Re: Question

The cardinality is then no longer aleph_0. But it is 2^aleph_0, the
cardinality of the reals.

Why? Well, the cardinality is aleph_0^aleph_0. As aleph_0 > 2,
aleph_0^aleph_0 >= 2^aleph_0. But we also have aleph_0 < 2^aleph_0,
hence aleph_0^aleph_0 <= 2^(aleph_0*aleph_0). Now aleph_0*aleph_0 =
aleph_0. So, aleph_0^aleph_0 <= 2^aleph_0 and aleph_0^aleph_0 >=
2^aleph_0, both of which can be simultaneously satisfied only if
equality holds. QED.

--OL

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