2

Most books on introductory logic seem to work on a metatheory where infinite sets are allowed to exist. This seems unnecessary: everything humans do is finite, so seems like it should be enough to assume existence of only finite things. The more we accept in our (meta)theory the more likely it is that something goes wrong.

However, in the book Lectures in Logic and Set Theory. Volume 1 by Tourlakis, the author says:

If it were not for Gödel’s incompleteness results, this position – that meta mathematical techniques must be finitary – might have prevailed. However, Gödel proved it to be futile, and most mathematicians have learnt to feel comfortable with infinitary metamathematical techniques, or at least with $\mathbb{N}$ and induction. Of course, it would be reckless to use as metamathematical tools “mathematics” of suspect consistency (e.g., the full naïve theory of sets).

It seems to me that Gödel's results don't make the finitary position any more futile than the infinitary ones. We can't know consistency of strong enough systems anyway, so the finitary position still seems safer, even if not as convenient. It's also strange that it seems to be very hard to find recent textbooks who work with finitary metatheory. (unless i'm searching wrongly?). Are there other disadvantages to a finitary metatheory or am i missing something?

RobPratt
  • 41,014
  • 3
  • 19
  • 50
Carla_
  • 413
  • 1
  • 8
  • 2
    Some forms of induction are permitted even when using a finitary metatheory. The generally accepted finitary metatheory is Primitive Recursive Arithmetic (PRA), and induction is the key nontrivial axiom of PRA. Most relative consistency metatheorems can be translated into PRA. I definitely agree with your analysis of the situation. – Mark Saving Jul 21 '21 at 20:55
  • 4
    I think you should think hard about the first two sentences of your question. A metatheory that allows infinite sets is weaker than one that disallows them. The more assumptions we make in our metatheory, the more mistakes we are likely to make. To say "it is enough to assume existence of only finite things" makes our metatheory harder to work with. – Rob Arthan Jul 21 '21 at 22:13
  • @RobArthan Not sure i understand your comment. By "the more likely that something goes wrong" i'm not talking about human mistakes. A finitist metatheory may be harder to work with, but if everything is done completely rigorously it won't have higher likehood of mistakes. However if there's something fundamentally wrong about accepting infinite sets, then the infinitist one might go wrong, while the finitist one won't. – Carla_ Jul 22 '21 at 13:36
  • What do you mean by "fundamentally wrong"? – Rob Arthan Jul 22 '21 at 21:20
  • @RobArthan Perhaps it could lead to a contradiction. – Carla_ Jul 22 '21 at 21:25
  • If that is what you mean, then I would argue that you can't meaningfully talk about the likelihood of the system being fundamentally wrong. I fail to see how adding axioms that outlaw infinite sets can decrease the chance of inconsistency. I take@MarkSaving's point that you can view a restricted theory like PRA as a theory of sets of some kind and use that as a metatheory, but that seems to me to be unnatural and to have its own weaknesses from a philosophical point of view. – Rob Arthan Jul 22 '21 at 21:32
  • @RobArthan In order to talk about infinite sets in the metatheory you need to assume infinite sets exist. This assumption could lead to a contradiction. So in a sense, not assuming existence of infinite sets seems less likely to go wrong than assuming it. So my point is that it should be better not to assume them, if we can go without them. It also seems to me philosophically unnatural to accept infinite things: i've never seen anything infinite, afaik only finite things exist. I'd like to know why you find PRA as a metatheory unnatural and what weaknesses it has. – Carla_ Jul 22 '21 at 21:57
  • 2
    It seems to be unnatural because what we want to do in a metatheory is talk about syntax. In particular, we want to quantify over formulas, so we can state and prove things like the deduction theorem. In PRA, we can give an algorithm that implements the deduction theorem, but we can't state clearly what it does. Please also note that you do not need to assume the existence of objects in order to talk about their properties and it is standard practice to reason about non-existent objects in mathematics (proof by contradiction). – Rob Arthan Jul 22 '21 at 22:05

0 Answers0