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This question was firstly asked in Mathematics Stack Exchange. Getting no answer, I copied it to Math Overflow, as Moishe Kohan commented.

For a vector space $V$ over $\mathbb R$, I say a subset $S$ of $V$ is "anti-convex" if $\forall a,b\in S (a\ne b)$, $\exists t\in \mathopen]0,1\mathclose[$, $b+t(a-b)\not\in S$. For example, all hollow circles $(x-a)^2+(y-b)^2=r^2$ on $\mathbb R^2$ are anti-convex.

Now I want to know if there is a path-connected, "anti-convex" subset of $\mathbb R^2$ containing $(\mathbb R\smallsetminus \mathbb Q)^2$.

My approach I've proven the case when $(\mathbb R\smallsetminus \mathbb Q)^2$ were replaced by a countable subset of $\mathbb R^2$ - Put them in order, and using hollow circles to connect the adjacent elements will work.

Further question (edited thanks to Sam Hopkins) Actually, I originally guessed that, if $\dim V\ge 2$, then for all anti-convex subset $S$, there's a path-connected anti-convex set containing $S$, but I'm already stuck at this special case.

Edit Since the original question was solved in Math Overflow, the one who answered my further question will receive the bounty.

yummy
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  • Do you mean $\mathbb{R}^2 \setminus \mathbb{Q}^2$? The case of $A = (\mathbb{R} \setminus \mathbb{Q})^2$ is trivial as any line between two points in $A$ must contain a point for which at least one of the coordinates is rational. – Comma May 08 '24 at 09:48
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    @Comma You're proving that $A$ itself is anti-convex, but $A$ itself isn't path-connected. Actually, any subset of an anti-convex set is anti-convex, so you can't find a supset of $\mathbb R^2\backslash \mathbb Q^2$ that is anti-convex. – yummy May 08 '24 at 10:05
  • Ah I missed the path connectedness requirement. – Comma May 08 '24 at 10:42
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    Cross-posted: https://mathoverflow.net/questions/471047/is-there-a-path-connected-anti-convex-subset-of-mathbb-r2-containing-m – Moishe Kohan May 11 '24 at 16:24

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