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This question is inspired by another question I had, where I wanted to prove something about the natural numbers.

Often in analysis books I see some proofs, where they use the natural numbers, but it is not really proved why they can argue like they do with the natural numbers, and I never even thought it was a problem, because it seemed so reasonable. Here are two different cases I have seen, it is the first one I am wondering about:

  1. We make a procedure where we look for something in a function that has the natural numbers as it domain. We start with 1, and if f(1) does not have the desired property we create an alement $a_1$, if f(2) does not have the property we want, we create an element $a_2$, and move on. Then we conclude that either we will have a finite number of elements $a_1,a_2,..,a_n$, or we will get an infinite sequence $a_1,a_2,...$, where every natural number k, will occur as $a_k$.

  2. We have a set S with the natural numbers, and we know that there is an n in the natural numbers that has this property. We want to show that there is a smallest element in here that has this property, we start by 1, and if one does not have this property we go to 2, if 2 not, we go to 3 etc., then we know that we will stop before n, or at n.

I have seen some books that have made arguments similar to these. Property 2 is the well ordering property, and is proved using induction. But can property 1 also be proved with induction?

Here is the example of the proof that uses property 1. It is from measure theory, and all the details are not important for this question, the only thing that is important for this question is why he can say what he says where I have marked in red:

enter image description here

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  • $\begingroup$ Is it accurate to call this induction in the way you have described, since the n case does not imply the n + 1 case? It is valid but it not induction unless one case implies the next. $\endgroup$ – theREALyumdub Aug 27 '15 at 1:21
  • $\begingroup$ @theREALyumdub: It is induction in the form of an inductive construction. $\endgroup$ – user21820 Aug 27 '15 at 1:53
  • $\begingroup$ Property (1) looks like it's essentially omega-consistency in disguise; effectively it's saying that either there is some specific $i$ for which $f(i)$ is true, or we have $\forall k\neg f(k)$. $\endgroup$ – Steven Stadnicki Aug 27 '15 at 2:58
  • $\begingroup$ I'm confused as to what you're asking. Any set is either finite or infinite, right? $\endgroup$ – Akiva Weinberger Aug 27 '15 at 3:09
  • $\begingroup$ My understanding of the question is: "We start by writing down $a_1$. At each step we either write down $a_k$ (where $k$ is the smallest number not yet written) or we stop. Then prove that we either end up with a finite sequence $a_1,\dots,a_n$ or the infinite sequence $a_1,a_2,\dots$." Is that the right understanding? $\endgroup$ – Akiva Weinberger Aug 27 '15 at 3:13
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Your property 1 is effectively the Recursion Theorem. It is often (sloppily) referred to as an "inductive procedure," which is to some extent reasonable, because we need the Principle of Mathematical Induction (or, equivalently, the well-ordering property of the natural numbers) to prove it.

The rough idea is that if we can start a procedure, and explicitly state (1) under what conditions the procedure must stop and (2) how specifically to proceed to the next step if said conditions aren't met in a given step, then either the process is well-defined, and proceeds for a (possibly finite) sequence of steps. (As user21820 aptly points out, we must do more than simply state the above. We must prove that we can start the procedure, and that we can continue so long as the conditions for stopping aren't met.)

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  • $\begingroup$ This is not enough. Just because you can extend a sequence to any arbitrary finite length does not mean that you can obtain the infinite sequence! Moreover, depending on how you do it, you may need some kind of choice axiom. See for example math.stackexchange.com/a/717963/21820 for some details. $\endgroup$ – user21820 Aug 27 '15 at 2:58
  • $\begingroup$ You are incorrect. Note the use of the word "deterministically." By this, I mean that there is a function $f$ such that, given some $a_n$ that does not meet the conditions for termination, we define $a_{n+1}:=f(a_n).$ This uses the Recursion Theorem, and no choice is required. $\endgroup$ – Cameron Buie Aug 27 '15 at 3:00
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    $\begingroup$ Okay sorry I misinterpreted your use of "deterministically" since you said "deterministically say" rather than "say ... how to proceed deterministically ...". The former might mean you "deterministically prove the ability to proceed". If you edit it I can change my vote. $\endgroup$ – user21820 Aug 27 '15 at 3:03
  • $\begingroup$ Does that change in wording make it less ambiguous? $\endgroup$ – Cameron Buie Aug 27 '15 at 3:05
  • $\begingroup$ Thank you, by the way, for pointing out the ambiguity! I had not considered that interpretation. $\endgroup$ – Cameron Buie Aug 27 '15 at 3:08

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