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.
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