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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.

A16
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  • No. The reals are densely ordered. You could always find a number between $a$ and $a+ \varepsilon$. – Gowexx Jul 23 '22 at 14:28
  • If $a$ and $\epsilon$ are fixed, then no, this doesn't work. If one for $\epsilon$ or $a$ is not fixed, then you don't need induction – Exodd Jul 23 '22 at 14:30
  • @Gowexx, thank you for your response. I understand why my method doesn't work now. – A16 Jul 23 '22 at 14:39
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    Perhaps a duplicate? Anyway, I think this should answer your question more generally: https://math.stackexchange.com/questions/4202/induction-on-real-numbers – legionwhale Jul 23 '22 at 14:48
  • 100 years ago or so, the "Heine-Borel theorem" was considered to be "induction for the reals". – GEdgar Jul 23 '22 at 14:54
  • See https://arxiv.org/pdf/1208.0973.pdf – B. Goddard Jul 23 '22 at 14:54
  • You can look at my answer to the similar question: https://math.stackexchange.com/questions/4465709/mathematical-induction-on-continuous-sets/4465731#4465731 – Mateo Jul 23 '22 at 21:22
  • This is mentioned elsewhere - but what my mind jumps to is connectedness arguments. "Let $U$ be a non-empty subset of the reals such that for each $x \in U$, there's an $\varepsilon > 0$ such that $(x - \varepsilon, x + \varepsilon) \subseteq U$ and for each $y \in \Bbb R \setminus U$, there's a $\delta > 0$ such that $(y - \delta, y + \delta) \subseteq \Bbb R \setminus U$. Then $U = \Bbb R$." You can use this to prove that any locally constant function is constant, for instance. This is also a good answer. – Izaak van Dongen Jul 23 '22 at 22:53

1 Answers1

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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).

Jean-Armand Moroni
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