# Are Cantor-like sets disjoint for $\xi,\eta$ with no common power?

Let $$\xi,\eta \in (0,\frac{1}{2})$$. Let $$C_\xi$$ (and analogously for $$C_\eta$$) be the perfect symmetric set built by iterating the transformation $$[0,1] \to [0,\xi]\cup [1-\xi, 1].$$

Will the sets $$C_\xi$$ and $$C_\eta$$ be disjoint, assuming $$\xi^n\neq\eta^m$$ for all $$n,m\ge1$$?

My approach has reduced the problem to this other question.

The answer is: no, they don't have to be disjoint (even after excluding trivial common elements $$0$$ and $$1$$).

Lets fix $$\xi$$ and let $$C_{\xi,n}$$ be the $$n$$-th step of the iterative process of creating $$C_\xi$$. If $$x\in C_{\xi,n}$$ then it is easy to see that $$\xi x\in C_{\xi,n+1}$$. On the other hand if $$x\in C_{\xi, n}$$ then $$1-x\in C_{\xi, n}$$ as well, because each set in the construction is symmetric. Lets solve the equality $$x=\xi(1-x)$$, which gives us $$x=\frac{\xi}{1+\xi}$$. Denote this number by $$A_\xi:=\frac{\xi}{1+\xi}$$.

By induction on $$n$$ (and previous properties) it is easy to see that $$A_\xi\in C_{\xi, n}$$ for any $$n$$, i.e. $$A_\xi\in C_{\xi}$$. An interesting case is when $$\xi=\frac{1}{3}$$ (the standard Cantor set) for which we have that $$A_\xi=\frac{1}{4}$$ belongs to $$C_\xi$$.

Intuitively we want a number that in each step we take its symmetry $$1-x$$ and then scale it down by multiplying by $$\xi$$. If we endup in the same number each time, then we can be sure that it belongs to the Cantor-like set.

This shows that $$A_\xi\in C_\xi$$ and for $$\eta=A_\xi$$ we have $$A_\xi\in C_\eta$$, because it is an endpoint of $$C_\eta$$. And so $$C_\xi$$ and $$C_\eta$$ have a common nontrivial point. Thus the only remaining question is whether $$\xi^n=\eta^m$$? Such equality would imply that $$(1+\xi)^m=\xi^{m-n}$$. By applying $$\log_{\xi}$$ we can conclude that this equation doesn't have solution when $$\log_{\xi}(1+\xi)$$ is irrational and it does otherwise. And since $$\log_\xi(1+\xi)$$, as a function of $$\xi\in(0,\frac{1}{2})$$, is continuous and not constant then there are infinitely (even uncountably) many such $$\xi$$.

For a concrete example consider $$\xi=\frac{1}{k}$$ ($$k>2$$ natural) for which $$\eta=A_\xi=\frac{1}{k+1}$$. For those it is easy to check manually that $$\xi^n\neq \eta^m$$, basically because $$gcd(k,k+1)=1$$.

All in all, there are uncountably many $$\xi$$ and $$\eta$$ such that $$\xi^n\neq \eta^m$$ and $$C_\xi$$ has a common (non-trivial, i.e. $$0$$ or $$1$$) element with $$C_\eta$$. Explicitly for $$\eta=\frac{\xi}{1+\xi}$$.

It is an interesting question whether there even are $$\xi$$ and $$\eta$$ such that $$C_\xi\cap C_\eta=\{0,1\}$$. I don't know that.

• Remarkable answer. I am afraid that there is a minor issue since the right-hand side can be greater than one if $n>m$. Nevertheless it is easy to see putting $x:=\frac{n}{m}$ that this equality has no solution for $x \in \mathbb Q$ Mar 3, 2022 at 13:56
• My next step is whether it is possible to prove the set of common points of $A_\xi$ and $A_\eta$ s countable. If it is I can construct uncountably many measures whose supports would still be disjoint Mar 3, 2022 at 13:58
• @MartinGeller ops, I totally forgot about negative powers. I fixed the answer. I think that whether it has a solution or not depends on $\log_\xi(1+\xi)$. Am I wrong? Mar 3, 2022 at 14:43
• @MartinGeller as for the set of common points: that I think is a lot harder. I don't even know where to start with it. Mar 3, 2022 at 15:04
• Oh yes you’re right. I was implicitly thinking that for $\xi= \frac{1}{4}$ this wouldn’t work Mar 3, 2022 at 20:17