It’s quite possible that what I’m encountering is the the momentary failure to understand Cantor’s theorem, or rather the mechanism it uses to enumerate the innumerable. So my math may just be old.
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|.
It’s quite possible that what I’m encountering is the the momentary failure to understand Cantor’s theorem, or rather the mechanism it uses to enumerate the innumerable. So my math may just be old.
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|.