A teacher once challenged the students to find a subset of $\mathbb R^2$ which intersects every line in finitely many points (not zero). The trick was to think about graphs instead of subsets. In fact, the subset $\{(x, x^3): x\in\mathbb R\}\subset\mathbb R^2$ intersects every line in $1$ or $3$ points.

Recently, in a set theory class, I've seen the construction of a subset of $\mathbb R^2$ that intersects every line in exactly two points. The construction used extensively the Axioma of Choice, so there was no hope for visualizing the set.

Thinking about it, I came to the conclusion that no curve could have this property: If a line cross a curve in two points, then one of the semi-planes determined by the line must contain a limited section of the curve, therefore, there is a parallel line in this semi-plane that doesn't cross the curve at all.

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But what if we have a set of curves? Can the union of several curves in the plane intersect every line in exactly two points? Or are sets with this property inherently complex?

  • $\begingroup$ By "several" do you mean "only finitely many" or also include "infinitely many"? $\endgroup$ Sep 23 at 5:45
  • 1
    $\begingroup$ "If a line cross a curve in two points, then one of the semi-planes determined by the line must contain a limited section of the curve" You are switching back and forth between "intersect" and "cross" here. Intersect is clear enough as a set theoretic term, whereas "cross" is much more difficult to define and gets into topology. And a curve can certainly intersect a line at exactly two points and still have both semi-planes contain an unbounded section of the curve (consider $y=0$ and $y=x^2(x-1)$). So more care would be needed to make that argument work. $\endgroup$
    – M W
    Sep 23 at 5:49
  • $\begingroup$ @MyMolecules I think a finite set of curves would be more impressive, but my curiosity extends to the infinite case as well. $\endgroup$ Sep 23 at 6:15
  • $\begingroup$ @MW I see, there is a flaw in my argument... The statement seems to be true still. $\endgroup$ Sep 23 at 6:20
  • $\begingroup$ I think if there was such a finite set of curves, no one would have bothered coming up with examples that needed the Axiom of Choice. Here's a discussion on MathOverflow: mathoverflow.net/questions/21470/… $\endgroup$ Sep 23 at 6:26

2 Answers 2


Two points sets, also known as Mazurkiewicz sets, are unfortunately not so well understood.

Let me start with the concluding question: are sets with this property inherently complex?

It is a wide open problem whether there exists a Borel Mazurkiewicz sets, it is known that any analytic Mazurkiewicz sets would have to be Borel and that there is a coanalytic Mazurkiewicz set assuming $V=L$.

In the opposite direction it is known, by a result of Larman there is no $F_\sigma$ Mazurkiewicz set, that is no Mazurkiewicz set can be written as a finite or countable union of closed sets, so in particular any finite or countable union of curves won't be a Mazurkiewicz set. As far as I know this is the best known lower bound in terms of Borel complexity for Mazurkiewicz sets.


We can also approach this problem through a path-connectedness argument. In particular, we can show through (relatively) elementary methods that a two-points set in the plane must be totally path disconnected, and therefore cannot be a union of any family of nonconstant curves, countable or otherwise.

Here is the argument.

First, recall that path connectedness in Hausdorff topological spaces is equivalent to arc-connectedness, so it suffices to show that a two points set contains no arcs.

To this end, let $S\subseteq \mathbb R^2$ be a two points set, and let $\gamma\colon [0,1]\to S$ be a parametrized arc. We lose no generality assuming $\gamma(0)=O$, $\gamma(1)=P$, where $O$ is the origin and $P$ is on the positive $x$-axis, and $\gamma(t)_x>0$ for all $t\in (0,1)$.

Consider two systems of polar coordinates, $(r,\theta)$ centered around $O$ and $(r',\theta')$ centered at $P$, each measuring angles in radians counterclockwise starting from the positive $x$ direction as usual.

Now the two point condition implies the map $\theta\circ \gamma|_{(0,1]}$ is a continuous injection into $[0,\pi)$, and therefore it maps bijectively onto some half-open interval $[0,\theta_0)$. Likewise, $\theta'\circ\gamma|_{[0,1)}$ maps bijectively onto some half-open interval $(\theta_0',\pi]$.

It is also apparent from the two point condition that $$S\cap (\{p\in\mathbb R^2\mid \theta(p)\in[0,\theta_0)\}\cup\{p\in\mathbb R^2 \mid\theta'(p)\in (\theta_0',\pi]\}=\gamma([0,1])\text{.}\tag{1}$$

Now if $\{ \theta(p)\in[0,\theta_0)\}\cup\{ \theta'(p)\in (\theta_0',\pi]\}$ is the entire upper half plane, then we have a contradiction, since any horizontal line that lies above the image of $\gamma$ will not intersect $S$.

On the other hand, if $\{ \theta(p)\in[0,\theta_0)\}\cup\{ \theta'(p)\in (\theta_0',\pi]\}$ is not the entire upper half-plane, then it is easily verified that the rays $\{\theta(p)=\theta_0\}$ and $\{\theta'(p)=\theta_0'\}$ intersect at some point $Q$, and $\gamma((0,1))$ is a subset of the open triangle $\Delta OPQ$.

But then by compactness of $\gamma([0,1])$, there is a horizontal line $L$ above $\gamma([0,1])$ and below $Q$. Observe that we have $$L\subseteq \{p\in\mathbb R^2\mid \theta(p)\in[0,\theta_0)\}\cup\{p\in\mathbb R^2 \mid\theta'(p)\in (\theta_0',\pi]\}\text{,}$$ and so by (1) we have that $L\cap S=\emptyset$, again a contradiction, and the proof is complete.


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