2 ms·

You're talking past one another. The other poster is discussing a correspondence between N and S, where S is the set of squares. So the function being discussed

by mb7733 3y ago

You're talking past one another. The other poster is discussing a correspondence between N and S, where S is the set of squares. So the function being discussed is f: N -> S, f(n) = n^2, which is obviously bijective. You seem to be talking about the function f: N -> N, which isn't surjective, but I don't see the relevance. Any surjective function can be made non-surjective by considering a larger range.

thaumasiotes 3y ago
> You seem to be talking about the function f: N -> N, which isn't surjective, but I don't see the relevance. It's worse than just irrelevant. The claim is that we have two sets, ℕ and S, that are the same size. The argument that they are the same size is the observation that we've demonstrated a bijection between them. No one ever mentioned a function from ℕ to ℕ, because that would tell you nothing about S. It would be like saying "in order to show that 5 is odd, observe that 2 and 2 are both even". The idea is gibberish.