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Let $b(X)$ be the boundary of any $X$ subset of the plane.

Does there exist $A,B$ convex compact sets of the plane, such that $C:=A\setminus B$ is simply connected and not empty, and such that there is no square inscribed in $b(C)$ that has a vertex in $b(C)\setminus B$ ?

If the answer is no, this would kind of mean that the Toeplitz conjecture is kind of linked to convexity (one of the first intersting resolved cases was the case of the boundary of a convex)

If the answer is yes... then it suggests a "fractal" way to built some counterexample.

Do not hesitate to say your "prognosis" in the comments, even if it is not argued so far, I have absolutally no intuition myself about the pronostic (I would say it must be "NO" but I would be influenced by the fact that according to me, if it is "YES", then there is a counterexample not far, and this would have been found - but I might as well underevaluate the difficulty of "YES=> counterexample" ...)

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  • $\begingroup$ “pronostic” is not an English word — what do you mean? Also it would help to explain this directly, rather than with quote marks that indicate some substantial unexplained part. $\endgroup$
    – user44143
    Commented Aug 2, 2021 at 1:18
  • $\begingroup$ The internet tells me pronostic means prediction/prognosis/forecast. I suppose frontiere means boundary, and connex means connected. $\endgroup$
    – Gerry Myerson
    Commented Aug 2, 2021 at 2:14
  • $\begingroup$ Thank you for your comments, I edited the grammar. @Matt F. the quote was not about the use of a foreign word (I did not know that it is not an english word), but to encourage some type of comments that would not be argued that much. As soon as it is not a very common behaviour, and as I still care about what is the intuition of other users, I put quote on this word. $\endgroup$
    – jcdornano
    Commented Aug 2, 2021 at 3:51
  • $\begingroup$ Let $\ A:=\{(x\ y): (x-11)^2+y^2\le 10^2\}\ $ and $\ B:=\{(x^2+y^2\le 20^2\}.\ $ It looks that there is no square inscribed into $\ b(A\setminus B).\ $ Thus, it seems that it is simpler than it is in the Question above. $\endgroup$
    – Wlod AA
    Commented Aug 2, 2021 at 4:33
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    $\begingroup$ @WlodAA, unless I did not understand your comment, $b(A\setminus B)$ is a Jordan curve (image of a continuous injection of the circle in the plane) that is the union of two smouth curves and in this case it is known that there is a square inscribed into it. $\endgroup$
    – jcdornano
    Commented Aug 2, 2021 at 4:43

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