Mathematical induction is often used to demonstrate proofs for integers, which is logical as $P(n) \implies P(n+1)$ creates a domino effect that extends infinitely to satisfy all integer cases (greater than the base case).

I am curious if this can be extended to continuous sets such as $[0, \infty)$, and under what conditions.

The main barrier to using induction on a continuous set is that there is no fixed separation between consecutive elements of a continuous set. In fact, "consecutive" is rather meaningless in this context. I see $2$ potential ways to get around this.

My first idea is that if we can prove $P(n) \implies P(n + a-b)$ where $a$ is some finite value and $0 \le b \le a$, i.e. if we prove that the implication holds over all values in an interval of the set, that should complete the proof (assuming the base case holds).

My second idea is to prove that $P(n) \implies P(n + \epsilon)$ where $\epsilon$ is the smallest number greater than $0$ (if that even makes sense!) somehow using calculus and then again have that complete the proof (assuming the base case holds).

I would like to know whether these ideas are correct and also whether they have any mathematical merit.

Another point of thinking is whether one can even define a base case. But for the purposes of this question, one may assume that the base case exists.