2
$\begingroup$

Let be $M$ a topological space, and $f: M\to M$ a dynamical system, i.e, a continuous map between from $M$ to $M$.

We say that a dynamical system, $f:M\to M$ is topologically transitive when, exists $x\in M$ such that, $Orb(x)=\{x,f(x),\ldots, f^n(x),\ldots\}$ is dense in $M.$

There is a problem in the book of Brin Stuck, An introduction to Dynamical Systems, that makes the following question: Is the product of two topologically transitive (minimal, topologically mixing) systems topologically transitive (minimal, topologically mixing)?

I already know that for minimal systems the answer is no, And as for mixing systems, the answer is yes.

But I have no intuition for the case of topologically transitive systems, so my question is:

Is the product of two topologically transitive maps, topologically transitive?

$\endgroup$
3
$\begingroup$

Let $M=\{0,1\}$ with the discrete topology, and let $f:M\to M:x\mapsto 1-x$. Clearly $\langle M,f\rangle$ is transitive. Let $$F=f\times f:M\times M\to M\times M:\langle x,y\rangle\mapsto\langle f(x),f(y)\rangle\;.$$ Then for each $p\in M\times M$, $|\operatorname{Orb}(p)|=2$, so $\operatorname{Orb}(p)$ is not dense in the $4$-point discrete space $M\times M$.

(In fact $M\times M$ is the union of the two disjoint $F$-orbits $\{\langle 0,0\rangle,\langle 1,1\rangle\}$ and $\{\langle 0,1\rangle,\langle 1,0\rangle\}$.)

$\endgroup$
  • $\begingroup$ Very elegant. Cool! $\endgroup$ – user27456 Apr 30 '12 at 19:01
1
$\begingroup$

Define $f:[0,2] \to [0,2]$ as $$f(x)=\begin{cases}2x+1,&\text{ if }x \in[0,\frac 12]\\-2x+3,&\text{ if }x \in[\frac 12,1]\\-x+2,&\text{ if }x \in[1,2]\end{cases}$$ Then it can be shown that $f$ is topologically transitive.

Let $U=(0,1)\times (0,1)$ and $V=(0,1)\times(1,2).$ Then $$(f\times f)(U)=f(0,1)\times f(0,1)\subseteq (1,2]\times(1,2]$$Therefore,$$(f\times f)^2(U)\subseteq f(1,2]\times f(1,2]\subseteq[0,1)\times[0,1)$$ Which implies $$(f\times f)^3(U)\subseteq [1,2]\times[1,2]$$So, $$(f\times f)^4(U)\subseteq [0,1]\times[0,1]$$

Continuing this way,$$(f\times f)^n(U)\cap V=\emptyset$$ for all $n\in \mathbb N.$ Thus, $f \times f$ is not topologically transitive.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy