This is my first time to ask a question here. Please tell me if I can improve it.
I would like to introduce the following definition inspired from a measure theory exercise.
Definition. A subset $K$ of $\mathbb{R}$ is called a finite universal set iff for any finite subset $S$ of $\mathbb{R}$, there exists real numbers $a\neq 0$ and $b$ such that $aS+b\subset K$.
Question. What's the relation between the finite universality and dimension 1 (in a suitable sense, e.g. Hausdorff)?
I will explain why I have this question. Begin from the following fact related to the Lebesgue measure $m$:
Claim. If $m(K)>0$, $K$ is finite universal.
It's not hard but a little tricky to prove this claim. Notice that such a property is invariant under stretching and translation. Let $|S|=n$, denote the points in $S$ as $q_1, q_2,... q_n$. Choose suitable $a$, $b_1$, $b_2$, such that $aS+b_1, aS+b_2\subset [0,1]$. Moreover, $[aq_i+b_1,aq_i+b_2]$ are asked to be $n$ disjoint closed intervals in $[0,1]$. By Lebesgue density theorem, one may choose an interval $I$ such that $m(I\cap K)>(1-\frac{b_2-b_1}{n})m(I)$. By the invariance of stretching, we may assume that $I=[0,1]$. With the measure assumption of $I$, we have: $$m\Big(([aq_1+b_1,aq_1+b_2]\cap K)+a(q_n-q_1)\cap ([aq_2+b_1,aq_2+b_2]\cap K)+a(q_n-q_2)\cap ... \cap ([aq_n+b_1,aq_n+b_2]\cap K)\Big)>n\Big((b_2-b_1)-\frac{b_2-b_1}{n}\Big)-(n-1)(b_2-b_1)=0.$$ Thus it's non-empty. Choose a point $q_0$ in it, one has $aS+(q_0-aq_n)\subset K$.
The positive measure is not a necessary condition of finite universality.
Example. Let $C_n$ be the Cantor set but digging out $1/n$ instead of $1/3$. Consider $$K=\bigcup_{n\in \mathbb{Z^+}}\bigcup_{s\in \mathbb{Z}} (C_n+s)$$ This is a measure 0 but finite universal set.
Oppositely we have:
Claim. If $K$ is finite universal, then the packing dimension $\mathrm{dim_P}(K)=1$.
All tuples of $n$ distinguished points in $\mathbb{R}$ moduling stretching and translation form an $(n-2)$-dim manifold $\mathcal{G}$. Consider the projection $\pi:\mathbb{R}^n\to \mathcal{G}$. By definition $\pi(K^n)=\mathcal{G}$. Thus $\mathrm{dim_H}(K^n)\geq n-2$ for any $n$. So $n\mathrm{dim_P}(K)\geq \mathrm{dim_H}(K^n)=n-2$. this implies the conclusion.
Now you can see, the finite universality is stronger than packing dimension 1, but weaker than positive Lebesgue measure. Naturally we are wondering if this purely algebraic condition is equivalent to some measure theory properties exactly.
By the way, if we slightly change the definition, we can get some similar properties:
Definition. A subset $K$ of $S^1$, the unit circle, is called a countable universal set iff for any countable subset $S$ of $S^1$, there exists a rotation $r$ such that $rS\subset K$.
Claim. If $m(K)=m(S^1)=1$, $K$ is countable universal.
Claim. Following two conditions are equivalent:
(1) There exists a countable universal set of measure less than 1.
(2) There exists a countable universal set of measure 0.
Question. What's the exact answer of the claim above?
I'm not sure if these questions have been studied in fractal geometry, at least I didn't find any reference. I would be super appreciated if someone can solve them or show me something to learn.
