From: utzoo!decvax!yale-com!nglasser
Newsgroups: net.math
Title: 2^aleph[0]
Article-I.D.: yale-com.753
Posted: Sun Jan 30 14:33:27 1983
Received: Mon Jan 31 03:27:04 1983

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