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Mathematical induction can be done using the axiom of induction, which is given as a formula written in the language of mathematical logic. Is there a way to express the ideas behind 'real induction' in mathematical logic as well?

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RandomDSdevel
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I believe that Theorem 3 of that paper, when unwound, is what you are looking for. That theorem just says that whenever a totally ordered set is Dedekind complete, then its only inductive subset is the whole thing. (Actually, it says the converse too, but I believe this direction is the one that interests you.)

Do you really want to see real induction written down as a sentence in the language of set theory? (The language of arithmetic isn't suitable for RI since RI talks about arbitrary total orderings.) This can be done, but it's uuuglyyy . . .

To give a sense of what I mean by "ugly," here's the predicate "$X$ has the greatest upper bound property" written out entirely in symbols:

$$\forall A(\exists y\forall a(a\in A\implies a<y)\implies\exists y(\forall a(a\in A\implies a<y)\wedge \forall z(z<y\implies \exists b\in A(z<b))).$$

(Quantifiers here range over $X$ or $\mathcal{P}(X)$ as is appropriate.) And this is just one tiny piece of the principle of real induction: we would need to translate the whole statement "for total orders, Dedekind complete iff principle of ordered induction" into symbols.

The point is: since induction on arbitrary totally ordered sets is much more general than induction on $\mathbb{N}$ alone, we should expect the statement of the former to be much longer than the statement of the latter; in particular, we shouldn't be surprised if it can't be written in one short string of symbols.

Noah Schweber
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  • Here's a good exercise: show that the usual principle of induction for $\mathbb{N}$ follows from the principle of ordered induction. First show that $\mathbb{N}$ is Dedekind complete, and then think about it . . . – Noah Schweber Jun 09 '15 at 22:49
  • I don't care if the 'string of symbols' is short, I just care to find out if it can be written! – RandomDSdevel Jun 14 '15 at 22:08
  • Also, what does proving that 'whenever a totally ordered set is Dedekind complete, then its only inductive subset is the whole thing' have to do with proving that a statement is true for all real numbers? Do I have to prove that the set of all elements $x$ for which a predicate $\mathrm{P}\left(x\right)$ is true is logically equivalent to the set $\mathbb{R}$ of all real numbers? – RandomDSdevel Jun 14 '15 at 22:21
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    If you want to show $P$ holds for every real number $x$, it's enough to show that ${x: P(x)}$ is inductive - since we then know that $P$ holds of every real, since the only inductive subset of $\mathbb{R}$ is $\mathbb{R}$ itself, and we know that since $\mathbb{R}$ is totally ordered and Dedekind complete. – Noah Schweber Jun 15 '15 at 02:43
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    Maybe this makes things clearer: Theorem 3 of Clark's paper shows that, if you want to prove that $P$ holds for every real number, it's enough to show that the set of reals for which $P$ holds is inductive, where the precise meaning of inductive is given on the top of page 3, just before Theorem 3. Now, let "$(*)$" denote the statement "if ${x: P(x)}$ is an inductive subset of $\mathbb{R}$, then $P(x)$ holds for every real $x$." This statement can be easily, if tediously, written in symbols by unwinding the definition of "inductive", and this is a good exercise. – Noah Schweber Jun 16 '15 at 02:09
  • OK, I've taken a stab at it: $((\exists \mathbb{S} \subset \mathbb{X}) \land (\exists a \in \mathbb{X} : {(({-\infty}), a]} \subset \mathbb{S}) \land (\forall x \in \mathbb{S} \ldotp (x = \mathcal{T}(\mathbb{X})) \lor (\exists y > x : [x, y] \subset \mathbb{S})) \land (\forall x \in {\mathbb{S} \setminus {\mathcal{B}(\mathbb{S})}} \ldotp {({(-\infty}), s)} \subset \mathbb{S} \Rightarrow x \in \mathbb{S})) \Rightarrow \mathbb{S} = \mathbb{X}$. Does that look right to you? – RandomDSdevel Jul 10 '15 at 20:08
  • Ack, crazy line breaking! Oh, well… – RandomDSdevel Jul 10 '15 at 20:09
  • Have you seen my latest comments above? – RandomDSdevel Aug 01 '15 at 17:28
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Your question can be interpreted in various ways. In a sense, the axiom of induction for cardinalities larger than countable is simply the axiom of choice, in the form of transfinite induction (or recursion as it is sometimes referred to). Along these lines you may find Transfinite induction on the real line to be of interest. It discusses numerous modern application of induction on the real numbers.

Ittay Weiss
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  • By 'real induction,' I meant the kind of induction discussed in this paper. – RandomDSdevel Jun 09 '15 at 21:47
  • I'm not sure what else you want then. The article you quote proves that what is called there the principal of real induction is equivalent to the completeness property (it even goes further to prove the same for general ordered sets). It is a restatement of some standard proof techniques in $\mathbb R$ and other posets. – Ittay Weiss Jun 09 '15 at 21:53