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 (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, 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 says about what it means to accept an answer.

  • $\begingroup$ +1 for the question, but -1 for the name "$2$-point set" (I'm not blaming you!). I was really confused by the third paragraph until I reread the first paragraph. $\endgroup$ Jul 9, 2015 at 16:50
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    $\begingroup$ Fair enough! I've always thought the name a bit unfortunate too. $\endgroup$
    – Will Brian
    Jul 9, 2015 at 16:51
  • $\begingroup$ I hope you don't mind, but I've edited the question to put quotation marks around "$2$-point set" in the third paragraph to protect any readers as dumb as me from that "Eh?" moment. $\endgroup$ Jul 9, 2015 at 17:03
  • $\begingroup$ That works for me. $\endgroup$
    – Will Brian
    Jul 9, 2015 at 17:04

3 Answers 3


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 $\omega$.

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.


The answer to the question is no.

Rather than typing a new answer, I've just edited my old one (which contained some partial progress). I think this is OK since the main idea of the old answer (or at least its main "trick" -- fixing a special line, and then rotating it slightly about a point not in $E$) is still present in this one. The proof I'm about to give combines this trick with the results from Terry Tao's answer.

Theorem: There is no subset of the plane meeting every line in two open intervals.

Proof: Suppose $E$ is such a set.

Following the terminology in Terry Tao's answer, we will say that $\omega \in S^1$ is a limit direction of $E$ if $E$ does not avoid an infinite open sector containing the direction $\omega$ in the limit. Terry Tao proves:

Lemma: Not every direction is a limit direction.

(Actually he proves more, but this is all we will need.)

In other words, there is an infinite open sector $S$ that misses $E$. Translating and rotating $E$ if necessary, we may assume that $S$ contains the origin $O$ and the positive $X$-axis. Consider the expression (using polar coordinates) $$(*) \ \ \ \ \ \ \ \ \ \ \ \{(r,\theta) : 0 < \theta < \phi, r \geq 0\} \cap E = \emptyset.$$

Clearly $\pi$ does not satisfy $(*)$ (otherwise $E$ does not meet the line $y = 1$). Let $$\phi_0 = \inf \{\phi \in [0,\pi] : \phi \text{ does not satisfy }(*)\}.$$ (Since $S$ is open, we will have $0 < \phi_0 < \pi$, although we won't actually need this fact.) Let $L$ be the line through the origin in the direction $\phi_0$.

$E$ is open (this observation is also due to Terry Tao; see his first comment on Igor Rivin's answer). It follows that the ray $\{(r,\phi_0) : r \geq 0\}$ contains no points of $E$ (if it did, then a slightly smaller value of $\phi_0$ would also fail to satisfy $(*)$). Therefore $E$ meets $L$ in two open intervals that are both on the same side of the origin $O$.

Let $U$ and $V$ be these two open intervals, labeled in such a way that $V$ is between $U$ and $O$. Fix a point $x \in L \setminus E$ with $x$ strictly between $U$ and $V$. For $\rho \in [0,\pi)$, let $R_\rho^x$ denote the rotation of the plane about the point $x$ by the angle $\rho$. Fix $u \in U$ and $v \in V$.

We have $u,v \in E$ and $O \in S$, with $E$ and $S$ both open. By the continuity (in $\rho$) of $R^x_\rho$, there is some $\rho_0$ small enough that if $0 < \rho < \rho_0$ then $R_\rho^x(u), R_\rho^x(v) \in E$ and $R_\rho^x(O) \in S$. We may assume $\rho_0 < \frac{\pi}{2}$.

By the maximality of $\phi_0$, there is some $\varepsilon < \rho_0$ such that $$\{(r,\phi_0+\varepsilon) : r \geq 0\} \cap E \neq \emptyset.$$ In other words, we may choose a point $e \in E$ such that, if we let $p = (r,\phi_0)$ for some $r > 0$, the angle $pOe$ is less than $\rho_0$. Since angle $pOe$ is acute (recall $\rho_0 < \frac{\pi}{2}$), and since $x$ is on the other side of $O$ from $p$, the angle $pxe$ is even smaller than the angle $pOe$.

Let $\rho$ be the measure of the angle $pxe$. The line $L' = R^x_\rho(L)$ does not meet $E$ in two open intervals. To see this, observe that $L'$ passes through $E$ near $u$, through $x$ (not in $E$), through $E$ again near $v$, through $S$ near $O$ (not in $E$), and then through $e$ (in $E$). QED

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    $\begingroup$ Very nice! Glad to hear the problem is finally solved. $\endgroup$
    – Terry Tao
    Jul 13, 2015 at 17:55
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    $\begingroup$ @TerryTao: Thanks! And thanks for not hesitating to post your partial results (which, as you can see, were a huge help). $\endgroup$
    – Will Brian
    Jul 13, 2015 at 20:34
  • $\begingroup$ mmm yes very clever $\endgroup$ Aug 20, 2016 at 5:46

It is a result of Frantz

   author={Frantz, Marc},
   title={On Sierpi\'nski's nonmeasurable set},
   journal={Fund. Math.},
   review={\MR{1141290 (93a:28006)}},

That such a set (actually, Frantz allows exceptional intersections, so that his result is clearly "nonempty", for example the union of two open infinite strips is fine) is measurable. I would guess that this means that such a set does not exist.

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    $\begingroup$ Actually, it seems to me that such a set (call it $E$) should be open. Let's say $(0,0)$ lies in $E$, then (up to reflection) we can find $a < 0 < b < c < d$ such that $(a,0), (b,0), (d,0)$ lie in $E$ and $(c,0)$ does not. Then we can find open vertical intervals $I,J,K$ around $(a,0), (b,0), (d,0)$ that lie in $E$. Any point $p$ sufficiently close to $(0,0)$ is collinear with a point in $I$, a point in $J$, $(c,0)$, and a point in $K$ and so does not meet $E$ in two intervals unless $p \in E$, giving the claim. $\endgroup$
    – Terry Tao
    Jul 9, 2015 at 19:00
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    $\begingroup$ This suggests that the problem should be solved by a topological argument (as opposed to, say, a descriptive set theory argument), and that the axiom of choice or other set-theoretic axioms should not be relevant. $\endgroup$
    – Terry Tao
    Jul 9, 2015 at 19:01
  • $\begingroup$ @TerryTao precisely my thought, although it is conceivable that some Sierpinski two point set can be thickened to give an example (in which case, there is still some descriptive set theory at the base). $\endgroup$
    – Igor Rivin
    Jul 9, 2015 at 19:13

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