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I'm looking for an open problem in analysis or number theory with just one "genuine" or "second order" quantifier.

E.g.

  • "Every continuous function $\mathbb{R} \rightarrow \mathbb{R}$ has the property $\theta$", where $\theta$ is expressible using only quantifiers over rationals.

  • "Every set $S$ of natural numbers has the property $\theta$", where $\theta$ is expressible using only quantifiers over rationals.

No cheat examples like "For every real number, Goldbach's conjecture holds"! That's an arithmetical problem.

In technical terms, I'm looking for a $\Pi^1_1$ sentence that we don't know how to reduce to an arithmetical sentence.

I'd also like it to be easy to state and obviously $\Pi^1_1$, so that it can be included in a logic paper without requiring much explanation.

Paul Blain Levy
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  • I don't know, but you could have a look at https://www.tandfonline.com/doi/abs/10.1080/0020739970280111?journalCode=tmes20 and https://www.math.ksu.edu/~ramm/papers/547.pdf and https://math.stackexchange.com/questions/1095743/innocent-looking-open-problems-in-real-analysis and https://math.stackexchange.com/questions/58638/open-conjectures-in-real-analysis-targeting-real-valued-functions-of-a-single-re and https://link.springer.com/article/10.1007/s00020-018-2460-8 and https://mathoverflow.net/questions/100265/not-especially-famous-long-open-problems-which-anyone-can-understand – Gerry Myerson Aug 08 '21 at 12:06
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    Problems of the form: is there a countable graph with such-and-such first-order property? Is there a countable structure with such-and-such first-order property? Negative answers would have the desired form $\Pi^1_1$. – Joel David Hamkins Aug 08 '21 at 12:18
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    $\forall (x,y).\zeta(x+iy)=0\implies x={1\over 2}$ is a natural thing to write, but RH turns out to have a $\Pi^1_0$ form. So your question has subtleties. – none Aug 08 '21 at 17:02
  • @none I think you mean to say that RH has a $\Pi^0_1$ form, that is, a purely arithmetic form as a universal arithmetic assertion. – Joel David Hamkins Aug 08 '21 at 19:14
  • Oops, yes, I can't edit the comment any more though. This thread's answers have some more possibilities. Is there a theorem of Takeuti that practically everything in classical analysis can be encoded in Peano arithmetic? That might make this question difficult. – none Aug 08 '21 at 21:21
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    I suspect the four exponentials conjecture and some other problems in transcendental number theory could qualify. – Wojowu Aug 09 '21 at 15:35
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    Sorry, what is meant by "genuine" quantifier? Doesn't the term "continuous function" implicitly contain quite a few quantifiers? (EDIT: Oh I see -- the condition that a given function be continuous is arithmetical, so the only analytic quantifier is over the function itself) – Tim Campion Aug 09 '21 at 15:52
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    @TimCampion a continuous function can be presented as a continuous function on rationals. – Paul Blain Levy Aug 11 '21 at 10:11
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    There are (open) special cases of the invariant subspace conjecture that are of the form requested by the OP, e.g., Conjecture 8 of https://terrytao.wordpress.com/2010/06/29/finitary-consequences-of-the-invariant-subspace-problem/ . – Terry Tao Aug 11 '21 at 15:10
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    @MattF. That's a fair point. Can you think of a way to phrase, say, Schanuel's conjecture in arithmetic way? – Wojowu Aug 12 '21 at 17:18
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    @Wojowu, I gave an arithmetic translation of Schanuel’s conjecture in my answer. –  Aug 15 '21 at 04:41
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    A quick suggestion: to avoid clutter, maybe the candidates which turned out to have arithmetical equivalents should be deleted? – Noah Schweber Aug 17 '21 at 19:48
  • @NoahSchweber done. Thanks for the suggestion! – Paul Blain Levy Aug 17 '21 at 20:41
  • I'm curious as to how this works exactly. For all analytic functions $F(z) : \mathcal{J} \to \mathbb{C}$ such that $F(e^z) = F(z)+1$ and $F(0) = -1$, then $F$ must have the form $\text{slog}(z)$. Where here $\text{slog}$ is Kneser's slog and $\mathcal{J}$ is the julia set of $\exp$ excluding periodic points. This is an open problem in iteration theory. I'm wondering if this is along the lines. There is only one quantifier as I understand it, but I'm not too sure. – Richard Diagram Aug 18 '21 at 01:45
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    @RichardDiagram this looks $\Pi^1_1$ to me, but I can't say for certain. – Paul Blain Levy Aug 18 '21 at 02:45
  • @PaulBlainLevy I agree, but I'm not the best at this sort of stuff, lol. – Richard Diagram Aug 18 '21 at 02:50

1 Answers1

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The Littlewood conjecture is an example that meets all my requirements.

It is easy to state and obviously $\Pi^1_1$. Furthermore this comment by Christian Reiher gives me confidence that it has no known reduction to an arithmetical sentence. (Hopefully it even lacks a known reduction to a $\Sigma^1_1$ sentence.)

Paul Blain Levy
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  • Thanks to everyone that helped, either by suggesting possible answers, or by pointing out that various $\Pi^1_1$ sentences suggested by me or others did in fact have a known reduction to an arithmetical sentence. – Paul Blain Levy Aug 17 '21 at 20:39