Re: For ANY sets A, B: If A is proper subset of B then set B has more elts than set A.

From: Andrzej Kolowski (akolowski_at_hotmail.com)
Date: 07/05/04 

Date: 5 Jul 2004 08:35:13 -0700

david.ferguson1@cox.net (David P. Ferguson) wrote in message news:<2cf2d21.0407020519.2ab6c354@posting.google.com>...
> Where "A < B" means "A is a proper subset of B" and "elts" is short
> for "elements"
>
> PROVE:
> For ANY sets A, B:(1) If A < B then set B has more elts than set A.
> (2) If A < B then no 1 to 1 correspondence f:A=>B
>
> To streamline this proof I will prove the proposition for a pair of
> example sets:
>
> BY DEFINITION OF A < B.
> Set A < B => B - A = non-empty C.
>
> SUPPOSE:
> Let set B be any {b1, b2, b3, b4, b5, b6... bn ...| bn elt of B}

  Here you are clearly assuming that B is denumerable.

> Again: If A < B then set B = set A + set C.
> Suppose that set B can be expressed as: {a1, c1, a2, c2, a3, c3 ...}
>
> I TAKE IT AS FACT THAT:
> There is a set of permutations of elts of set B:
> phi (B) = B x (a1, b1) x (2, b2) x (a3, b3) x ...(an, bn) x ... |"an"
> elt B}
> Such that: phi (B) = {a1, a2, a3, a4... c1, c2, c3 ...}
>
> SINCE:
> B x (a1, b1) = B1 = {a1, c1, a2, c2, a3, c3, a4, c4, a5, c5, ...}
> B1 x (a2, b2) = B2 = {a1, a2, c1, c2, a3, c3, a4, c4, a5, c5, ...}
> B2 x (a3, b3) = B3 = {a1, a2, a3, c2, c1, c3, a4, c4, a5, c5, ...}
> B3 x (a4, b4) = B4 = {a1, a2, a3, a4, c1, c3, c2, c4, a5, c5, ...}
> B4 x (a5, b5) = B5 = {a1, a2, a3, a4, a5, c3, c2, c4, c1, c5, ...}
> UNTIL SET A IS "EXHAUSTED":(See "INEXHAUSTIBLE", below)
> Bw-1 x (bw, bw) = Bw = {a1, a2, a3, a4, .., c1, c2, c3, c4, c5, ...}
>
> The number of permutations in phi(B) is determined by the number of
> elements in set A.
>
> THIS HOLDS FOR ANY SUBSET A OF ANY SET B.
>
> The only possible exception to the above is the case where set A is
> claimed to be "INEXHAUSTIBLE".(THIS IS A COMMON PERCEPTION OF
> "INFINITE").
>
> This presents no obsticle since: For all ax an elt of A: The
> permutation (ax, bx) exists as surely as the set A exists.
>
> That "There might exist an element ax of set A for which (ax,bx) does
> not exist." would make a good discussion all by itself!
>
> The proof that set B has MORE elements than set A still holds. And the
> argument above still demonstrates the absolute impossibility of ANY
> one to one correspondence being established between sets A and B.[BE
> THEY FINITE OR INFINITE] I know that this is bound to be an unpopular
> viewpoint but I am dead serious about this. I am well aware of the so
> called proofs that one to one correspondences have been demonstrated
> between some infinite sets and some of their proper subsets. My
> research has revealed that all of the proofs (that I am aware of)
> contain fatal flaws and hidden assumptions. I reject their idea that
> denumerably INFINITE sets can be put into 1 to 1 correspondence with
> many of their proper subsets.
> david.ferguson1@cox.net

  The basic error here is in assuming that the composition of an
infinite number of permutations is a permutation.

  Say you are considering permutations of N = {1, 2, 3, ... }. Let
pi_n be the permutation that switches the n-th term and the
(n + 1)st term. For example, pi_2(1) = 1, pi_2(2) = 3, and
pi_2(3) = 2. Thus

  pi_2(N) = (1, 3, 2, 4, 5, ... }.

  Now consider the composition of pi_1, pi_2, ... That is,
first apply pi_1, then follow it by pi_2, then pi_3, etc.

  Call the result pi_w.

  Note that

  pi_w(N) = {2, 3, 4, 5, 6, ...}.

  Note that pi_w(1) is not in N.

  Note also that 1 is not in pi_w(N).

  Therefore pi_w is not a permutation of N.

  Therefore the composition of an infinite number of permutations
need not be a permutation.

  The same holds for your construction. To see this, you need
only note that the position of "c1" after your string of
permutations does not correspond to any positive integer. In
a sense it corresponds to "infinity", but "infinity" is not an
element of B (which by your notation is denumerable).

  Andrzej

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