Forward and backward iterations of a homeomorphism

Question

Let $X$ be a compact metric space. We say a homeomorphism $h:X\to X$ is transitive if there exists $x\in X$ such that $\{h^n(x):n\in \mathbb N\}$ is dense in $X$.

According to the answer in Transitive homeomorphism is positively transitive, I think if $h$ is transitive, then so is $h^{-1}$.

Now I want to ask about a related property. Suppose that for every pair of non-empty open sets $U,V\subseteq X$ there exists $N\in \mathbb N$ such that $h^n[U]\cap V\neq\varnothing$ for all $n\geq N$. We say $h$ is mixing. Does this imply $h^{-1}$ is also mixing?

score 0 · Accepted Answer · answered Oct 21 '19 at 20:57

0

Yes, let $U$ and $V$ be open sets and $N\in \mathbb{N}$ one chosen such that for all $n\ge N$: $h^n (U)\cap V \ne \emptyset $. Then we also have $\emptyset \ne h^{-n}(V)\cap U=h^{-n}( h^n (h^{-n}(V))\cap h^n(U))=h^{-n}(V\cap h^n(U))$ and hence also $h^{-1}$ is mixing.

answered Oct 21 '19 at 20:57

Florian

736

dont we want to prove $h^{-n}[U]\cap V\neq\varnothing$? – Mr Math Oct 21 '19 at 21:05
it works if at the beginning you write $h^n(V)\cap U\neq\varnothing$ for all $n\geq N$... – Mr Math Oct 21 '19 at 21:08

Forward and backward iterations of a homeomorphism

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