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Definition of incommensurability:

  • Supposing $a,b \in \mathbb{R}^*$, $a,b$ are called incommensurable if and only if $\frac{a}{b}\notin\mathbb{Q}$.

From the lectures of KAM theory concerning the small divisor problem, I wonder:

  1. Is it possible to construct a non denumerable subset $S$ of $\mathbb{R}^\ast$ such that $\forall \left(a,b\right) \in S^2, a,b $ incommensurable?
  2. Same question without admitting the Axiom of Choice.

The problem is inspired by the fact that every integrable Hamiltonian system of $N$ degrees of freedom in classical mechanics, shall have a $N$-periodic motion. In the case which two fundamental frequencies of such system are commensurable (called the degeneracy), there will exist additional physical conserved quantities. So it would be interesting to see if one's able or not to construct a non denumerable set of incommensurable real frequencies, for which is the necessary condition of the existence of a classical integrable system with non denumerable degrees of freedom. (Normally the classical smooth fields in finite dimensions are systems with only denumerable degrees of freedom)

Xander Henderson
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TheoVereka
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  • Don't you mean $\frac ab\notin\mathbb Q$ in the definition? – Hagen von Eitzen Nov 04 '23 at 13:13
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    You've got it backwards. To say that $a,b$ are *commensurable* means that $a/b \in \mathbb Q$, and to say they are *incommensurable* means that $a/b \not\in\mathbb Q$. Besides that, posts consisting of bare problem statements without any context can easily attract close votes on this site. – Lee Mosher Nov 04 '23 at 13:15
  • Thank you, Hagen and Lee, for pointing this error out, and for informing me the good manner of posting questions in this community. I edited the post. – TheoVereka Nov 07 '23 at 19:22
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    See https://mathoverflow.net/questions/23202/explicit-big-linearly-independent-sets for explicit uncountable subsets with a stronger property. – Eric Wofsey Nov 07 '23 at 19:50
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    This question often arises in set theory when one considers the real numbers as a vector space over the rationals. There is a “Hamel basis” $H$ of reals with the property that each real $r$ has a unique representation as a finite sum $r = \sum q_ih_i$ where the $q_i$ are rational and $h_i\in H$. Such a basis always has cardinality ${\mathfrak c} = |\Bbb R|$. (See for example). However, the existence of the Hamel basis is equivalent to the axiom of choice. – MJD Nov 07 '23 at 20:08

1 Answers1

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Given some $x \in \mathbb{R}$, let $[x]$ denote the equivalence class of $x$, which is defined to be $$ [x] = \{ y \in \mathbb{R} : x/y \in \mathbb{Q} \}. $$ Note $[x]$ is precisely the set of real numbers numbers which are commensurable with $x$. Note, also, that $[x]$ is countable: the function $$ \mathbb{Q} \to [x] : q \mapsto qx$$ is bijective, and the rationals are countable.

The set of real numbers can be written as the union of all of these equivalence clases. Via the axiom of choice, choose a representative from each one of these equivalence classes, and let $S$ denote the set of these chosen representatives. Then $$ \mathbb{R} = \bigcup_{x\in S} [x]. $$ Each equivalence class $[x]$ is countable, and so $\mathbb{R}$ has been expressed as a union of sets, each of which is countable. But $\mathbb{R}$ is uncountable, so there must be uncountably many sets in that union. Therefore $S$ is uncountable.

Note that this argument does rely on the axiom of choice, and I do not see any way of "constructing" such an $S$ without reliance on it (which does not mean that no such method exists, I just don't see any "obvious" way to get around the axiom of choice).

Xander Henderson
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    Since it is consistent with ZF that the real numbers is a countable union of sets with 2 elements each, I don't think you can avoid AC. – Arturo Magidin Nov 07 '23 at 19:46
  • @ArturoMagidin Yeah, that is my intuition, but I lack the formal knowledge / experience / training to feel confident about getting the details right. I live in a world where the AoC is taken for granted. – Xander Henderson Nov 07 '23 at 19:50
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    +1, just beware to restrict everywhere to nonzero real numbers. Note for the OP: this answer essentially says that the projective space (= set of one-dimensional subspaces) of the $\Bbb Q$-vector space $\Bbb R$ is uncountable. – Anne Bauval Nov 09 '23 at 15:21
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    You can avoid AC: follow this link, given yesterday in @EricWofsey's comment. – Anne Bauval Nov 09 '23 at 15:31
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    @Arturo: It is consistent that the reals are a countable union of countable sets. Yes. But every countable union of finite sets of reals is countable, so not equal to the whole set. The reason is simple: since the reals are linearly ordered we can uniformly enumerate all the finite subsets. – Asaf Karagila Nov 09 '23 at 16:55