Is there a path-connected, "anti-convex" subset of $\mathbb R^2$ containing $(\mathbb R\smallsetminus \mathbb Q)^2$?

Question

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.

Answer 1

It seems an path-connected anti-convex subset of $\mathbb{R}^2$ containing $(\mathbb{R}\setminus\mathbb{Q})^2$ exists.

Firstly, let $A$ be a countable, dense subset of $\mathbb{R}^2$, and let $B$ be the union of a family of arcs $\gamma_{a,a'}$, for $a\neq a'\in A$, such that $\gamma_{a,a'}$ is an arc from $a$ to $a'$ with diameter $\leq 2d(a,a')$.

Note that any subset of $\mathbb{R}^2$ containing $B$ is path-connected (you can use the arcs in $B$ to go from any point of $\mathbb{R}^2$ to any other point). Also note that the set $(\mathbb{R}\setminus\mathbb{Q})^2$ is obtained by removing from $\mathbb{R}^2$ a countable sequence of lines, which I will call $(l_n)_{n\in\mathbb{N}}$, such that every nontrivial segment in $\mathbb{R}^2$ intersects infinitely many of the lines $l_n$.

In the following, for any points $x,y\in\mathbb{R}^2$, I will say $\gamma$ is an `arc of circle' from $x$ to $y$ if $\gamma$ is the shorter of the two arcs going from $x$ to $y$ inside some circle containing $x,y$ (in particular, $\gamma$ has diameter $\leq2d(x,y)$).

Claim: If $A$ is a dense set of $\mathbb{R}^2$, there is a family of arcs of circle $\gamma_{a,a'}$ from $a$ to $a'$, with $a\neq a'\in A$, such that if $B=\bigcup_{a,a'}\gamma_{a,a'}$, then $B\cap \cup_nl_n$ has no three collinear points, except if all three inside one of the lines $l_n$.

If we prove the claim we will be done: indeed, the set $S:=B\cup(\mathbb{R}\setminus\mathbb{Q})^2$ is path connected, as we noted above, and no segment $[x,y]$ can be contained in $S$; this is obvious if $[x,y]$ is contained in one of the lines $l_n$ (then its intersection with $S$ is countable), and if not, then $[x,y]$ contains infinitely many points outside $(\mathbb{R}\setminus\mathbb{Q})^2$, and $B$ contains at most $2$ points of $[x,y]$.

To prove the claim, let $((a_n,a_n'))_{n\in\mathbb{N}}$ be a numbering of the pairs of points $a,a'$ in $A$ with $a\neq a'$, and we construct the arc $\gamma_{a_n,a_n'}$ recursively.

In the step $n$, we want to choose an arc of circle $\gamma_n$ from $a_n$ to $a_n'$ such that the points of $\gamma_n\cap\cup_nl_n$ are not collinear (collinear inside some line which is not one of the $l_n$) with points in $C:=\left(\cup_{i=1}^{n-1}\gamma_{a_i,a_i'}\right)\cap\cup_nl_n$ (note that $C$ is countable). So let $E$ be the set of points in $\cup_nl_n$ which are collinear (inside some line which is not one of the $l_n$) to two points of $C$. Note that $E$ is countable, as there are only countably many pairs of points in $C$. Note that all arcs $\gamma$ from $a_n$ to $a_n'$ except countably many do not contain any points of $E$ (except maybe $a_n$ and $a_n'$, but we can ensure that $a_n,a_n'\not\in E$ by choosing $A$ to be disjoint from $\cup_nl_n$); if we choose any of those arcs $\gamma$, then no point of $\gamma\cap\cup_nl_n$ can be collinear with two points of $C$.

We also want no points of $C$ to be collinear with two points of $\gamma_n\cap\cup_nl_n$; it seems that will also happen for all choices of $\gamma_n$ except countably many. To see why, we only have to prove that for each fixed $m,n\in\mathbb{N}$ and each $c\in C$, the point $c$ and two points in $\gamma_n\cap l_n$, $\gamma_n\cap l_m$ respectively can only be collinear for countably many choices of $\gamma$. It seems that will be the case except if the point $c$ is collinear with $a_n,a_n'$. But we can avoid that: before starting the induction, choose the set $A$ in such a way that, for all $n$, no two points of $A$ lie inside $l_n$, so that the set $F=(\cup_{a\neq a'\in A}\text{line}(a,a'))\cap\cup_nl_n$ is countable , and then choose the curves $\gamma_n$ so that they contain no points in $F$; thus, no point in $C$ can be aligned with two points in $A$.

Answer 2

Here is a more concrete construction. Let $X=\left\{\frac{n}{2^m};n\in\mathbb{Z},m\in\mathbb{N}\right\}\subseteq\mathbb{Q}$ and consider the set

$$S=(\mathbb{R}\setminus\mathbb{Q})^2\cup\{(x,y)\in\mathbb{R}^2;\text{ either $x$ or $y$ are in $X$, but not both}\}.$$

Note that $S$ is anti-convex: indeed, suppose for contradiction that $S$ contains some segment $[p,q]$, for some $p,q\in\mathbb{R}^2$. It is easy to see that $[p,q]$ cannot be horizontal or vertical, so it contains two points $(x_1,y_1)$ and $(x_2,y_2)$, where $x_1\neq x_2$ are in $\mathbb{Q}\setminus X$. This implies that $y_1,y_2\in X$. But $[p,q]$ also contains for all $n\in\mathbb{N}$ the point $\left(\frac{x_1+(3^{n}-1)x_2}{3^n},\frac{y_1+(3^{n}-1)y_2}{3^n}\right)$, which is not in $S$ for big enough $n$.

But $S$ is path-connected. Indeed, suppose we have points $(x_1,y_1)$ and $(x_2,y_2)$ in $S$, let's find a path between them. We can assume that $x_1<x_2$ and $y_1<y_2$. Let $A_1=\{x\in X;x_1<x<x_2\}$, $B_1=\{y\in X;y_1<y<y_2\}$, $A_2=\{x\in\mathbb{Q}\setminus X;x_1<x<x_2\}$, $B_2=\{y\in\mathbb{Q}\setminus X;y_1<y<y_2\}$.

Claim: There is some order isomorphism $f:[x_1,x_2]\to[y_1,y_2]$ such that $f(A_1)=B_2$ and $f(A_2)=B_1$.

Sketch of proof: One can first construct an order isomorphism $g:A_1\cup A_2\to B_1\cup B_2$ so that $g(A_1)=B_2$ and $g(A_2)=B_1$, using a back and forth method. And then one can note that this order isomorphism extends to an order isomorphism (so, a homeomorphism) $f:[x_1,x_2]\to[y_1,y_2]$.$\square$

And then, the graph of the function $f$ is an arc from $(x_1,y_1)$ to $(x_2,y_2)$ contained in $S$. Indeed, for any irrational $x$ in the interval $(x_1,x_2)$, $f(x)$ is irrational, so $(x,f(x))\in S$. And if $x$ is rational, then either $x\in A_1$ or $x\in A_2$, and in both cases $(x,f(x))\in S$.