Cantor’s theorem says the power set of X has a strictly larger cardinality than X.
When |X| is a natural number, the power set of X has cardinality 2^(|X|), since you can think of an element of the power set as a choice, for each element of X, of “in the subset” vs “not in the subset.” Hence the notation 2^X for the power set of X.
Cantor’s theorem applies to all sets, not just finite ones. You can show this with a simple argument. Let X be a set and suppose there is a bijection f : X -> 2^(X). Let D be the set { x in X : x is not in f(x) }. (The fact that this is well defined is given by the comprehension axiom of ZFC, so we aren’t running into a Russell’s paradox issue.) Since f is a bijection, there is an element y of X so that f(y) = D. Now either:
y is in D. But then by definition y is not in f(y) = D, a contradiction.
y is not in D. But then by definition, since y is not in f(y), y is in D.
Thus, there cannot exist such a bijection f, and |2^(X)| != |X|. It’s easy enough to show that the inequality only goes one way, i.e. |2^(X)| > |X|.
Cantor’s theorem says the power set of X has a strictly larger cardinality than X.
When |X| is a natural number, the power set of X has cardinality 2^(|X|), since you can think of an element of the power set as a choice, for each element of X, of “in the subset” vs “not in the subset.” Hence the notation 2^X for the power set of X.
Cantor’s theorem applies to all sets, not just finite ones. You can show this with a simple argument. Let X be a set and suppose there is a bijection f : X -> 2^(X). Let D be the set { x in X : x is not in f(x) }. (The fact that this is well defined is given by the comprehension axiom of ZFC, so we aren’t running into a Russell’s paradox issue.) Since f is a bijection, there is an element y of X so that f(y) = D. Now either:
y is in D. But then by definition y is not in f(y) = D, a contradiction.
y is not in D. But then by definition, since y is not in f(y), y is in D.
Thus, there cannot exist such a bijection f, and |2^(X)| != |X|. It’s easy enough to show that the inequality only goes one way, i.e. |2^(X)| > |X|.