Re: Cantor's proof that #(Evens) = #(Naturals) is inconsistent

From: David McAnally (D.McAnally_at_i'm_a_gnu.uq.net.au)
Date: 09/25/04 

Date: 25 Sep 2004 14:48:50 GMT

David P. Ferguson, you seem to mentally endow sets with extra qualities
which they do not have. You seem to want to endow sets (like N) with an
intrinsic order.

You have a mental picture of a set being affected by its construction.
When you define E as the set of even natural numbers, you mentally leave
gaps in the set where all the odd elements have been taken out, but when
you define E* = {2n : n in N}, you do not imagine any such gaps. The fact
that sets like E do not have gaps in them seems to have escaped you.
Sets do not come with this sort of additional structure.

Similarly, sets do not automatically come with an intrinsic order. Let A
be a set. There is no order per se on A. An order on A is a subset \rho
of A^2 such that

        (1) for all x in A for all y in A for all z in A ([(x,y) in \rho
            and (y,z) in \rho] => (x,z) in \rho),

        (2) for all x in A for all y in A ([(x,y) in \rho and (y,x) in
            \rho] => x = y).

Condition (1) states that \rho is transitive, and condition (2) states
that \rho is antisymmetric.

Similarly, a set N does not come with an intrinsic order on it. A model
of Peano's axioms is actually a triple, (N,0,S), where N is a set, 0 is
an element of N, and S is a function S : N -> N, i.e. S is a subset of N^2
such that

        (1) for all m in N exists n in N ((m,n) in S),

        (2) for all m in N for all n in N for all p in N ([(m,n) in S and
            (m,p) in S] => n = p).

These two conditions on S can be summed up in the one statement that

    for all m in N exists n in N for all p in N ((m,p) in S <=> p = n).

N does not, itself, have an intrinsic order, but there is a natural order
associated with a model (N,0,S) of Peano's Axioms (where not only N is
given, but also an element 0 of N, and a function S : N -> N, are given as
well). The natural order on a model (N,0,S) is the unique order \rho on N
(so that \rho is a subset of N^2) such that

        for all n in N ((n,n) in \rho and (n,S(n)) in \rho).

The unique order \rho on N satisfying this condition is a well-ordering,
and \rho has order type \omega. Note specifically that \rho requires S in
order to be defined. This means that \rho is not intrinsic in N.

There is a theorem in chaos theory which imposes an order on the positive
integers. The imposed order is of type \omega^2 + \omega*, where \omega*
is the type of the order (N,>), i.e. \omega* is the type of the reverse
ordering to an order of type \omega. Specifically, the ordering of the
positive integers in the theorem is determined by

3, 5, 7, 9, ..., 2 x 3, 2 x 5, 2 x 7, 2 x 9, ..., 2^2 x 3, 2^2 x 5,
2^2 x 7, 2^2 x 9, ..., 2^3 x 3, 2^3 x 5, 2^3 x 7, 2^3 x 9, ..., . . .,
..., 2^4, 2^3, 2^2, 2, 1.

Specifically, for k and m in N, and for odd positive integers n and p,
the ordering determines that 2^k n precedes 2^m p in the ordering iff one
of the following holds:

        (1) k = m and 1 < n < p,

        (2) k < m and 1 < n and 1 < p,

        (3) 1 < n and p = 1,

        (4) k > m and n = p = 1.

Does the fact that the theorem discusses an order on the positive integers
of type \omega^2 + \omega* mean that the positive integers are
intrinsically endowed with this order type? And does that mean the
cardinality of the positive integers with this ordering is infinitely
times larger than the cardinality of N? N-{0} (the set of positive
integers) has the same cardinality as N, and that is independent on the
order imposed on N-{0}, whether it be the normal ordering or the ordering
imposed on N-{0} by the theorem in chaos theory, or any other ordering.

Similarly, a bijection between sets A and B is just a subset f of A x B
such that

        (1) for all x in A exists y in B ((x,y) in f),

        (2) for all x in A for all y in B for all z in B ([(x,y) in f and
            (x,z) in f] => y = z),

        (3) for all y in B exists x in A ((x,y) in f),

        (4) for all x in A for all y in A for all z in B ([(x,z) in f and
            (y,z) in f] => x = y).

These four conditions can be summarized as the two conditions

    for all x in A exists y in B for all z in B ((x,z) in f <=> z = y),

    for all z in B exists x in A for all y in A ((y,z) in f <=> y = x).

For sets N and E, the set f = {(n,2n) : n in N} satisfies all the
necessary conditions for a bijection, so N and E are bijective, and so
have the same cardinality.

David

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