This question was firstly asked in mathematics stack exchange. Getting no answer, I copied it to here.

For a vector space $V$ over $\mathbb R$, I say a subset $S$ of $V$ is "anti-convex" if $\forall a,b\in S (a\ne b)$, $\exists t\in ]0,1[$, $b+t(a-b)\not\in S$. 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 $(\mathbb R\backslash \mathbb Q)^2$.

My approach I've proven the case when $(\mathbb R\backslash \mathbb Q)^2$ 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.

This question was firstly asked in mathematics stack exchange. Getting no answer, I copied it to here.

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.

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

I've proven the case when $(\mathbb R\backslash \mathbb Q)^2$ 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.

Actually, I originally guessed that, 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.

This question was firstly asked in mathematics stack exchange. Getting no answer, I copied it to here.

For a vector space $V$ over $\mathbb R$, I say a subset $S$ of $V$ is "anti-convex" if $\forall a,b\in S (a\ne b)$, $\exists t\in ]0,1[$, $b+t(a-b)\not\in S$. 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 $(\mathbb R\backslash \mathbb Q)^2$.

My approach I've proven the case when $(\mathbb R\backslash \mathbb Q)^2$ 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.