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That proof assumes that "s(n)=n+1 is clearly an injection" - ie that every natural number has a unique successor, and uses it to prove something about the size
by roywiggins 3y ago
That proof assumes that "s(n)=n+1 is clearly an injection" - ie that every natural number has a unique successor, and uses it to prove something about the size of the set. But it doesn't prove there is "always a number following another number", it assumes it.
imho it's part of the definition of what the set of natural numbers even is. Peano arithmetic just has S(n) always exists and is injective as axioms.
- anon291 3y agoCorrect. All descriptions of the natural numbers of which I'm aware require the assumption that one can take some object and apply some operation 'S' on that object to produce another object that is neither the original object nor any of the objects produced before by repeated applications of 'S' to some root object. Whether you simply define a number as 0 and S(x) where x is a natural number or you define natural numbers as nested sets in ZFC or as church-encoded lambda terms, etc, you assume the existence of such a transformation.