Comment by math_comment_21

If I had to write again I might say "same theorems about natural numbers" and capitalize ROUGHLY. It is a conversation, what exactly I am weaseling around (not just nonstandard model theoretic issues), and I take your caveat about consistency strength - with that said would you still call it misleading? Why is it that eg x+y=y+x for x y given takes exponential length proof in Robinson compared to PA? For the reason stated, which is true in a very broad sense.

>> If for all n we can prove P(n), then `∀n P(n)` should be an acceptable proposition

> But how can you prove that P(n) for all n without induction?

Just to be clear to all readers, the axiom of COUNTABLE choice is uncontroversial. Nobody is disturbed by induction.

The issue it that when you allow UNCOUNTABLE choice - choices being made for all real numbers (in a non-algorithmic way, I believe - so not a simple formula) - there are some unpleasant consequences.

> But how can you prove that P(n) for all n without induction? Maybe I misinterpret what you're saying, or I'm naive about something in formal languages, but if we can prove P(n) for all n. then `∀n P(n)` just looks like a trivial transcription of the conclusion into different symbols.

Of course it is likely that an interesting result about all positive integers, that is "really" about positive integers, is proved by induction, but you certainly don't need induction to prove P(n): n = 1.n, or, more boringly, P(n): 0 = 0. (These statements are obviously un-interesting, both in the human sense of the word and in the sense that they are just statements about semi-rings, of which the non-negative integers are an example.)

My understanding is that the difference between "For every positive integer n, I can prove P(n)" and "I can prove ∀n.P(n)" is that the former only guarantees that we can come up with some terrible ad hoc proof for P(1), some different terrible ad hoc proof for P(2), and so on. How could I be sure I have all these infinitely many different terrible ad hoc proofs without induction? I dunno, but that's all that the first statement guarantees. Whereas the second statement, in the context of computability, guarantees that there is some computable function that takes a positive integer n and produces a proof of P(n); that is, there is some sort of guaranteed uniformity to the proofs.

I think it may be easier to picture if connected with math_comment_21's analogy in https://news.ycombinator.com/item?id=44269153: the analogous statements in the category of topological spaces (I think one actually has to work about topos, but I don't know enough about topos theory to speak in that language) about a map f : X \to Y are "every element of Y has a pre-image under f in X" versus "I can continuously select, for each element of Y, a pre-image of it under f in X", i.e., "there is a continuous pre-inverse g : Y \to X of f."

Rejecting induction could be quite useful if you want to be very precise about the implications of your constructions wrt. computational complexity. This is of course only a mildly strengthened variant of the usual arguments for constructivism.

As I pointed out at https://news.ycombinator.com/item?id=44271589, there are systems that can accept Cantor's argument, without concluding that there are more reals than rational numbers.

As you point out, there are many mathematical systems that contain all of the numbers in the high school mathematics concept of "reals". Since those with a high school understanding of reals do not know which of those systems they would agree with, they should not be asked to accept as true, any results that hold in only some of those systems.

And that is why I don't like mathematicians telling lay audiences that there are more reals than rationals.

For sure there would have to be a meta theory of some sort. But, I think that many classical mathematicians would be happy with that meta theory to be weaker, rather than stronger. I don't think there is a need for that theory to have AC. After all, thanks to its independence of from the other axioms, AC does not add logical strength to a classical theory.

(For die-hard constructivists, such as myself, the story is of course different. But that story is for a another time. I am presenting the classical view here.)