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Topic: VIRGIL CANNOT ANTI-DIAGONALISE THIS POWERSET(N)! <<<<<
Replies: 5   Last Post: Dec 30, 2012 2:10 AM

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 Virgil Posts: 8,833 Registered: 1/6/11
Re: VIRGIL CAN ANTI-DIAGONALISE ANY POWERSET(N)! <<<<<
Posted: Dec 29, 2012 10:46 PM

In article
camgirls@hush.com wrote:

> 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).
--

Date Subject Author
12/29/12 camgirls@hush.com
12/29/12 Virgil
12/29/12 INFINITY POWER
12/30/12 Virgil
12/30/12 INFINITY POWER
12/30/12 Virgil