Proof
Two sets are equinumerous (have the same cardinality) if and only if there exists a one-to-one correspondence between them. To establish Cantor's theorem it is enough to show that, for any given set A, no function f from A into, the power set of A, can be surjective, i.e. to show the existence of at least one subset of A that is not an element of the image of A under f. Such a subset, is given by the following construction:
This means, by definition, that for all x in A, x ∈ B if and only if x ∉ f(x). For all x the sets B and f(x) cannot be the same because B was constructed from elements of A whose images (under f) did not include themselves. More specifically, consider any x ∈ A, then either x ∈ f(x) or x ∉ f(x). In the former case, f(x) cannot equal B because x ∈ f(x) by assumption and x ∉ B by the construction of B. In the latter case, f(x) cannot equal B because x ∉ f(x) by assumption and x ∈ B by the construction of B.
Thus there is no x such that f(x) = B; in other words, B is not in the image of f. Because B is in the power set of A, the power set of A has a greater cardinality than A itself.
Another way to think of the proof is that B, empty or non-empty, is always in the power set of A. For f to be onto, some element of A must map to B. But that leads to a contradiction: no element of B can map to B because that would contradict the criterion of membership in B, thus the element mapping to B must not be an element of B meaning that it satisfies the criterion for membership in B, another contradiction. So the assumption that an element of A maps to B must be false; and f can not be onto.
Because of the double occurrence of x in the expression "x ∉ f(x)", this is a diagonal argument.
Read more about this topic: Cantor's Theorem
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