Is there such a thing as continuous induction? If so has it ever been used to prove anything? Context:
"I am currently wondering if induction could be extended to the reals instead of just integers and the method I developed for this is say there exists some $\epsilon>0$ and some statement S that holds for f(a) at $x=a$.
Assert that statement S holds for f(a + $n*\epsilon$) and f(a - $n*\epsilon$) where n is some integer. Then use the assertions to prove that S holds for f(a + $(n+1)*\epsilon$) and f(a - $(n+1)*\epsilon$) and the resulting interval of x for which S does hold for f(x) would be all real numbers."
Question: Does this kind of continuous induction work, is it really able to prove statement S holds for all real values of x? If this kind of induction does work, is there ever been a problem where this is used? If so please give a simple example. Thanks.
 A: In general we could say induction is:

*

*$f$ is a predicate on $E$.

*We prove that if $f$ is true on a subset $A$ of $E$, it is also true on a subset $g(A)$ of $E$. (Actually we could have "a set of subsets" instead of just "a subset").

*Define the following partial order relation on $\mathcal{P}(E)$: transitive closure of:
$A \le (A \cup g(A))$,
$A \subset B \Rightarrow B \le A$.

*Usage of the induction: if $f$ is true on $A$, then $f$ is true on any $B \ge A$. In particular, if $\{B | B \ge A\}$ has a greatest element $C$, $f$ is true on $C$.

Classical induction is when $E = \mathbb{N}$ and (for example) $g([0, n]) = \{n+1\}$. The partial order is the transitive closure of $[0, n] \le [0, n+1]$, whose greatest element is $\mathbb{N}$.
An induction on $\mathbb{R}$ would be for example (taking this from a paper in the comments):
$\forall x, \exists y \gt x, g(\{x\}) = [x, y]$, and
$g([a, b[) = [a, b]$.
This induction can be used to prove the following: if $f$ is continuous, $f(0) > 0$, and $\forall x, f(x) \ne 0$, then $\forall x>0, f(x) > 0$:

*

*$f$ being continuous, if $f(x) >0$
then $\exists y>x,$
$\forall z \in [x, y], |f(z) - f(x)| < f(x)/2$,
hence $f(z) > 0$. So this proves $g(\{x\}) = [x, y]$.

*And similarly if $f$ is strictly positive on $[a, b[$ it is also on $[a, b]$ (we cannot have $f(b) < 0$ because then, by continuity, $f$ would be strictly negative on some interval on left of $b$; we cannot have $f(b) = 0$, by hypothesis) so $g([a, b[ = [a, b]$.

In topology we could probably prove (I have not checked) by this induction the following result: if $f$ is a continuous function from $E$ to a discrete set $F$, then $F$ is constant on each connected component of $E$.

That being said, there are cases where the discrete induction is transformed into something different in the continuous case, e.g.:

*

*A classical induction:
$a(0) = 0,$
$\forall n \in \mathbb{N}, n \gt 0, a(n) = a(n-1) + n$,
we deduce $a(n) = \frac {n(n+1)} 2$ by induction.

*The continuous equivalent would be:
$f(0) = 0,$
$\forall x \in \mathbb{R}, x \ge 0, f'(x) = x$,
we deduce $f(x) = \frac {x^2} 2$ by integration.

Actually here the integration theory may hide some classical discrete induction (at least for the Riemann integral, which uses discretization).
But we may also say that induction is a specific case of integration. Example with discrete induction and discrete integration:
$f(n) \Rightarrow f(n+1)$ may be interpreted as $\Delta(f)(n) = f(n+1) - f(n) = 0$.
So $f(n)$ is the antiderivative of $0$, i.e. a constant, and if $f(0)$ is true this constant is true.
(Actually we do not have $f(n) \text {false} \Rightarrow f(n+1) \text {false}$, so it is slightly less direct than that).
