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Russell Easterly says...
>Let set B be equal to any ordinal.
>A_0 = Union(B)
>A_i = A_(i-1) after removing the smallest element of A_(i-1) for i>0.
>Let A be the set of all A_i.
The problem here is that with the usual definition of "ordinal",
if you take an ordinal, and remove the smallest element, then the
result is no longer an ordinal. In set theory, by convention an
ordinal is the set of all smaller ordinals. For example,
0 = empty set (because there are no smaller ordinals)
1 = { 0 } (because 0 is the only smaller ordinal)
2 = { 0, 1 } (because 0 and 1 are the ordinals smaller than 2)
etc.
Let's take a particular example: B = { 0, 1, 2, 3, ... }
(which is the ordinal omega)
A_0 = { 0, 1, 2, 3, ... } = omega
A_1 = { 1, 2, 3, ... } (which is *not* an ordinal)
A_2 = { 2, 3, 4, ... } (which again is not an ordinal)
A_3 = { 3, 4, 5, ... }
etc.
In general, A_n = { n, n+1, n+2, ... }
>R1 lets me prove A has a smallest element.
No. It says that A has an element x such that
x has no elements in common with A. Since all
the elements of A are infinite sets, and
A_0 contains only finite sets, then
A_0 intersect A = empty set
But A_0 is not the "smallest" under inclusion. There is
no smallest under inclusion.
>Since A is partially ordered by inclusion, the intersection of A
>equals the smallest element of A.
No, it does not. The intersection of A is
{ x | forall n, x is an element of A_n }
That is equal to the empty in our case. But the empty set
is *not* an element of A. So it is false that the intersection
of A is equal to the smallest element.
The intersection of A will only be equal to the smallest element
in the special case in which A is finite.
>We know the intersection of A is the empty set.
Correct.
>This means the empty set is a member of A.
No, it doesn't mean that.
--
Daryl McCullough
Ithaca, NY
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