# Existence or otherwise of a set of “sufficiently intricate” open cells

In my question Existence or otherwise of a set of "sufficiently intricate" open sets, I asked about whether it is possible to partition Lebesgue-almost all of $\mathbb{R}^d$ into a finite number of connected open sets in such a manner that for some vector $\mathbf{v} \in \mathbb{R}^d$, none of these open sets contains two points whose difference is $\mathbf{v}$.

If $d=3$, it is possible to achieve this with just two open sets: Imagine a three-dimensional "solid grid" where the "nodes" are cubes of side-length 1 and the "connections between nodes" are also cubes of side length 1. We take one of the open sets to be the interior of this "grid", and we take the other open set to be the interior of the complement of this "grid"; and we let $\mathbf{v}$ be a vector that takes you from one corner of a length-1 cube within this grid to the diagonally opposite corner.

Since such a partition can be found for $d=3$, it immediately follows that such a partition can be found for $d > 3$. (E.g. just take the preimages of the above two sets under the natural projection $\mathbb{R}^d \to \mathbb{R}^3$.)

However, if $d=2$ then such a partition does not exist!

I now wish to slightly "strengthen" my question:

Fix $d \in \mathbb{N}$. Do there exist mutually disjoint open sets $V_1,\ldots,V_n \subset \mathbb{R}^d$, each homeomorphic to $\mathbb{R}^d$, and $\mathbf{v} \in \mathbb{R}^d$ such that

• $\mathbb{R}^d \setminus (\bigcup_{i=1}^n V_i)$ is a Lebesgue-null set, and
• for each $i \in \{1,\ldots,n\}$, for every $\mathbf{x} \in V_i$, $\,\mathbf{x}+\mathbf{v}$ does not belong to $V_i$?

If so, I am also interested in knowing about the structure of the set of possible vectors $\mathbf{v}$, particularly whether it just consists of isolated points:

Given mutually disjoint open sets $V_1,\ldots,V_n \subset \mathbb{R}^d$, each homeomorphic to $\mathbb{R}^d$, such that $\mathbb{R}^d \setminus (\bigcup_{i=1}^n V_i)$ is a Lebesgue-null set, define $$N_{[V_1,\ldots,V_n]} \, := \, \{ \mathbf{v} \in \mathbb{R}^d \, : \, \forall \, i \in \{1,\ldots,n\}, \ \forall \, \mathbf{x} \in V_i, \ \mathbf{x}+\mathbf{v} \not\in V_i \, \}.$$ (So the previous part just asks whether it is possible for $N_{[V_1,\ldots,V_n]}$ to be non-empty.)

If $N_{[V_1,\ldots,V_n]}$ is non-empty, is it possible that $N_{[V_1,\ldots,V_n]}$ contains a bounded infinite set, or even a curve [image of a continuous function on an interval] of points?

Update: As in the answer of Sebastian Goette below (which was given before I added my additional part to the question), the answer to the first part is yes: in the "solid grid" example, one can just cut out plane segments from the grid and from its complement (which is also a grid) in such a manner that the resulting two sets are homeomorphic to $\mathbb{R}^3$.

Now in this grid example, I believe the set $N_{[V_1,V_2]}$ is precisely given by $\mathbf{v}+G$ where $\mathbf{v}$ is a vector taking one corner of the unit cube to the diagonally opposite corner, and $G$ is the additive group generated by the vectors taking each corner of the length-2 upright cube to the diagonally opposite corner. [Someone can correct me if this is wrong!]

It still remains open as to whether there exists an example of a situation where $N_{[V_1,V_2]}$ contains a bounded infinite set.

• The second question seems to be most interesting for $d=3$. If for higher dimensions, you simply take preimages under projections, then $N_{[V_1,V_2]}$ will admit a free $\mathbb R^{d-3}$-action. – Sebastian Goette Nov 19 '15 at 13:05
• I also believe the second question has a positive answer once you construct $k>2$ sets. Maybe you want to try yourself, by regarding remainders of $\lfloor x_i\rfloor$ mod $k$? – Sebastian Goette Nov 19 '15 at 13:18
• @SebastianGoette Thank you for this - yes, I'm pretty sure you're right, this should work. You've been really helpful, thank you. – Julian Newman Nov 19 '15 at 15:56
• @SebastianGoette: Actually, constructing the $k>2$ sets by considering remainders of $\lfloor x_i \rfloor$ mod $k$ seems not to be so straightforward (if it is even possible at all); $k=2$ has the special property that $3k-2=k^2$ [the only other number with this property is 1]. ($3k-2$ is the volume of the repeating unit in a "solid grid" where the side-length of the repeating unit is $k$ times the volume of a "node".) I'll carry on trying to think about it - but I wonder if you already have some more detailed idea of how to do the construction? – Julian Newman Nov 20 '15 at 17:30
• I noticed that problem. That is why I stopped - it would have taken too much time. I think for $k=3$ you can get an interval and its translates, and I though about trying $k=5$ to get a whole cube - but that looked indeed more difficult. – Sebastian Goette Nov 20 '15 at 18:04

Take the three-dimensional example, which comes from a grid. Find a maximal tree inside the grids corresponding to $V_1$ and $V_2$, so pick a subset of the set of edges that connects each node to every other one, but creates no loops. Now, "cut" the cubes corresponding to the remaining edges in $V_1$, $V_2$ by taking away a piece of a hypersurface. What remains are two thickened trees, hence homeomorphic to $\mathbb R^3$.