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With "normal" induction you usually proof a statement for the natural numbers $\mathbb{N}$. But I think that you can extend this idea to any series/ordered set of numbers when using the natural numbers $\mathbb{N}$ as the index of the series elements. But I am still unsure how such a proof (over the integers $\mathbb{Z}$ for example) would look like.

Can you prove all of these example statements with induction?

(1): $\forall n \in \mathbb{N}: n+1>n$ (This is obviously possible)

(2): $\forall k \in \mathbb{Z}: k+1>k$

(3): $\forall x \in \mathbb{R^+}: x+1>x$

I think for (3) you could use the intervall ]0,1] as the base case and then the normal induction step. But would it still be possible with only one real number as the base case?

And I think all series/ordered set have cardinality $\aleph_0$ or lower. So can you generally say that induction only works for sets of cardinality $\aleph_0$?

Qwox
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1 Answers1

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What you are looking for is called "Transfinite induction". The set can be arbitrarily large, but has to be "Well-ordered".

Simon
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  • So you can't use induction over the real numbers because those aren't well-ordered? – Qwox Oct 19 '21 at 13:17
  • @Qwox You can, if you use the well-ordering theorem to order it in a well-order. Be warned that this order is very very different than the on you are used to. – Rushabh Mehta Oct 19 '21 at 13:23
  • (1) every set, indcluding the real numbers, can be well-ordered (using the axiom of choice), but the usual natural ordering of the real numbers is not a well-ordering. So in practice no. (2) but your approach of using an interval like $]0,1]$ does of course work with the normal induction – Simon Oct 19 '21 at 13:23
  • @Qwox If you want to use the same order, what you proposed with using $[0,1)$ as a base case would work. Or, you could do something like $P(x)\to P(x+1)$ for all naturals $x$, $P(0)$ and $P(x)$ implies $P(y)$ for all $y<x$. – Rushabh Mehta Oct 19 '21 at 13:24
  • @DonThousand How would a well-ordered subset of the real numbers look like? My problem is, I think, that I don't get how a set can be well-ordered and uncountable infinite at the same time. For example in the intervall ]0,1] the lower bound is the problem. I know that you could approach it with a function like 1/2^n, but that misses so many other real numbers. – Qwox Oct 19 '21 at 15:42
  • @Qwox Great question. In fact, you can't even write down what it would look like. That's part of the challenge of the axiom of choice. The well-ordering of the reals is so messy that there is no formula that can describe it. – Rushabh Mehta Oct 19 '21 at 15:55
  • @DonThousand For Example some Property is easy proofable for x=0 but you want to proof it for $x \in [0,1]$. That wouldn't be possible with induction, right? Edit: As a side note: I just want to write something about induction and I am not sure whether induction is only for sets of cardinality $\aleph_0$ – Qwox Oct 19 '21 at 16:01
  • @Qwox Not with the standard order, no. – Rushabh Mehta Oct 19 '21 at 16:02