5 Mar
2009
5 Mar
'09
9:23 a.m.
Am Donnerstag, 5. März 2009 15:12 schrieb Daniel Fischer:
Yes, but the continuum hypothesis is 2^Aleph_0 == Aleph_1, which is quite something different from 2^Aleph_0 == card(R).
You can show the latter easily with the Cantor-Bernstein theorem, independent of CH or AC.
Just to flesh this up a bit: let f : P(N) -> R be given by f(M) = sum [2*3^(-k) | k <- M ] f is easily seen to be injective. define g : (0,1) -> P(N) by let x = sum [a_k*2^(-k) | k in N (\{0}), a_k in {0,1}, infinitely many a_k = 1] and then g(x) = {k | a_k = 1} again clearly g is an injection. Now the Cantor-Bernstein theorem asserts there is a bijection between the two sets.