Go backward to Infinite Sets Go up to Top Go forward to Permutations |
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
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.
f: S ->injective P(S) f(x) := {x}.
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.