Virgil
Posts:
8,833
Registered:
1/6/11


Re: VIRGIL CAN ANTIDIAGONALISE ANY POWERSET(N)! <<<<<
Posted:
Dec 29, 2012 10:46 PM


In article <49449aa441e24c5bb12032cfcc328d5c@px4g2000pbc.googlegroups.com>, camgirls@hush.com wrote:
> USE YOUR ANTIDIAGONAL 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). 

