Existence or otherwise of a set of "sufficiently intricate" open sets Fix $d \in \mathbb{N}$. Do there exist mutually disjoint connected open sets $V_1,\ldots,V_n \subset \mathbb{R}^d$ 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 $V_i$?


[The answer is clearly no if $d=1$, but I have not managed to work out the answer for $d \geq 2$.]
 A: The answer is "yes" for $d\ge 3$ and "no" for $d\le 2$.
For $d=3$, let $V_1$, $V_2\subset\mathbb R^3$ be the open interiors of the sets
$$U_1=\{\,(x_1,x_2,x_3)\mid\lfloor x_i\rfloor\text{ is even for at most one }i\,\}\;,$$
$$U_2=\{\,(x_1,x_2,x_3)\mid\lfloor x_i\rfloor\text{ is even for at least two }i\,\}\;.$$
Then $V_1$, $V_2$ are open, and because $\mathbb R^3=U_1\mathbin{\dot\cup} U_2$, the remainder $\mathbb R^3\setminus(V_1,V_2)$ is a Lebesgue $0$-set. It is not too hard to see that $V_1$, $V_2$ are connected. Now take $\mathbf v=(1,1,1)$.
Sets like those above are used in Puppe's proof of the Blakers-Massey theorem, see tom Dieck's book Algebraic Topology, section 6.9.
One can even arrange $V_1$, $V_2$ to be contractible. To this end, regard both sets as thickened graphs. Pick a maximal tree in each of them, then take away open unit squares from $V_1$ and $V_2$ that "cut" the remaining edges. What remains are thickened trees.
For $d>3$, take the preimages under the projection $\mathbb R^d\to\mathbb R^3$
and $\mathbf v=(1,1,1,0,\dots,0)$.
For $n>2$, cut small holes into the sets above, so that $V_3, \dots, V_n$ are very small balls inside $U_1$ or $U_2$. Or, if $d=4n-1$, take
$$U_k=\{\,(x_i)\mid\lfloor x_i\rfloor\text{ is even for exactly $2k$ or $2k+1$ indices }i\,\}$$
for $k=0, \dots, 2n-1$, and continue as above.
For $d=2$, no solutions exist. For assume there was a solution. After a linear transformation, we may assume that $\mathbf v=(1,0)$ and that $(0,0)\in V_0$. Then $(t,0)$ cannot lie in the closure of $\overline{V_1}$ for any $t> 1$, because otherwise there would be a path $\gamma$ from $(0,0)$ to a point close to $(0,t)$ inside $V_1$. This path would keep a certain distance $\epsilon$ from $(1,0)\notin V_1$. Now we continue the path to a point of distance less than $\epsilon$ from $(0,t)$.
Then $\gamma$ has to intersect $\gamma+\mathbf v$.
If $V_1$ stays away from $(1,\infty)\times\{0\}$, then $(2,0)$ will be inside another set, say $V_2$, or at least in $\overline{V_2}$. We can repeat the argument above and show that $V_2$ then must stay away from $(3,\infty)\times\{0\}$. Continuing this argument, we see that finitely many sets $V_i$ do not suffice.
