# Significance of topological transitivity in chaotic systems

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.

• To know what topological transitivity means, a good starting point is how it is defined. Do you know the definition? Commented Oct 28, 2022 at 13:09
• @LeeMosher I am aware of the formal definition yes: a map $f:X\to X$ is said to be topologically transitive if, for non empty, open subsets $A,B\subset X,\,\exists n\in\Bbb{N}^+,\,s.t\,f^n(B)\cap B=\emptyset$ Commented Oct 29, 2022 at 15:42
• I think you got the definition of topological transitivity a little wrong, it should say that for nonempty, open subsets $A,B \subset X$ there exists $n \in \mathbb N^+$ such that $f^n(A) \cap B \ne \emptyset$. Commented Oct 29, 2022 at 19:40

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:

1. (topological version) There does not exist a disjoint pair of nonempty open subsets $$A,B \subset X$$ such that $$f(A)=A$$ and $$f(B)=B$$.
2. (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:

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