Significance of topological transitivity in chaotic systems

Question

I am currently undertaking a research project on chaos theory during my final year of undergraduate study. I came across the three conditions in order for a system to be chaotic. Both having dense periodic orbits and sensitive dependence made intuitive sense to me as to why these both must be present but I do not understand what is meant by topological transitivity and hence, the significance of it. This field of mathematics is entirely new to me so whilst I am used to seeing definitions anything too technical may go over my head. If anyone could help explain this to me I would be very grateful.

Answer 1

There is a general sense of transitivity in dynamical systems, which goes like this:

Given some kind of dynamical system on some kind of phase space, i.e. some mathematical object $X$ and a self-isomorphism $f : X \to X$, to say that $f$ is transitive means that you cannot find a disjoint pair of nontrivial, invariant sub-objects of $X$. A bit more formally, this means there are no subobjects $A,B \subset X$ such that $A \ne \emptyset$, $B \ne \emptyset$, $A \cap B = \emptyset$, $f(A)=A$, and $f(B)=B$.

I consciously tried to be vague in that formulation, so as to allow many types of transitivity.

The intuition here is that if $f$ is not transitive then you should be able to analyze $f$ by breaking it down into two (or more) independent invariant pieces. Transitive dynamical systems cannot be analyzed in that fashion.

For example, in the realm of bare set theory, consider $f$ to be a permutation of the finite set $X = \{1,2,3,...,n\}$. One definition of transitivity of $f$ is that the cycle decomposition of f consists of only a single cycle. For example the permutation $(1 2 3 4 5 6)$ is transitive, but the permutation $(1 3 5)(2 4 6)$ is not transitive. One can easily checks that the following are equivalent:

1. The cycle decomposition of $f$ consists of only a single cycle.
2. There does not a disjoint pair of nonempty subsets $A,B \subset X$ such that $f(A)=A$ and $f(B)=B$.
3. For any nonempty subsets $A,B \subset X$ there exists $n$ such that $f^n(A) \cap B \ne \emptyset$.

Consider now the case that $X$ is a topological space and $f : X \to X$ is a self-homeomorphism, i.e. a continuous self-map with continuous inverse. As you can see, topological transitivity of $f$ as you defined it in your comment (corrected so that $f^n(A) \cap B \ne \emptyset$) is quite similar to item 3, the main difference being that instead of letting $A,B$ be any old nonempty subsets, you choose them to be elements of the actual topology on $X$, i.e. open subsets. Furthermore, you should be able to use your knowledge of topology to prove that topological transitivity is equivalent to the following version of 2:

2. (topological version) There does not exist a disjoint pair of nonempty open subsets $A,B \subset X$ such that $f(A)=A$ and $f(B)=B$.
3. (topological version) For any pair of nonempty subsets $A,B \subset X$ there exists $n \ge 1$ such that $f^n(A) \cap B \ne \emptyset$.

Finally, since you included the ergodic-theory tag in your post, let me give the ergodic theoretic version of transitivity. Let $X$ be a measure space, for example a topological space equipped with a Borel measure. Let $f : X \to X$ be a measure preserving bijection. To say that $f$ is ergodic means that (either one of) the following two equivalent statements holds:

2. (measure theoretic version) There does not exist a disjoint pair of positive measure subsets $A,B \subset X$ such that $f(A)=A$ and $f(B)=B$.
3. (measure theoretic version) For any pair of positive measure subsets $A,B \subset X$ there exists $n \ge 1$ such that $f^n(A) \cap B \ne \emptyset$.