[net.math] 2^aleph0

nglasser (01/30/83)

In a recent article someone said that you can PROVE that 2^aleph[0] is equl
to the power set of aleph[0]. But 2^S, where S is a set, is DEFINED to be the
power set of S. The name 2^S comes from that fact that for finite sets S,
|power set of S| = 2^|S|.
                               Not contradicting, merely elucidating
                               - Nathan Glasser
                               ..decvax!yale-comix!nglasser