# Example of a simple dynamical system with no equicontinuity points but not sensitive

I am looking for an example of a simple discrete dynamical system with no equicontinuity points but not sensitive.

Found this example in the book of Kurka. But could not really understand it.

$$X=\left\{ \left( x,y,z\right) \in \mathbb{R}^{3}:x^{2}+y^{2}\leq 1,z=0\right\} \cup \left\{ \left( x,y,z\right) \in \mathbb{R}^{3}:\left( x-1\right) ^{2}+z^{2}=1,y=0\right\} .$$

The function $$f$$ is defined by : $$\begin{eqnarray*} f\left( r\cos \left( t\right) ,r\sin \left( t\right) ,0\right) &=&\left( r\cos \left( 2t\right) ,r\sin \left( 2t\right) ,0\right) \\ f\left( 1-\cos \left( t\right) ,0,\sin \left( t\right) \right) &=&\left( 1-\cos \left( 2t\right) ,0,\sin \left( 2t\right) \right) \end{eqnarray*}$$

Thank you.

That's about the simplest example I know, but I can explain it.

First, there's the doubling map, defined on any circle :

$$X_r = \{(x,y,z) \in \mathbb{R}^3 : x^2+y^2=r^2, z=0\},$$

$$f_r(r \cos(t), r \sin(t), 0) = (r \cos(2t), r \sin(2t), 0).$$

For any positive $$r$$, the system $$(X_r, f_r)$$ is sensitive, and has no equicontinuity point. Now, do the same on a disc:

$$X_D = \{(x,y,z) \in \mathbb{R}^3 : x^2+y^2\leq 1, z=0\},$$

$$f_D(r \cos(t), r \sin(t), 0) = (r \cos(2t), r \sin(2t), 0).$$

Then $$(X_D, f_D)$$ is not sensitive anymore: the point $$(0,0,0)$$ is stable. However, for the same reason, it is has an equicontinuity point, $$(0,0,0)$$.

The next step is to add a circle, on which we act via the doubling map, and glue it to $$X_D$$ at the point $$(0,0,0)$$. That's how you get $$(X,f)$$. Since all points belong to a circle on which we act via an angle-doubling map, $$(X,f)$$ has no equicontinuity point.

That said, $$(0,0,0)$$ is not stable anymore. The catch, however, is that $$(X,f)$$ is still not sensitive! The notion of sensitivity is global, i.e; uniform in $$x$$. I don't remember the definition exactly, but I think it is something like "For all $$\varepsilon >0$$, there exists $$\delta >0$$ such that, for all $$x$$...".

Now, the systems $$(X_r, f_r)$$ are each sensitive, but not uniformly so. If you start on $$D$$ and within a distance $$\varepsilon$$ from $$(0,0,0)$$, then you will always stay on $$D$$ and within a distance $$\varepsilon$$ from $$(0,0,0)$$, which contradicts sensitivity.