by the wording of the set's
description, there are still 81 elements. If, however,
the set is described as the set of rational numbers
of the form p/q etc., then there's no reason to count
any elements of the same value more than once.

Excellent example. One is a set of ordered pairs; the other an equivalence
class of ordered pairs, called rational numbers.

Thank you for the point.

.

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