3
$\begingroup$

Question: Does there exist a compact path-connected set $A\subseteq\mathbb C$ such that:

  1. $A\cap(A+1)=A\cap(A+i)=\emptyset$,
  2. $A\cap(A+1+i)\neq\emptyset$, and
  3. $A\cap(A+1-i)\neq\emptyset$?

Remarks: If such $A$ exists, let $B$ denote the union of the 4 translates $A\pm1$, $A\pm i$. Then $B$ is path-connected and disjoint from $A$. Intuitively one might want to say that $B$ "surrounds" $A$ (hence the question title), but I can't actually prove that there exists a loop in $B$ with nonzero winding number around $A$.

One may assume that $A$ is the union of two intersecting paths, whose endpoints are $1+i$ and $1-i$ apart, respectively.

Passing to the torus $\mathbb C/(1+i)\mathbb Z[i]$ shows that any $A$ satisfying (2) and (3) must intersect some translate of the form $A+m+ni$, with $m+n$ odd.

Improve this question
$\endgroup$
2
  • 2
    $\begingroup$ Here's a strategy to try to prove this can't happen: the images of your two paths in your quotient torus should give it a cellular decomposition. Your hypotheses should imply that the Euler characteristic of this decomposition is non-zero, which is a contradiction. $\endgroup$
    – HJRW
    Commented Nov 11, 2020 at 14:46
  • $\begingroup$ I'm not sure that works, for exactly the reason in the last paragraph of my question; the four translates map to one copy of $A$ in the torus, but I don't know that this is disjoint from the image of the original copy of $A$. $\endgroup$
    – chronondecay
    Commented Nov 13, 2020 at 2:32

0

Reset to default

You must log in to answer this question.