Date: Dec 29, 2012 10:46 PM
Author: Virgil
Subject: Re: VIRGIL CAN ANTI-DIAGONALISE ANY POWERSET(N)!  <<<<<

In article 
<49449aa4-41e2-4c5b-b120-32cfcc328d5c@px4g2000pbc.googlegroups.com>,
camgirls@hush.com wrote:

> USE YOUR ANTI-DIAGONAL METHOD
> ON THIS SET OF *ALL* SUBSETS OF N!



NO! I see no evidence of any surjection from any set to its power set

S P(S)
--- ------
{} {{}} No bijection
{a} {{a},{}} No bijection
{a,b} {{a,b}, {a}, {b}, {}}

AS the size of the set S increases s does the difference in size between
S and P(S), so the less possible bijection becomes.

Supposing that you have any function from |N to its power set, 2^|N,
the set of all subsets of N, i.e.,
f:|N -> 2^|N: f: n -> f(n)

For what n in |N is f(n) equal to the set {m in |N: not m in f(m))}
which is a subset of |N, thus a member of P(|N)
Unless you can show that there must be such an n, which you cannot, you
do not have a surjection from |N to P(|N).
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