Sets whose intersection with line segments have finite components 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?
 A: 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 $2k\pi$.
A: 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.  
http://en.wikipedia.org/wiki/O-minimal_theory
http://en.wikipedia.org/wiki/Wilkie%27s_the orem
The model-theoretic condition is stronger than the question for one space, because definability allows projection from higher-dimensional sets.
