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

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 !

• What is an example of such an A? Commented Apr 2, 2018 at 21:08
• @WilliamElliot: A simple example is the complement of the set $\{(x,x^2):x\neq 0\}$. Commented Apr 2, 2018 at 22:16
• Good Q. Deceptively simple.... I suspect the A is "Yes". Commented Apr 3, 2018 at 0:28
• Is not the topologist sin curve also an example? @EricWofsey Commented Apr 3, 2018 at 3:25

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 $$U=\{(x,y,z)\in\mathbb{R}^3:x>0, x^2/2 and $$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 $$0$$.

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.

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.

• Very nice! In retrospect, I feel silly for not having thought of something like this. :) Commented Apr 12, 2018 at 20:31
• Congrats! Nice examples ! (both of them) :) Commented Apr 13, 2018 at 7:05