|
That's a good question -- this result is arguably the one that made Georg Cantor a legend. (It's called Cantor's theorem, after all!) I had to refresh myself on the proof, but it boils down to a less obvious kind of "diagonalization" (this time to construct a counterexample). We can show that S has a strictly smaller cardinality than 2^S by showing that no function from S to 2^S can hit every value in 2^S. That is, there's going to be some element of 2^S that no element of S gets mapped to. Suppose, as one does, to the contrary: that there is a function, call it F, that hits every element of 2^S. We'll construct a particularly pathological element of 2^S, call it B, the set {s in S | s not in F(s)} of values of S that are not members of the subset F picks out for them. (Remember that 2^S is the set of subsets of S, so it is meaningful to ask whether any s is a member of F(s).) If this smells like Russell's paradox, good -- follow your nose! Now, since B is a subset of 2^S, and F is presumed to hit every subset, there's going to be some element b such that F(b) = B. We ask the critical question: is b in B? If it is, then b is not in F(b) -- but F(b) is B, contradicting our assumption. If it isn't, then b is in F(b) -- but F(b) is B, contradicting our assumption. Since we end up at a contradiction no matter which way we resolve "is b in B", and we know this question must be answered one way or the other, we must admit that our supposed function F must not have the property we asked for; it must not hit every element of 2^S. So 2^S really is "bigger" than S, in that no matter how we pair every element of S with an element of 2^S, we run out of elements of the former before we've exhausted the latter. |
I guess I should read the book on infinity and aleph numbers linked in the other thread.
It’s both satisfying and unsettling that you kinda reinvent numbers with cardinality of infinities. Like wtf is going on.