10
$\begingroup$

The question is essentially the one in the title.

Question. What are some examples of (major) problems in set theory which are solved using techniques outside of mathematical logic?

$\endgroup$
16
  • 17
    $\begingroup$ Getting tenure? :) $\endgroup$
    – Asaf Karagila
    Aug 15, 2018 at 8:08
  • 4
    $\begingroup$ Can you give an example problem for which this could conceivably happen? I have a feeling that if someone asks, for example, a consistency question about Banach spaces, and then it turns out to have a ZFC answer via methods internal to functional analysis, then we would just say it wasn't a set-theoretical problem after all. $\endgroup$ Aug 15, 2018 at 8:35
  • 4
    $\begingroup$ More of a speculation than an answer: It's conjectured that every Suslin ccc forcing adds a Cohen real or a random real. Shelah showed that if such a forcing adds an unbounded real then it adds a Cohen real. Farah and Zapletal showed that if $\mathbb P$ is Suslin ccc and $\omega^{\omega}$-bounding then $RO(\mathbb P)$ is a Maharam algebra. Therefore, the above problem reduces to the following: Does every Maharam algebra adds a random real? Talagrand showed that there is a Maharam algebra which is not a measure algebra, solving an old problem by Von Neumann... $\endgroup$
    – Haim
    Aug 15, 2018 at 19:55
  • 11
    $\begingroup$ I don't think logic or set theory have such sharp boundaries that it is possible to provide a definitive answer. Huge parts of set theory, such as Borel equivalence relation theory or set-theoretic topology, are deeply connected with other related areas, and it could sometimes be difficult to describe a method as existing in only set theory or the companion area. $\endgroup$ Aug 17, 2018 at 22:38
  • 6
    $\begingroup$ If you had asked for logic problems outside of logic, rather than set theory problems outside of set theory, then maybe Tarski's problems about the first order of free groups theory fit the bill, having been solved by methods of geometric group theory. $\endgroup$
    – Lee Mosher
    Aug 17, 2018 at 22:43

2 Answers 2

8
$\begingroup$

Inspired by Joel David Hamkin's comment---Simon Thomas has provided applications of various super-ridigity theorems (from the ergodic theory of group actions) to the theory of the Borel complexity of countable equivalence relations, for example he shows that the universal countable equivalence relation is not essentially free

Thomas, Simon, Popa superrigidity and countable Borel equivalence relations, Ann. Pure Appl. Logic 158, No. 3, 175-189 (2009). ZBL1162.03029.

$\endgroup$
0
6
$\begingroup$

I think the question could in principle have good answers (not turning too much on what is a method from logic) even though no one has found any yet. As a counterfactual hypothetical example, Cantor invented the ordinal numbers motivated by considerations from Fourier analysis. Maybe he also tried using Fourier analysis in his attempts to prove CH. He didn't succeed, but imagine that he did. Fourier analysis is classical mathematics from well outside of logic, so that would be a clearly satisfactory answer. One can imagine more such examples, so the question is really: has anything like that ever actually worked? Maybe not, but could it happen in principle? I dunno. I remember that it's possible to prove the Banach-Tarski paradox using the Hahn-Banach theorem (which is grounded in a weak form of AC), though that probably doesn't "count" since it doesn't actually recast BT as a functional analysis problem.

Dropping down from set theory, in computability theory, there's a famous and surprising theorem of Barrington that the complexity class NC1 can be solved by branching programs of fixed width 5, but not width 4 or less. The reason 5 is the minimum is because S5 (the symmetric group on 5 letters) is not solvable but S4 and smaller are. So that's group theory finding its way into a computability problem. No reason such things can't happen in set theory.

(Too long for a comment, I guess).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.