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Define a set $S \subset \mathbb{R}^d$ as a transcendental distance set if the distance between any pair of distinct points of $S$ is transcendental. For example, $S = \{ k \, \pi \;\mid\; k=1,2,\ldots \}$ is such a set in $\mathbb{R}^1$.

Q1. How large can a transcendental distance set be in $\mathbb{R}^d$?

The above $S$ is countably infinite. What is an example of an uncountably infinite $S$ in $\mathbb{R}^1$, and more generally in $\mathbb{R}^d$? It would seem that random sets should be transcendental distance sets.

Suppose one further filters, in $\mathbb{R}^d$ requiring that the area of all triangles determined by $3$-point subsets be transcendental, the volume of all tetrahedra determined by $4$-point subsets be transcendental, and so on up to the measure of $d+1$ simplices. Call such a set a transcendental measure set.

Q2. How large can a transcendental measure set be in $\mathbb{R}^d$?

I know there is work on "Avoiding rational distances" (e.g., Ashutosh Kumar. Real Analysis Exchange 38.2 (2012): 493-498.) that may answer my questions, but I have not yet seen direct implications.

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    $\begingroup$ If you assume axiom of choice, I believe it's straightforward to use transfinite induction to construct transcendental-distance set of size continuum. I'm not sure, but I'd expect it to generalize to transcendental measure sets for any $d$. $\endgroup$
    – Wojowu
    Jul 26, 2015 at 14:26
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    $\begingroup$ If you believe that all algebraic irrationals are normal numbers, then if you take the set of all real numbers in [0,1] whose decimal digits are 5's and 6's say, then differences of this set are not normal numbers (they have no 3s in their expansion for example) and so cannot be algebraic. On the other hand, you cannot have a positive measure set because differences of positive measure sets contain intervals. $\endgroup$
    – Anthony Quas
    Jul 26, 2015 at 14:27
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    $\begingroup$ You can use Zorn's lemma and measure theory. Suppose $S$ is a maximal subset of $\mathbb{R}^n$ with no algebraic volumes - one exists by Zorn's lemma. Let $D$ be the set of all points $z$ of $\mathbb{R}^n$ such that $z$ forms a simplex of algebraic volume together with some $n$ points of $S$. Since the set of $n$-tuples of points of $S$ is countable, and the set of algebraic numbers is countable, you see that $D$ is a union of a countable collection of hyperplanes, thus has measure zero. This means that there is a point in the complement to $D\cup S$, so $S$ is not maximal - contradiction! $\endgroup$
    – Dmitry Vaintrob
    Jul 28, 2015 at 5:25
  • $\begingroup$ The same sort of argument can be used if you're avoiding any countable collection of algebraic relations. $\endgroup$
    – Dmitry Vaintrob
    Jul 28, 2015 at 5:25

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Answer to this question provides an algebraically independent set of real numbers of size continuum. Enumerate it as $\alpha_{i,r}$ where $i\in\{1,...,d\},r\in\Bbb R$. Now define set $S=\{(\alpha_{1,r},...,\alpha_{d,r}):r\in\Bbb R\}\subseteq\Bbb R^d$. Now, measure of any simplex with vertices on these points can be expressed using coordinates of the vertices and radicals. If any of these turned out to be an algebraic number, we would have a non-trivial algebraic relation between $\alpha_{i,r}$, which is not possible, as they are algebraically independent.

As $|S|=\frak{c}$, we see that transcendental measure set can have size $\frak{c}$. By taking subsets of $S$, we can get transcendental measure set of any size at most continuum, and, clearly, not of any greater cardinality.

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    $\begingroup$ Thank you, especially for the link to the $\frak{c}$-sized algebraically independent set of reals. $\endgroup$
    – Joseph O'Rourke
    Jul 26, 2015 at 16:46

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