Using the Axiom of Choice, it is possible to construct a subset of the plane that meets every line in two points (these are called "$2$-point sets"). What if, instead of points, we ask for two open intervals?


Some related ideas/constructions were explored in [this paper](https://wrbrian.files.wordpress.com/2012/01/slicesets1.pdf) (for example, it is shown that there is a subset of the plane meeting every line in a copy of the Cantor set).


If you want one open interval instead of two, then $\mathbb{R}^2$ itself suffices (and this is the only solution). If you want three open intervals, then just take the complement of a "$2$-point set" (and this idea can be modified to give you $n$ open intervals for every $n > 2$). Infinitely many open intervals is even easier (take the union of the interiors of the hexagons in [this tiling](http://mathworld.wolfram.com/images/eps-gif/HexagonalGrid_700.gif), for example). The case of two intervals seems more stubborn.


**UPDATE:** This question has now been answered: there is no such subset of the plane.


Terry Tao posted some "partial progress" on the problem, which later turned out to be the first half of a complete proof. The second half can be found in my answer below. Neither post stands alone as a complete answer, but both taken together do the job.


I can't accept both answers, so I've accepted Terry's, since it came first and ended up being the first half of a correct proof. This seems to me to be in line with what the [help center](https://mathoverflow.net/help/accepted-answer) says about what it means to accept an answer.



Is there a subset of the plane that meets every line in two open intervals?
Let $E$ be a set of the claimed form. Call a direction $\omega \in S^1$ a *limit direction* of $E$ if there exists a sequence $p\_n$ of points in $E$ going to infinity whose argument goes to $\omega$, or equivalently if $E$ does not avoid an infinite open sector containing the direction $\omega$ in the limit. I can show the following:



> 
> **Proposition**. The set of limit directions is a closed non-empty subset of $S^1$ which is not all of $S^1$. Furthermore, if $\omega$ is a limit direction, then every line parallel to $\omega$ meets $E$ in the union of two open intervals, one of which is half-infinite in the direction $\omega$.
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This doesn't settle the question yet, but may be useful partial progress towards a complete solution. One consequence of this proposition is that if one line meets $E$ with a half-infinite interval (plus another interval), then all parallel lines do also (since the direction of the half-infinite interval is clearly a limit direction).


**Proof**: It is clear that the set of limit directions is closed. Since $E$ meets every line at least once, it is unbounded. Thus there is a sequence of points $p\_n$ in $E$ going to infinity. By Bolzano-Weierstrass this shows that there is at least one limit direction.


Suppose that the direction $(1,0)$ was a limit direction, thus we have a sequence $p\_n = (x\_n,y\_n)$ in $E$ where $x\_n \to +\infty$ and $y\_n/x\_n \to 0$. By reflection and passing to subsequence we may assume that the $y\_n$ are all nonnegative; by shifting $E$ upwards slightly we may assume they are all strictly positive.


By hypothesis, $E$ meets the $x$-axis in two disjoint intervals $(a,b) \times \{0\}$ and $(c,d) \times \{0\}$ with $a < b < c < d$. Suppose that $d$ was finite. If we choose $a < x < b < y < c < z < d < w$, then there are vertical open intervals $I\_x, I\_z$ around $(x,0)$ and $(z,0)$ that respectively lie in $E$, while $(y,0)$ and $(w,0)$ do not lie in $E$. 


Consider the vertical line through $(w,0)$. This meets $E$ in two open intervals, neither of which contains $(w,0)$. Thus there is an open interval $J\_w = \{w\} \times (0,\varepsilon)$ that either lies completely outside of $E$, or completely inside $E$. But for $n$ large enough, we can find a line that meets $I\_x$, $(y,0)$, $I\_z$, $J\_w$, and $p\_n$ in that order. Since this line has to meet $E$ in two intervals, this forces $J\_w$ to lie completely inside $E$.


We conclude: if $d$ is finite, then for sufficiently large $w$, there exists $\varepsilon>0$ such that $(w,0)$ lies outside of $E$ but $\{w\} \times (0,\varepsilon)$ lies in $E$.


Of course, for $d$ infinite, we have $(w,0) \in E$ for all sufficiently large $w$.


Shifting $E$ upwards (which keeps $y\_n$ positive and $y\_n/x\_n$ going to zero), we conclude that for any $t \leq 0$, either $(w,t) \in E$ for all sufficiently large $w$, or else for sufficiently large $w$ (depending on $t$), there exists $\varepsilon>0$ such that $(w,t)$ lies outside of $E$ but $\{w\} \times (t,t+\varepsilon)$ lies in $E$.


If the second option holds true for at least three values of $t \leq 0$, then we conclude for sufficiently large $w$ that the indicator of $E$ on the vertical line $\{w\} \times {\mathbf R}$ changes value at least five times, and so $E$ does not meet this line in two intervals, a contradiction. Thus the first option must hold for at least one $t \leq 0$. In particular we now have a sequence $(x\_n,y\_n)$ of points with $y\_n = t < 0$ and $y\_n/x\_n \to 0$, so by reflection all the results we had for $E$ also hold for the reflection of $E$ across the $x$ axis. In particular, if $d$ is finite, it is now true that for sufficiently large $w$, there is an interval $K\_w = \{w\} \times (-\varepsilon,\varepsilon)$ such that $K\_w$ meets $E$ at every point of $K\_w$ except for the midpoint $(w,0)$. 


Now by intersecting $E$ with ${\mathbf R} \times \{1\}$ we may find $f < g$ such that $(f,1) \in E$ and $(g,1) \not \in E$. Using the vertical line $\{f\} \times {\mathbf R}$ we may find a vertical interval $L\_f$ around $(f,1)$ that lies in $E$. But one can then find arbitrarily large $w\_1 < w\_2 < w\_3$ close together such that there is a line passing through $L\_f$, $(g,1)$, $K\_{w\_1}$, $(w\_2,0)$, $K\_{w\_3}$ in that order, contradicting the fact that $E$ has to meet this line in two intervals. (To find $w\_1,w\_2,w\_3$, one can for instance use the Lebesgue differentiation theorem to locate an arbitrarily large real $w\_0$ where the length of $K\_w$ is bounded from below for set of $w$ of asymptotic density $1$ near $w\_0$, then set $w\_1,w\_2,w\_3$ to be sufficiently close generic points near $w\_0$.) We conclude that $d$ must be infinite. Translating this up and down, we now conclude that $E$ meets every horizontal line in two intervals, one of which is half-infinite to the right.


Finally one has to show that not every direction is a limit direction. This is an observation (now deleted) of Robert Israel: if every direction was a limit direction, then every line meets the complement of E in a closed interval, so the complement of E is convex. But by the Hahn-Banach separation theorem we can then find a line that separates a point of E from its complement, so E meets all of that line rather than meeting it in two intervals, a contradiction. $\Box$


One can say a bit more about the limit directions. If $\omega\_1,\omega\_2,\omega\_3,\omega\_4,\omega\_5$ lie in a semicircle in that order, one cannot have $\omega\_1,\omega\_3,\omega\_5$ a limit direction and $\omega\_2,\omega\_4$ not, since in that case there would be rays in the directions $\omega\_3,\omega\_5$ that were in $E$ and rays in the directions $\omega\_2,\omega\_4$ that lay outside of $E$, which contradicts the proposition in the direction $\omega\_1$. I think this means that the set of non-limit directions consists of at most three open intervals.


