limit cycles of dynamical systems Consider $2D$ dynamical systems $X' = F(X)$ where $X=(x,y)$ is a 2-vector,  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 $x$-axis 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.]   
 A: $L$ need not be closed.  Consider a vector field $X$ with the following properties:


*

*$X(x,0)=(0,x)$ for all $x>0$.

*For $x\in (0,1)$, the trajectory starting at $(x,0)$ follows a vertical line until it passes through the point $(0,\frac 1{1-x})$, and then travels counter-clockwise around the origin until it returns to $(x,0)$.

*The trajectory starting at $(1,0)$ never leaves the vertical line $x=1$.

*The trajectory starting at $(x,0)$ for any $x>0$ is non-periodic.


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 "north-south" 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 non-closed 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.
