# Does anyone understand this comment about the continuum hypothesis?

At 31:37 in his lecture titled What is a manifold? posted on Youtube, Mikhail Gromov states that if we do not allow generic functions to exist then the continuum hypothesis is "obviously" true, and that if we do allow generic functions to exist, then the continuum hypothesis is "obviously" false.

What concept is he referring to? It sounds extremely important.

Note: this is a follow-up to a question I posted on Math.SE: https://math.stackexchange.com/questions/1887350/what-is-a-generic-genetic-geometric-map-in-the-study-of-manifolds

Also I'm not sure how to tag this question, so please feel free to fix the tags as appropriate.

EDIT: When Gromov mentions generic functions for the first time in the lecture, around 29:30 if I remember correctly, it seems like it is with regards to generic points (which are generic by Sard's Theorem I believe) for which the Implicit Function theorem can be used to generate a manifold from the equation $f(x)=0$; the functions $f$ for which this holds are what he refers to as "generic". I don't know if this helps anyone at all; I don't really understand Gromov's terminology.

• It may mean generic in the sense of forcing. See also Generic filter. – Mohammad Golshani Aug 18 '16 at 6:28
• Given that Gromov is being very philosophical about what we can and can't know about what 'arbitrary' manifolds look like, I would take such remarks at best as being indicative, rather than referring to an actual mathematical principle. I edited to put the title of the talk in your actual post, to save people having to load up a YouTube video to check what it was (I've watched that talk, so instantly knew the context of the quote once I knew it was that one). – David Roberts Aug 18 '16 at 9:26
• Yet another interpretation: If we could randomly and independently choose a real number for each real number, the random functions obtained might be the generic functions in question. And the resulting functions would verify the premise behind Freiling's argument against CH. – Michael Greinecker Aug 19 '16 at 16:13
• I believe what is meant here is different, simply that CH is true for "explicit" sets/functions (Borel or, if you accept large cardinals, even projective and somewhat more complicated things), while if you allow "typical" sets/functions into your picture, then "many more" show up. – Andrés E. Caicedo Aug 20 '16 at 14:29
• I sat through a talk of Matteo Viale where he proved that if you add a Cohen real, then as a real, it's generic in the algebraic geometry sense also (he then used this to give a neat proof to some known statement related to Schanuel's conjecture). This lends itself to what @Andrés wrote. And that would be my interpretation of the comment as well. – Asaf Karagila Aug 21 '16 at 5:35

## 3 Answers

I remember reading in Proof from the Book (starting p. 119) that Erdős proved the equivalence between the continuum hypothesis and the following problem : Let $(f_\alpha)$ be a pairwise distinct family of analytic functions over $\mathbb{C}$, sur that for any $z\in \mathbb{C}, (f_\alpha(z))$ is countable. Is the family $f_\alpha$ is itself countable? (Edit : more precisely, $c > \aleph_1$ iff any such family is countable, otherwise you have such a family with cardinality $c$). It looks like a good way to interpret the CH towards an explication of Gromov's statement.

• Sure, answer edited. – vfmmk Aug 20 '16 at 12:40
• How is this related to Gromov's statement? On its face Erdos's result says there are more functions under CH. Of course it really means there are more points in $\mathbb{C}$ where the family could be uncountable --- but I don't see what that has to do with the question. – Nik Weaver Aug 20 '16 at 16:27
• *fewer points in C – Nik Weaver Aug 21 '16 at 23:18

You might read "if we do not allow generic functions to exist then the continuum hypothesis is obviously true" as a reference to the fact that Borel sets satisfy CH, or possibly to the fact that CH holds in Godel's constructible universe. For the "obviously false" part, I suspect a nod to Paul Cohen's comment:

A point of view which the author feels may eventually come to be accepted is that CH is obviously false. The main reason one accepts the axiom of infinity is probably that we feel it absurd to think that the process of adding only one set at a time can exhaust the entire universe. Similarly with the higher axioms of infinity. Now $\aleph_1$ is the set of all countable ordinals and this is merely a special and the simplest way of generating a higher cardinal. The set $c$ is, in contrast, generated by a totally new and more powerful principle, namely the power set axiom. It is unreasonable to expect that any description of a cardinal which attempts to build up that cardinal from ideas deriving from the replacement axiom can ever reach $c$. Thus $c$ is greater than $\aleph_1$, $\aleph_\omega$, $\aleph_\alpha$ where $\alpha = \aleph_\omega$, etc. This point of view regards $c$ as an incredibly rich set given to us by one bold new axiom, which can never be approached by any piecemeal process of construction.

I feel disingenuous not adding that my personal view is that $\aleph_1$ and the real line are both proper classes, so that CH is not a meaningful question.

• have you expanded on that personal view anywhere? – Will Sawin Aug 18 '16 at 13:09
• @WillSawin: yes, for instance in my answer to this question. – Nik Weaver Aug 18 '16 at 13:12
• Why not support Pocket set theory for the 2016 presidency, then? – Asaf Karagila Aug 18 '16 at 13:36
• @NikWeaver, viewing N as a proper class was also a theme in the work of Vopenka. That this should rile traditionalists goes without saying. +1 – Mikhail Katz Aug 18 '16 at 13:51
• @ToddTrimble: I don't have the exact reference, but it's in the last chapter of Set Theory and the Continuum Hypothesis. (I got the quote just now by Googling some keywords that I remembered.) – Nik Weaver Aug 18 '16 at 14:24

See this answer by Hamkins on multiverses and switches. What Gromov probably means is that if we allow the full power of set-theoretic methods like forcing then there is no reason whatsoever to assume that the set-theoretic universe we are working in should necessarily satisfy CH (an example of a switch in the sense of Hamkins).

• And by that same token, there is no reason whatsoever to assume that the set-theoretic universe we are working in should necessarily satisfy $\neg$ CH. That doesn't seem to jibe with Gromov's "obviously false". – Todd Trimble Aug 18 '16 at 14:20
• @ToddTrimble It seems that a possible reading of generics/no generics is $\mathsf{CH}$ vs forcing axioms. For instance, PFA implies $\neg\mathsf{CH}$. – Pedro Sánchez Terraf Aug 24 '16 at 18:06