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We know that for $\mathbb{N}$ the Well-Ordering Principle is equivalent to Induction.

But what about $\mathbb{R}$? The induction principle does exist for $\mathbb{R}$ as well, but is it equivalent to the well-ordering of $\mathbb{R}$ in that formulation? If not what makes $\mathbb{R}$ special? And if we were to reformulate it for $\mathbb{Q}$ only would it be equivalent to the well ordering of $\mathbb{Q}$ (my intuition says yes since it has the same cardinality as the naturals)?

(Principle of Real Induction) Let $a<b$ be reals, and $S\subset[a,b]$, if:

  1. $a\in S$,
  2. $\forall x\in S \ x\neq b \implies \exists y>x : [x,y]\subset S$,
  3. $\forall x\in\mathbb{R}\ [a,x)\in S\implies x\in S$,
    then $S=[a,b]$.
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Stupid Questions Inc
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    I'm not sure I understand your question. Can you point to a page in the linked PDF where it discusses induction on the reals? You can do it, using what is called "transfinite induction" by well ordering the reals, which might answer your question. I'm not sure though, because induction on the reals is a somewhat nonstandard topic – HallaSurvivor Dec 04 '19 at 10:09
  • @HallaSurvivor I already put the page that I meant in the URL but certain browsers apparently don't redirect you there, anyway it's page 117 i will put it directly in the post (i don't know if it's equivalent to transfinite induction *cough*zorn's lemma*cough*) – Stupid Questions Inc Dec 04 '19 at 10:11

1 Answers1

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The linked method of "Real Induction" is not actually induction (though as a logician, I might be being a pedant about the use of the word "induction" here...). Instead, it is an application of the topology of the reals.

Loosely, what the "Principle of Real Induction" is saying is "if I have the bottom of $[a,b]$, and I can always get bigger, then I know I have $[a,b)$. But if I also have closure, I must have $[a,b]$." This is a slick way of proving properties about intervals, if we know we can always make our set of "good" points grow upwards. It is not induction, however, because we are adding LOTS of points at a time. When we make the move from $[a,x]$ to $[a,y]$, we are adding in all the uncountably many points inside $(x,y]$.

"True" Induction (the logician in me wants to remove the scare quotes, but the decent human wants to leave them in) is characterized by doing things one step at a time. Think about the natural numbers: We prove it for 0. That lets us prove it for 1. That lets us prove it for 2. That lets us prove it for 3. and so on.

The way to actually induct on the reals (actually the real polynomials, but it's the same idea) might be as follows. This isn't necessarily a good example, it's just the first I came up with.

Say we have $\mathbb{R}$, and we want to make it so that every polynomial factors into linear pieces (and pretend we don't know what $\mathbb{C}$ is...). One overkill way of doing this might be to take the set of all polynomials $\mathbb{R}[x]$ and well order it (using the axiom of choice). This gives us a good notion of the "first polynomial", followed by the "second polynomial", and so on. Then we'll go one at a time, and add in all of the roots of each polynomial. At the end of time, we'll be left with a field in which every polynomial has a root. I'm glossing over a fair number of technical details, most notably we have "limit stages" in the induction, but these are not much of a bother.

This is not to say that the "Real Induction" you've linked to isn't useful! Quite the opposite. In fact, here is an excellent PDF with applications of this (badly named, imo) technique. It can clean up a lot of analysis in a very pleasing way. Indeed, it seems like many mathematicians enjoy using this tool, and think it is aptly named.

More philosophically, it is a fact of mathematics that different people will have different opinions about what is or isn't an intuitive name for an object or technique. Often, the names we choose align with how we view the relationships between different branches of math, and this can shed some light on how we think. I would never call this tool "induction", because it doesn't rely on a well ordering for its use. However I can see the similarities, and how somebody with a more analytical background might view it as a very similar tool in spirit, which justifies the name. This is part of the beauty (and, occasionally, pain) of working with mathematics.

Let me know if you have any follow-up questions. I'll be happy to explain more, or clarify anything that you feel I haven't addressed.

I hope this helps ^_^

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HallaSurvivor
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  • While your answer has cleared that up a few things, the question still stands: is there a version of induction for the reals (for example the intervals have natural numbers so we'll be dealing with sets that can be bijected to $\mathbb{N}$) such that it is equivalent to the well-order of the reals? – Stupid Questions Inc Dec 04 '19 at 10:43
  • Yes, but it has nothing to do with $\mathbb{N}$. Instead, we would put $\mathbb{R}$ in bijection with $\omega_1$ (https://en.wikipedia.org/wiki/First_uncountable_ordinal) and use the fact that $\omega_1$ is well ordered. – HallaSurvivor Dec 04 '19 at 10:45
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    @HallaSurvivor you are assuming continuum hypothesis. – Slup Dec 04 '19 at 10:51
  • @Slup, yes (for convenience). There's no need to get bogged down in subtleties in a question like this. You're right, though, that I should have pointed it out! – HallaSurvivor Dec 04 '19 at 17:09
  • @HallaSurvivor Another question, what is real induction equivalent to? – Stupid Questions Inc Dec 05 '19 at 05:29
  • I'm not sure it's "equivalent" to anything. It relies on the ordering (but not a WELL ordering) $<$ of the reals, as well as the topology. I'm not sure if there's much more to say, unfortunately – HallaSurvivor Dec 05 '19 at 05:40
  • @HallaSurvivor here's another question that I had, assume $N$ has the same cardinality as $\mathbb{N}$ then we construct a well-ordering of $N$ that is the "same" as the well-ordering of $\mathbb{N}$ by using a bijection between them (they have the same cardinality so it exists), hence can we construct a principle of induction for any set $N$ that has the same cardinality of $\mathbb{N}$? – Stupid Questions Inc Dec 13 '19 at 20:20
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    Yes! Indeed this is exactly how transfinite induction works. You fix a bijection between your set of interest and some well ordered set (often using the axiom of choice). Then you transfer the well order along this bijection. If your set is countable, then you can use $\mathbb{N}$, though it is usually called $\omega$ in this context. – HallaSurvivor Dec 13 '19 at 20:24
  • @HallaSurvivor So if I had a countable set with a bijection $\varphi$ all I'll need to show is that $P(\varphi(0))$ is true and $P(\varphi(n))\implies P(\varphi(n+1))$ ? :) – Stupid Questions Inc Dec 14 '19 at 09:30
  • Yup! Because then you can conclude $P(\varphi(n))$ for every $n$, and by bijective-ness, that's $P(x)$ for every $x$ in your other countable set. – HallaSurvivor Dec 14 '19 at 18:02