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Can you find an open connected subset $\Omega$ in $\mathbb{C}$ such that it has the following property?

If $a,b,c \in \Omega$ are the vertices of a right triangle, then the rectangle given by $a,b,c$ lies in $\Omega$.

I suspect that the only such set is $\mathbb{C}$ itself, but I am unable to prove this, beyond some handwaving about taking points near the boundary.

Any suggestions?

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Assume $a\in\Omega\ne\mathbb C$. The complement is closed hence there exists $b\notin\Omega$ that minimizes $d(a,b)$. Consider the square with $ab$ as diagonal. Then the other two vertices are closer to $a$ than $b$, hence are in $\Omega$, hence this is a square with three vertices in $\Omega$, contradiction! (We see that general rectangles are not needed, squares suffice)

  • So it is not necessary for $\Omega$ to be connected? – miracle173 Feb 17 '14 at 07:44
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    @miracle173 In fact it is sufficient that the complement $\mathbb C-\Omega$ is not dense in $\mathbb C$ (i.e $\Omega$ has at least one interior point $a$. If $r:=\inf{,d(a,x)\mid x\notin \Omega,}>0$ pick $b\notin \Omega$ with $d(a,b)<\sqrt 2 r$ and proceed as above. – Hagen von Eitzen Feb 17 '14 at 18:41