3
$\begingroup$

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 $d \geq 2$.]

$\endgroup$

1 Answer 1

5
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ Thank you very much for this. Now since posting the question, I've come to realise that for the problem I'm working on, I should take the sets $V_1,\ldots,V_n$ to be not just connected, but actually homeomorphic to $\mathbb{R}^d$! Do you have any thoughts as to whether, under this additional constraint, a vector $\mathbf{v} \in \mathbb{R}^d$ with the stated property can exist (with $d \geq 3$)? [Or maybe I should post this as a new MathOverflow question?] $\endgroup$ Nov 18, 2015 at 4:50
  • $\begingroup$ Thank you; I certainly wasn't just going to replace my original question with the new one. Having your answer to my current version of the problem is by no means irrelevant for me. $\endgroup$ Nov 18, 2015 at 13:07
  • $\begingroup$ I've just asked my new question: mathoverflow.net/questions/223925/… $\endgroup$ Nov 18, 2015 at 13:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.