Consider 2d dynamical systems X' = F(X) where X is a 2vector, and there is an equilibrium point at the origin. Let L be the set of numbers x > 0 such that a limit cycle of the system meets the xaxis at (x,0). Is L necessarily closed? More generally what sets L can arise in this way? [The reason I want to know is that I'm teaching dynamical systems.]

$L$ need not be closed. Consider a vector field $X$ with the following properties:
Then $L=[0,1)$ is not closed. In the other direction, given any closed $L$, it is possible to construct a vector field realising that $L$. Indeed, begin by constructing a map $f\colon [0,\infty)\to [0,\infty)$ such that $L = \{ x \mid f(x)=x \}$, and then construct a flow whose first return map to the positive $x$axis is the map $f$. To construct the map, just observe that the complement of $L$ is a countable union of open intervals, and so you can define $f(x) = x$ for all $x\in L$, and then define $f$ on each open interval $(a,b)$ as a "northsouth" map, that is, a map of the interval $[a,b]$ that fixes $a$ and $b$ and moves every other point towards $b$. To construct the flow from the map, just draw circles of radius $x$ for every $x\in L$, and then spirals connecting $(x,0)$ to $(f(x),0)$ for every $x\notin L$. This still leaves open the question of which nonclosed sets can be realised. Clearly there's a topological obstruction to using the procedure at the beginning of my answer to remove lots of boundary points from $L$; indeed, once you remove a single boundary point that way, then the trajectory starting at any larger value of $x$ cannot wind around the origin, and so life becomes a little more complicated... Edit: Here's a more general construction. Let $X(x,y) = (y,x)$ so that the flow is along circles centred at the origin. Fix any closed set $E\subset (0,\infty)$ and let $\phi(x,y) = 0$ precisely when $x=0$ and $y\in E$, with $\phi>0$ everywhere else. Then the flow for the vector field $\phi X$ has the property that the set of $x>0$ for which $(x,0)$ is contained in a closed orbit is precisely the set of $x$ not contained in $E$. Thus you can get any open set as your set $L$. Indeed, you can combine the two constructions to get any set that can be written as the intersection of an open set and a closed set. There's probably other stuff you can do too, but this illustrates some of the things that can happen. 

