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The goal of this question is to recall a certain mathematical fact -not in my field- that I was once briefly told and that I have fogotten, and also to collect similar results.

The fact, I think, was about the undecidability (read: independence from ZFC or maybe ZF axioms) of a certain seemingly very "natural" sentence about the convergence of sequences of holomorphic functions in one variable.

So the question is:

What are some natural undecidable sentences about holomorphic functions? Where by "undecidable" I mean independent from the ZFC or the ZF axioms of set theory, and by "natural" I mean something that is not manifestly designed to be an independence result and possibly that arised quite autonomously from Logic.

Edit: I'm aware there are some independence results (I think by Kranz and Di Biase) related to the boundary behaviour of holomorphic functions. The "fact" I wanted to recall is not part of this theory, though independence examples related to this theory are well accepted in the answers.

edited Feb 20 at 14:39

asked Feb 20 at 14:34

Qfwfq
8,473●33●85

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Since apparently there is no "unique" answer here, but this rather asks for a list of results of a certain flavor I thinnk this should be CW. – quid Feb 20 at 14:58

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I don't agree that a question like this needs to be CW, in part because I don't think we will have such a huge list of answers. Why shouldn't those who can come up with good examples be rewarded for doing so? – Joel David Hamkins Feb 20 at 16:16

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I don't really understand your remarks on up-and-down voting, since I think CW does not affect one's ability to vote in either direction. I had read your first comment as basically saying that the question should be CW, lest someone get too many unearned points. But I guess you have some more complicated reasoning about CW going on. – Joel David Hamkins Feb 21 at 2:22

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@Joel: I think he is saying that he would like to be able to downvote some answers to more freely reorganize the rankings, but he does not feel comfortable doing so if someones points are on the line. It is more appropriate to downvote a CW answer. – Steve Feb 21 at 9:58

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Thanks, Steve, but does this make any sense? To avoid the pain of -2, one omits future +10s? I don't really understand the perspective. My own view is that mathematically substantive questions should not be put into CW mode simply because they have several answers. Indeed, I think that very few MO questions actually admit a unique answer, perhaps none do, since one can often contribute ideas from different perspectives or additional gloss and so on. – Joel David Hamkins Feb 22 at 0:15

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quid · Accepted Answer · 2013-02-20 18:06:01Z UTC

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Very likely the fact that you are trying to remember is the interpolation problem solved by Erdős.

http://www.renyi.hu/~p_erdos/1964-04.pdf

edited Feb 20 at 18:06

quid

answered Feb 20 at 17:09

Nik Weaver
4,417●1●7●20

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You beat me to it! But to be fair, I just vaguely remembered there was such a result, and then applied creative googling to find, e.g. Problem 16 of dpmms.cam.ac.uk/study/III/2012-13/… – Adam Epstein Feb 20 at 17:13

I also spent well over 3 minutes trying to get an umlaut, any style umlaut, over the o in "Erdos". :) – Adam Epstein Feb 20 at 17:15

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well I spent at least 10 minutes staring blankly into space, so I feel I should get the credit – Nik Weaver Feb 20 at 17:22

So now I am going to have to read this paper. – Adam Epstein Feb 20 at 17:26

You should. It's really easy and cute. – Nik Weaver Feb 20 at 17:27

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quid · Answer 2 · 2013-02-20 18:03:17Z UTC

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There is a famous result of Erdős in An interpolation problem associated with the Continuum Hypothesis concerning families $\cal F$ of entire functions $f:\mathbb{C}\rightarrow\mathbb{C}$ such that for every $\zeta\in\mathbb{C}$ the set {$f(\zeta): f\in\mathcal{F}$} is countable.

Theorem

(1) If $2^{\aleph_0}>\aleph_1$ then every such family is countable.

(2) If $2^{\aleph_0}=\aleph_1$ then every such family has cardinality $2^{\aleph_0}$.

edited Feb 20 at 18:03

quid

answered Feb 20 at 17:12

Adam Epstein
946●2●12

I added the right type of o, since you seem to care. (A way, not super convenient but less than three minutes, is copy-paste from Wikipedia.) – quid Feb 20 at 18:04

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Quid has fixed the spelling in the text, but for the record, the TeX command for ő (which wouldn’t work here, as MathJax only supports math mode) is not \"o, but \H{o}. – Emil Jeřábek Feb 20 at 18:10

Emil - I actually tried that first, and it also came out wrong, at least in the previewer. Quid -thanks. – Adam Epstein Feb 20 at 18:58

Yes, this was a remark on general TeX usage. As I wrote, this site does not support LaTeX text-mode commands, such as diacritics. – Emil Jeřábek Feb 20 at 19:40

The proof is very elegant, and it appears in "Proof from the Book" by Aigner and Ziegler. – Malik Younsi Feb 20 at 20:24

Rodrigo A. Pérez · Answer 3 · 2013-02-20 14:54:07Z UTC

The paper Constructing Non-Computable Julia Sets by Mark Braverman and Michael Yampolsky gives examples of quadratic polynomials with non-computable Julia sets. In particular, the question of deciding whether some point belongs to the Julia set is intractable.

Undecidability and holomorphic functions (Reference request)

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