An example of a similar universal cover for 5 points.

Question

Let's say that $A\subset\mathbb{R}^2$ is a similar universal cover for $n$ points if:

1. $A$ is closed.
2. The interior of $A$ is empty.
3. For every finite set $B\subset\mathbb{R}^2$ containing exactly $n$ points there is a set $A'\subset\mathbb{R}^2$ which is similar to $A$ such that $B\subset A'$.

Let $S_n=\{A\subset\mathbb{R}^2|A$ is a similar universal cover for $n$ points$\}$.

Examples

1. Any line segment is an element of $S_2$.
2. A letter "T" is an element of $S_3$.
3. A circunference with a radius is an element of $S_4$: if the four points are colinear, you can put them on the radius. If they're not, then you can form a circunference with three of them so that the fourth is inside the circle. Then you can rotate the radius until you cover that last point.

Can you find an example of an element of $S_5$?

Answer 1

I am not sure what happens with your definition, but here is an answer when we modify (2) to

(2') $A$ has Hausdorff dimension $\le 1$ (e.g. $A$ is a countable union of smooth curves).

Such sets $A$ have empty interior but the converse is false.

I will use the notation $S_n'$ for the collection of corresponding $n$-universal subsets of $R^2$ (by the way, you should not use the name "universal cover" for this, it is already taken).

Here is a proof that $S_n'=\emptyset$ for $n\ge 5$.

Consider the $n$-fold product $A^n$ of $A\subset R^2$ satisfying (2'); $A^n\subset R^{2n}$. The group $G$ of similarities of the Euclidean plane is a Lie group of dimension $4$. The group $G$ acts smoothly on $R^2$ and we obtain the corresponding diagonal action of $G$ on $R^{2n}$: $g(z_1,...,z_n)=(gz_1,...,gz_n)$, where $z_k\in R^2$ for each $k$. In other words, we have the smooth map $\mu: G\times R^{2n}\to R^{2n}$, $$ \mu(g, z_1,...,z_n)= (gz_1,...,gz_n). $$ The statement that $A$ satisfies your universality condition is equivalent to the property that $\mu(G\times A^n)=R^{2n}$. However, $\mu$ is locally Lipschitz (since it is smooth) $G$ has Hausdorff dimension $4$, $A^n$ has Hausdorff dimension $\le n$, hence, $\mu(G\times A^n)$ has Hausdorff dimension $\le 4+n$. For $n\ge 5$, $4+n< 2n$ and $2n$ is the Hausdorff dimension of $R^{2n}$. (For $n=4$ we have the equality.) Hence, $\mu(G\times A^n)$ cannot equal $R^{2n}$. Thus, $A$ cannot belong to $S'_n$.

The same argument works if (2') is replaced by

(2'') $A$ has Hausdorff dimension $<2$.

I let $S_n''$ denote the corresponding collection of universal subsets of $R^2$.

Then the result is that $$ \bigcap_{n\ge 1} S_n'' =\emptyset. $$

I do not know what happens for your original definition of $S_n$. One can attempt to replace Hausdorff dimension with topological dimension. Then $int(A)=\emptyset$ means that $dim(A)\le 1$, hence $dim(A^n)\le n$. However, there are examples of smooth maps which raise topological dimension. For instance, one can take a topological curve $P\subset {\mathbb R}^3$ (the graph of a Peano curve), such that for the action $\mu$ of ${\mathbb R}$ on ${\mathbb R}^3$ by translations along the $x$-axis, $\mu({\mathbb R}\times P)$ is 3-dimensional, equal to the product of the unit square and the real line.

Edit. The fact that $$ \bigcap_{n\in {\mathbb N}} S_n \ne \emptyset $$ is proven in Theorem 1.12 in

C. G. Wastun, Universal covers of finite sets. J. Geom. 32 (1988), no. 1-2, 192–201.

In fact, he proves even more: There exists a closed subset $A$ with empty interior in $E^2$ such that for each finite subset $F\subset E^2$ there exists a horizontal translation $T$ of $E^2$ such that $T(F)\subset A$. His sets $A$ are products of certain closed, perfect, totally disconnected subsets of the x-axis with the y-axis.

Answer 2

I think I found an example, but I'm curious because I think it has Hausdorf dimension <2.

Let $C\subset[0,1]$ as the Cantor set obtained by removing open intervals formed by numbers whose base 16 expansion does not have the "digit" 15. Ends of this intervals, like $0.(15)000\dots=0.(14)(15)(15)(15)\dots$ stay on $C$. Now define $B=\bigcup_{i=0}^4(i+C)$, so that $B\times B\subset[0,5]\times[0,5]$. $B\times B$ is closed and has empty interior.

I'll show that $B\times B$ is in $S_5$: Let $A=\{z_1,z_2,z_3,z_4,z_5\}\subset\mathbb{R}^2$. Let $Q$ be a similarity transformation such that $Q(A)\subset[0,5]\times[0,5]$. Now I show that there is a translation $L$ such that $L(Q(A))\subset B\times B$.

Let $p_1(x,y)=x$ and $x_j=p_1(Qz_j)$ and assume, without loss of generality, that $x_1<x_2<x_3<x<4<x_5$. Now, define $d_i=x_{i+1}-x_i$ for $i=1,\dots,4$. I show now that we can realize this four distances in $B$:

Let $d_i-[d_i]=0.d_{i1}d_{i2}\dots$ in base 16, where $[\cdot]$ stands for the floor function. To show that we can realize the four distances in $B$, we only need to prove that we can built a number $p=0.p_1p_2\dots$ (in base 16) so that for every $j$ none of the numbers

$p_j$,

$p_j+d_{1j},p_j+d_{1j}+1$,

$p_j+d_{1j}+d_{2j},p_j+d_{1j}+d_{2j}+1,p_j+d_{1j}+d_{2j}+2,$

$\dots$,

$p_j+d_{1j}+\dots+d_{4j},p_j+d_{1j}+\dots+d_{4j}+1,\dots,p_j+d_{1j}+\dots+d_{4j}+4$,

is congruent with 15 mod 16. Noticing that this list of numbers has at most 15 distinct numbers, we can choose $p_j$ so that none of them is congruent with 15 mod 16. In this way, we can realize the four distances in $B$, so that we can translate $Q(A)$ so that all the projections into the $x$-axis of its elements lie in $B$. Analogously, we can do the same for the projections to the $y$-axis, so that there is a translation $L$ such that $L(Q(A))\subset B\times B$. So, $B\times B$ is in $S_5$.

It is not so difficult to see that this process can be generalized for $n\leq6$, so that $S_n\neq\emptyset$ $\forall n\in\mathbb{N}$.

However, $B\times B$ has Hausdorff dimension <1, so that $S'_5\neq\emptyset$ and therefore $S''_5\neq\emptyset$. Please chech this, because I can't find the mistake on the previous proof nor in mine.