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Size of Powersets

Proposition: Every set is smaller than its powerset:

forall S: S is smaller than P(S).

Proof: Take arbitrary S. S is not larger than P(S) because we can define

f: S ->injective P(S)
f(x) := {x}.
Assume that S and P(S) are of the same size, i.e., there exists some f: S ->bijective P(S). We show a contradiction.

Take A := {x in S: x not  in f(x)}. Since f is surjective and A subset S, i.e., A in P(S), we have some a in S with f(a) = A. But then we know a in A <=> a not  in f(a) <=> a not  in A.

Author: Wolfgang Schreiner
Last Modification: December 7, 1999

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