Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Certain subsets $A$ of $\mathbb{R}^n$ satisfy the property that given any line segment $L$, $A \cap L$ has a finite number of connected components. For instance, $A$ can be any finitary union of convex sets. Is the converse true? Is this property studied in literature?

share|cite|improve this question
I think it can be a finite union of disjoint compact sets too. – Tunococ Jan 31 '13 at 10:35
Would any closed subset $B$ of the unit interval that has an infinite number of components serve as a counterexample, since $B$ would be compact? Or do all compact subsets of the real line have a finite number of components? Sorry, I'm a novice in topology. – Herng Yi Jan 31 '13 at 10:44
Yes, the Cantor set ( is compact but not a finite union of intervals. On the other hand, the unit circle has the desired property but is not a finite union of convex sets. – Colin McQuillan Jan 31 '13 at 11:05
Ok, here's my another attempt. What about the algebra generated by convex sets? You allow finite union, finite intersection and complement. Use convex sets as basis. – Tunococ Jan 31 '13 at 11:21
@Tunococ: I think this is a counterexample (i.e. has the property but isn't in the algebra): $\{(x,y)\mid \text{$x>0$ and $0<y<1/x$ and $\lfloor x \rfloor$ is even}\}$. Maybe your condition is right for bounded sets though. – Colin McQuillan Jan 31 '13 at 11:28
up vote 1 down vote accepted

Any set in real-algebraic geometry satisfies this: the locus defined by some polynomial equations and inequalities. I do not know if they can all be expressed as finite unions of convex sets. Adding exponents and logarithms to the equations does not change the finite intersection property. Infinite intersections appear only when you allow oscillatory functions.

Considering one space only, $\mathbb{R^n}$, any class of continuous functions for which the hypersurfaces $f(x_1,\dots,x_n)=0$ intersect any line in finitely many intervals defines sets with the same property (allowing inequalities and finite logical combinations using union, intersection and complement). These are curved generalizations of polyhedral regions with finite number of faces.

In model theory there is an area called $o$-minimal structures that deals with "finite type" topology for sets with nice definitions by formulas. The proof that a class of functions gives an $o$-minimal structure implies finite-type intersections with lines, and with any finite-type loci defined using functions from that class. orem

The model-theoretic condition is stronger than the question for one space, because definability allows projection from higher-dimensional sets.

share|cite|improve this answer

The converse is not true, as mentioned in the comments. I don't know if the property has been studied. Tunococ mentioned the sufficient condition that $A$ is in the algebra generated by convex sets. This is (trivially) also a necessary condition in dimension 1.

Here is an argument that Tunococ's condition is not necessary, even if we restrict to very nice sets $A$: bounded open sets that are the interior of their closure. I don't have any proofs - this is just plausible reasoning.

Let $A$ be the union, over each positive integer $n$, of the open disc with centre $(\cos 1/n,\sin 1/n)$ and radius $1/9n^9$. I think the only line that could intersect $A$ with infinitely many intervals is the tangent $x=1$, but that doesn't intersect $A$ at all. I think some construction like this would even make $A\cap L$ uniformly bounded.

If $A\subseteq\mathbb R^2$ is a union of an infinite number of disjoint discs, then $A$ is not in the algebra generated by convex sets. I am not sure how to prove this, but the boundary of a non-empty convex set has total curvature $2\pi$, so the boundary of a set in the algebra generated by $k$ convex sets has total curvature at most $2k\pi$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.