15
$\begingroup$

I found this problem, and I can't get an answer to it:

Prove that there are subfields of $\Bbb{R}$ that are

a) non-measurable.

b) of measure zero and continuum cardinality.

I can't seem to imagine how to construct such subfields of $\Bbb{R}$.

  • 0
    For (b), my idea would be to use the [cantor set](http://en.wikipedia.org/wiki/Cantor_set) and have a look at the subfield of $\mathbb{R}$ generated by this set.2011-06-22
  • 0
    @Beni Bogosel: Old result of Kuratowski, but may have used CH, I don't remember.2011-06-22
  • 0
    Are there some extension fields of $\Bbb{Q}$ which may turn out to be non-measurable?2011-06-22
  • 0
    The standard cantor set $E$ certainly doesn't work because $E - E$ contains an interval. You'll need to remove a lot more than middle thirds.2011-06-22
  • 0
    @Beni Bogosel: they would have to be extensions of $\mathbb Q$, because every subfield of $\mathbb R$ contains $\mathbb Q$.2011-06-22
  • 0
    @Robert Israel: I realize that. I was going to something like $\Bbb{Q}[\sqrt{3}]$, or an extension like $\Bbb{Q}[\sqrt{2},\sqrt{3},\sqrt{5},...,\sqrt{prime\ number}...]$. How can we study measurability of this fields?2011-06-22
  • 0
    For (b) you could consider the set of reals which, up to a rational multiple, have a very low density of nonzeros in their binary expansion. Consider the set of $x\in\mathbb{R}$ which can be expressed as $x=q\sum_{n=0}^\infty2^{-n}a_n$ for rational $q$ and integers $a_n$ satifying $\sum_{n=0}^N\vert a_n\vert=O(N^\epsilon)$ (all $\epsilon > 0$).2011-06-22
  • 0
    ...that just came to mind, but is it really a field? Or just a ring?2011-06-22
  • 0
    @George Lowther: Yes. The question is about fields. The answer given below by JDH has a positive answer to both question, but in a complicated manner. I hope some more easier examples will surface...2011-06-22
  • 0
    Unless I'm mistaken, any set of Hausdorff dimension 0 should generate a field of Hausdorff dimension 0. This is because, if $A$ and $B$ have Hausdorff dimension 0, so does $A \times B$, and $A + B$ and $A B$ are images of $A \times B$ under locally Lipschitz functions.2011-06-22
  • 0
    If you replace $O(N^\epsilon)$ by $o(\log N)$ then the field generated by such numbers are either rational or are Liouville numbers, so has measure zero.2011-06-22
  • 0
    @Beni: "Easy" is rather subjective. I was thinking about explicit examples, which do not require using the axiom of choice.2011-06-22
  • 0
    @Robert: Very nice. So the field generated by a "thin" Cantor set (Hausdorff dimension 0) will give an example of (b).2011-06-22
  • 0
    @Robert, what about inverses?2011-06-23
  • 0
    @JDH: $z \to 1/z$ is a Locally Lipschitz function, so it also preserves Hausdorff dimension.2011-06-24
  • 2
    @Robert, @George, so if this is a solution, then why not post a fully explained answer?2011-06-24
  • 0
    @BeniBogosel How have you "found" this question ? Did you come up with it on your own or have you found it in a book/article etc. ? (Sorry for the late commentary on a question know 5 years old.)2016-04-21
  • 0
    @temo The question is quite old, so I don't really remember. I guess it was in some book on linear algebra or on some other topics in real analysis2016-04-22
  • 0
    @BeniBogosel Ok, thanks anyway for taking the time to answer after all this time.2016-04-26

3 Answers 3

9

I expect direct constructions, but while we are waiting, let me at least offer a proof that both assertions are consistent with ZFC, since they can be obtained by forcing.

Statement (b) is true of the ground-model reals after adding a Cohen real. In the answer to this MO question, Martin Goldstern explains that the set of reals in the ground model $\mathbb{R}^V$ has measure $0$ in the forcing extension $V[c]$, but it is easy to see that it has size continuum there. So this is a subfield of $\mathbb{R}$ in the universe $V[c]$ which has measure $0$ and size continuum.

Statement (a) is true in the models mentioned in Andreas Blass' comment to my answer to this MO question, where it is explained that the set of constructible reals, which is certainly a subfield, can be non-measurable afte forcing over the constructible universe.

Perhaps one can turn both of these arguments into actual ZFC constructions by considering partially generic filters. Or perhaps there are easier direct constructions.

  • 0
    Thank you very much for your quick answer. I will wait for some simpler direct constructions, if they are possible.2011-06-22
  • 0
    I will accept your answer, although it is not very simple to understand. Have you found any other simpler solutions?2011-06-27
  • 0
    Beni, thanks for accepting, although I still expect someone to post a direct construction (such as the Hausdorff dimension argument mentioned in the comments). So please feel free to unaccept later.2011-06-28
4

Rather bizarrely, here's an answer to one of the questions, but I'm not sure which. But maybe it's a useful starting point.

Pick a transcendence basis of $B$ of $\mathbb{R}$ over $\mathbb{Q}$, any $b \in B$, and look at the field $F$ generated by $B\setminus \{b\}$. Since the translates of $F$ by elements of $\mathbb{Q}b$ are pairwise disjoint, $F$ is either null or nonmeasurable.

I suspect (hope?) that this is nonmeasurable, maybe after taking algebraic closure. If this is so, then a strategy for getting a null field could be tossing out more elements of $B$.

2

You can try the answers (including mine) for a question at MathOverflow... https://mathoverflow.net/questions/27352/a-question-about-fields-of-real-numbers/27358#27358