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Jan 26, 2017 at 10:49 comment added Joel Adler I agree that the statement for the special case $\mathcal{N}$ has appeared already. But properties of the real numbers were used in its proof, and the general statement $|M|<|P(M)|$ does not follow.
Jan 25, 2017 at 16:09 comment added Noah Schweber Yes, but $(0,1)$ is immediately in bijection with $2^\mathbb{N}$; I wouldn't consider that a nontrivial fact. IMO the two results are "morally equivalent." (Of course, that's not a very strong opinion - I didn't downvote! I'm just saying the statement has arguably appeared already.)
Jan 25, 2017 at 16:06 comment added Joel Adler In his first answer he addressed Russell's antinomy, didn't he? And didn't he show in his second answer that the interval (0,1) of reals is uncountable, not the power set of the natural numbers?
Jan 24, 2017 at 20:55 comment added Noah Schweber Actually, Joel David Hamkins already gave this answer. (Technically he only gave the answer for the powerset of the naturals versus the naturals, but the proof is identical to the general case.)
S Jan 24, 2017 at 20:48 history answered Joel Adler CC BY-SA 3.0
S Jan 24, 2017 at 20:48 history made wiki Post Made Community Wiki by Joel Adler