The second-order variables in the second order Induction Axiom of (second order) Peano Arithmetic range over the set of all subsets of the natural numbers, that is, it has uncountable cardinality. Does the induction scheme on first order Peano Arithmetic has also uncountable number of instances?
2026-03-25 15:39:24.1774453164
Mathematical Induction: First vs. Second order Induction Axiom
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No, it has countably many instances: the set of finite strings from a finite (or countable) alphabet is countable. In that sense, there are "fewer".
But note that there is only one second-order induction axiom: it just has a variable ranging over all subsets, but that does not make it uncountably many separate axioms any more than $\forall n(S(n) \neq 0)$ is infinitely many axioms.