4
$\begingroup$

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.

$\endgroup$

1 Answer 1

5
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ 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 $x \in \mathbb Q$ $\endgroup$ Mar 3, 2022 at 13:56
  • $\begingroup$ 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 $\endgroup$ Mar 3, 2022 at 13:58
  • $\begingroup$ @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? $\endgroup$
    – freakish
    Mar 3, 2022 at 14:43
  • $\begingroup$ @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. $\endgroup$
    – freakish
    Mar 3, 2022 at 15:04
  • $\begingroup$ Oh yes you’re right. I was implicitly thinking that for $\xi= \frac{1}{4}$ this wouldn’t work $\endgroup$ Mar 3, 2022 at 20:17

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .