0

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.

  • 2
    Well your second idea fails since there is no smallest positive real. – coffeemath Jun 04 '22 at 21:23
  • @coffeemath Hmm, I thought limits could help somehow, but maybe not. (Perhaps I need a more formal introduction to calculus using epsilon-delta, which I haven’t learnt yet). –  Jun 04 '22 at 21:24
  • 1
    Your first idea works with a fixed $a$, but it leads back to ordinary induction regarding the claims $\forall t: a_n\le t<a_{n+1}\implies P(t)$ for subintervals $[a_n,a_{n+1})$ of length $a$. – Berci Jun 04 '22 at 21:42
  • This basically becomes a question for topology. E.g. For connected sets, the corresponding argument seems to be: $P(t_0)$ is true for some $t_0$ (base case) and for any $t$, there is some neighborhood $N$ of $t$ such that $P(t) \iff P(s)$ for all $s$ in $N$ (inductive step). Therefore $P(t)$ is true for all $t$ – Brian Moehring Jun 04 '22 at 21:54
  • 2
    Does this answer your question? Induction on Real Numbers – Patrick Stevens Jun 04 '22 at 21:55
  • @BrianMoehring Wow, math is truly interlinked. I don't know much about topology, but thanks for sharing. –  Jun 04 '22 at 21:58
  • @BrianMoehring That fact is used in Bernstein's "continuity method" of solving PDE, where a "family" of PDE is parametrized by a sequence and you prove that each PDE has a solution by using exactly the logic you propose. Typically, the method of extension involves the implicit function theorem. – Sarvesh Ravichandran Iyer Jun 05 '22 at 13:41

1 Answers1

2

There are some proofs more or less of this kind, but not precisely this kind. I know the two approaches. One close to the compactness of closed bounded sets and the other -- the connectedness of intervals. This can be used in a more general situation than just $\Bbb R$. In both cases we can use the fact that some property is satisfied locally and prove that it is satisfied globally.

Two examples:

  • We know that continuous functions are locally bounded but in compact sets we can conclude that they are globally bounded.
  • If a function is locally constant on a connected space then it's globally constant.

The problems with your ideas are the following:

  • As @coffeemath pointed out, there is no smallest positive real number. Even if we try to use the idea of infinitesimals, we are stuck.
  • We'd like to prove that $x<1$ for all $x\in [0,\infty)$. First we see that $0$ satisfies this inequality. Now, if some $x$ satisfies this inequality, that is if $x<1$, then all the numbers from $[x,x+\varepsilon)$ also does for some small $\varepsilon$. This doesn't prove that each positive number is smaller than 1.
Mateo
  • 5,216
  • +1 for examples and for "This can be used in a more general situation than just $\mathbb{R}$"; that is interesting! –  Jun 04 '22 at 22:05