Connectedness and path connectedness of a set whose intersection with lines is open

Question

Let $A \subseteq \mathbb{R^2}$ a not open set with the property:

Its intersection with every line L is open in L with the induced Euclidean topology.

If the set is connected is it necessarily path connected ?

Am not sure if is it true , i tried with the standard technique :

Let $\alpha \in A$ and define $\Pi = \{\omega \in A : \text{there is a path from }\alpha \text{ to } \omega\}$ and tried to show that $\Pi$ is clopen by using the property of $A$ but i couldn't prove anything , it only works when $A$ is open.

Any help would be nice , thanks !

Answer 1

Define $$B=\mathbb R^2\setminus\{(x,y)\mid x>0\text{ and }x^2\leq y\leq 4x^2\}$$ $$C=\{(x,y)\mid x>1\text{ and }2x^2< y< 3x^2\}$$ $$D_n=\{(x,y)\mid x>2^{-n}\text{ and }(2+2^{-2n+1})x^2< y< (2+2^{-2n})x^2\}$$ and $$A=B\cup C\cup\bigcup_{n\geq 1} D_n.$$

$B$ is open along each line - the only problem would be at $(0,0),$ but horizontal lines are ok and non-horizontal lines are ok. $B$ is path-connected. $C\cup\bigcup_{n\geq 1} D_n$ is open and path-connected and has $(0,0)\in B$ as an accumulation point. So $A$ is connected and open along each line.

But $A$ is not path-connected. Suppose for contradiction that there was a path from $(0,0)$ to any point in $C.$ Let $t$ be the time it first touches the boundary of $C.$ Before this time it cannot have been in any $D_n$ because each $D_n$ is a bounded distance away from $B$ and each of $D_m,$ $m\neq n.$ So the path can only have been in $B,$ but $B$ and $C$ are not connected to each other.

This is a variation on the deleted comb space.

Answer 2

I do not know how to answer your question, but I can answer it in the negative if you replace $\mathbb{R}^2$ by $\mathbb{R}^n$ for $n>2$. I will give a counterexample in $\mathbb{R}^3$; for general $n>2$ you can then just take the product of my example with $\mathbb{R}^{n-3}$. I will discuss a bit about why $n=2$ is harder at the end.

Let $C\subset\mathbb{R}^3$ be the curve $$C=\{(x,x^2,\sin 1/x):x>0\}.$$ Note that any line in $\mathbb{R}^3$ intersects $C$ at only finitely many points. The idea is to construct two thickenings $U\subset B$ of $C$ where $U$ is open, $B$ is closed except as $x\to 0$, and such that the thickenings narrow fast enough as $x\to 0$ so that no line can "detect" that $B$ is accumulating at points with $x=0$. We will then define $A$ to be $U\cup(\mathbb{R}^3\setminus B)$, and $A$ will be connected but not path-connected because $U$ approaches $\mathbb{R}^3\setminus B$ but cannot reach it with a path.

In detail, we define $$U=\{(x,y,z)\in\mathbb{R}^3:x>0, x^2/2<y<3x^2/2, \text{ and } \sin 1/x-1/2<z<\sin 1/x+1/2\}$$ and $$B=\{(x,y,z)\in\mathbb{R}^3:x>0, x^2/2\leq y\leq 3x^2/2, \text{ and } \sin 1/x-1/2\leq z\leq\sin 1/x+1/2\}.$$ It is clear that $U$ is open. I claim furthermore that any line $L\subset\mathbb{R}^3$ has closed intersection with $B$. Indeed, note that the projection of $B$ onto the $xy$-plane is the space between the two parabolas $y=x^2/2$ and $y=3x^2/2$ in the open right half-plane. No line in the plane intersects this set at points accumulating at the origin. Going back to $\mathbb{R}^3$, this means there exists $\epsilon>0$ such that the intersection $L\cap B$ is contained in $\{(x,y,z):x\geq\epsilon\}$. Since $B$ is closed in $\{(x,y,z):x\geq\epsilon\}$ (it only fails to be closed as $x$ approaches $0$), this means $B\cap L$ is closed.

So now we define $A=U\cup (\mathbb{R}^3\setminus B)$. Since $U$ is open and the intersection of $B$ with any line is closed, the intersection of $A$ with any line is open. Also, it is clear that $U$ and $\mathbb{R}^3\setminus B$ are both connected, and so $A$ is connected since $U$ accumulates at points of $\mathbb{R}^3\setminus B$ (namely, all the points $(0,0,z)$ for $-3/2\leq z\leq 3/2$). However, $A$ is not path-connected: to get a path from a point in $U$ to a point in $\mathbb{R}^3\setminus B$, you would need to traverse a path inside $U$ along which the $x$-coordinate approaches $0$, but then the $z$-coordinate must oscillate and so the path will fail to be continuous when the $x$-coordinate reaches $0$.

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I do not see a way to make an example like this work in $\mathbb{R}^2$. The problem is that if you want a path to oscillate infinitely as you approach a point, it will need to pass through some lines through that point infinitely many times. In particular, for instance, if you took my example and just dropped the $y$ coordinate everywhere (so we just have a thickened topologist's sine curve in the plane), the intersection of $B$ with any non-vertical line through the origin would fail to be closed (it contains points approaching the origin but not the origin itself).

So a counterexample in $\mathbb{R}^2$ would have to use some different idea, and I would not be surprised if it is impossible.