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It is well known that every number $x$ in the unit interval $[0,1]$ is the arithmetic mean of two elements of the (triadic) Cantor set $C$. The way to see it I like the most: the Cantor set is the unique fixed point of a self-map on the set of non-empty closed subsets of $[0,1]$, namely $\kappa: F\mapsto \frac13F+\big\{0,\frac23\big\}$; this map $\kappa$ is a contraction with respect to the Hausdorff distance, which is complete, and the subset of all closed sets $F$ such that $F+F=[0,2]$ is a closed non-empty $\kappa$-invariant set, so it contains the fixed point. (So, incidentally, if $C':=C-\frac12$, we have $C'-C'=[-1,1]$, which gives a counterexample to the converse of the Steinhaus theorem, stating that $A-A$ is a neighborhood of $0$ provided that $A$ has positive Lebesgue measure.)

The analogous result for products of two elements of $C$ is not true: already $C_1:=[0,\frac13]\cup[\frac23,1]$ happens to give a set of products with a gap: $C_1\cdot C_1=[0,\frac13]\cup[\frac49,1]\subsetneq[0,1]$. Things seem to go better for products of three Cantor set's numbers, and I would believe that they actually cover the whole unit interval, since it turns out that $C_1\cdot C_1\cdot C_1=[0,1]$. However, this is not sufficient to conclude the result for $C$: I'm not sure if e.g. the set of all closed sets $F\subset[0,1]$ such that $F+F=[0,2]$ and $F\cdot F\cdot F=[0,1]$ is actually $\kappa$-invariant.

Questions: is every $x\in [0,1]$ a product of (finitely many) Cantor set's elements? If so, is there a uniform bound on the set of factors? And, if so, what is the minimum $n$ such that any $x\in [0,1]$ can be expressed as a product of $n$ numbers in the Cantor set?

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    $\begingroup$ It may be useful to call it , at least the first time, "the triadic Cantor set". $\endgroup$
    – YCor
    Commented Dec 22, 2023 at 13:15
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    $\begingroup$ I'd suggest "every" in place of "any" in your title and questions. The word "any" is famously ambiguous between $\exists$ and $\forall$. In questions especially, it often seems more to carry the $\exists$ meaning, such as in "Was any student able to solve the problem?" or "Is any doctor on the plane?". But I believe you want the $\forall$ meaning. $\endgroup$
    – Joel David Hamkins
    Commented Dec 22, 2023 at 13:32
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    $\begingroup$ Hmm. I could be wrong, but it looks like the numbers that are products of $2$ members of the Cantor set should be a closed nowhere dense subset of $[0,1]$. If this were true for all $n \geq 2$, then you'd have your answer. $\endgroup$
    – Will Brian
    Commented Dec 22, 2023 at 13:42
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    $\begingroup$ Yes. arxiv.org/abs/1711.08791, "Cantor Set Arithmetic" by Athreya, Reznick, and Tyson, in the Monthly in 2019: every element in $[0,1]$ can be written as $x^2y$ for some Cantor set elements $x,y$. $\endgroup$
    – Zach Teitler
    Commented Dec 22, 2023 at 13:47

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Yes, every real number $u \in [0,1]$ can be written as $u = x^2 y$ where $x,y \in C$ are in the Cantor set $C$. In particular, every real number in $[0,1]$ is a product of three Cantor set elements. This is a result of Jayadev S. Athreya, Bruce Reznick, and Jeremy T. Tyson, in Cantor Set Arithmetic, Amer. Math. Monthly 2019, https://doi.org/10.1080/00029890.2019.1528121 (and on the arxiv: https://arxiv.org/abs/1711.08791).

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  • $\begingroup$ Nice paper… Am I wrong or theorem 3.1 (stating that $$C+\lambda C=[0,1+\lambda]$$ holds for every $1/3\le \lambda \le3$) in fact is also true for all $\lambda\ge0$ ? $\endgroup$
    – Pietro Majer
    Commented Dec 23, 2023 at 18:27
  • $\begingroup$ @PietroMajer Well, it doesn't work for $\lambda=0$. For $0<\lambda<1/3$ I don't think it works either: Let $C_1 = [0,1/3] \cup [2/3,1]$. For $\lambda < 1/3$ it seems to me that $C_1 + \lambda C_1 = [0,1/3+\lambda] \cup [2/3,1+\lambda]$, so it misses $(1/3+\lambda,2/3)$. $\endgroup$
    – Zach Teitler
    Commented Dec 23, 2023 at 23:29
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    $\begingroup$ Ops, sorry, I forgot to delete this naive comment. Yes, i realized immediately after, $1/3\le \lambda\le3$ it is exactly the condition to have the property $F+\lambda F=[0,(1+\lambda)]$ invariant for the set-contraction $F\mapsto \frac13F+\{0,\frac23\}$, therefore, to have the fixed point $C$ there! $\endgroup$
    – Pietro Majer
    Commented Dec 24, 2023 at 5:54

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