Does every connected set that is not a line segment cross some dyadic square?

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?

flag	asked Jan 31 at 14:07 Kevin Johnson 101●2

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. – Günter Rote Jan 31 at 17:01

I think that every path-connected component of such a non-crossing set should be a line segment. – Marco Golla Jan 31 at 19:44

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. – Kevin Johnson Feb 7 at 8:54

There used to be an interesting answer posted here. Where is it? Was it wrong? – Günter Rote Feb 17 at 13:17

Hi Gunter, I pointed out that there were two mistakes in this answer. A day later it was removed by somebody. – Kevin Johnson Feb 18 at 9:01

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