17
$\begingroup$

A dyadic square is a subset of $R^2$ of the form $x + 2^{-n} [0,1]^2$ with $x \in 2^{-m} Z^2$, for integers $m,n \geq 0$. We say that a set $A$ crosses a square $S$ if there exists a connected subset of $A \cap S$ which intersects two opposite sides of the square $S$. Clearly, the 45 degree line $\{ (\pi + t, t) : t \in R \}$ does not cross any dyadic square. Does every non-trivial closed, connected set that is not a line segment cross some dyadic square?

Improve this question
$\endgroup$
12
  • 1
    $\begingroup$ I would have expected the condition $m\le n$. Is that what you mean? Or is this another version of the problem? Every line segment that is not $\pm45$ degrees, and every sufficiently smooth curve that is not a $\pm45$ degree line segment must cross a dyadic square (even with the strong definition of dyadic square). So if there are counterexamples they are pretty pathological. $\endgroup$
    – Günter Rote
    Commented Jan 31, 2013 at 17:01
  • $\begingroup$ I think that every path-connected component of such a non-crossing set should be a line segment. $\endgroup$
    – Marco Golla
    Commented Jan 31, 2013 at 19:44
  • 1
    $\begingroup$ Gunter Rote, I do not mean the condition $m\leq n$. E.g. the square $[1/8,5/8]\times [1/4,3/4]$ is a dyadic square with my definition. Perhaps you would like to call such a square a translated dyadic square. $\endgroup$
    – Kevin Johnson
    Commented Feb 7, 2013 at 8:54
  • $\begingroup$ There used to be an interesting answer posted here. Where is it? Was it wrong? $\endgroup$
    – Günter Rote
    Commented Feb 17, 2013 at 13:17
  • $\begingroup$ Hi Gunter, I pointed out that there were two mistakes in this answer. A day later it was removed by somebody. $\endgroup$
    – Kevin Johnson
    Commented Feb 18, 2013 at 9:01

0

Reset to default

You must log in to answer this question.