I'm studying something about the Conjugation and my reference propose to prove that the property of being topologically mixing is preserved by conjugation. Obviously the definition of topologically mixing is clear while about the conjugation I have that:
Give two metric spaces $X,Y$ and $f:X\rightarrow X$, $h:Y\rightarrow Y$ we can say that $f$ and $h$ are conjugate if there exists a homeomorphism $\theta:X\rightarrow Y$ such that $\theta\circ f=h\circ \theta$.($f\sim h$)
Shortly which I have to prove is: given X and Y two metric spaces and the two applications $f:X\rightarrow X$ and $h:Y\rightarrow Y$, $f$ is conjugated with $h$ by $\theta$ then $f$ is topologically mixing $\iff$ $h$ is too.
TOPOLOGICALLY MIXING: A function $f$ is topologically mixing if $\forall A,B$ open and not empty set $\exists n\in \mathbf{N}$ such that: $f^{a}(A)\cap B\not=\emptyset,\ \forall a\ge n.$
First show by induction that from $\theta f = h \theta$ it follows that
$$\forall n \in \Bbb N: \theta f^n = h^n \theta\tag{1}$$
and from that
$$\forall n \in \Bbb N: f^n \theta^{-1} = \theta^{-1} h^n \tag{2}$$
Now, if $h$ is top. mixing, let $U,V$ be non-empty open in $X$. Then $\theta[U], \theta[V]$ are non-empty open in $Y$ (homeomorphisms are open maps) so we have some $n \in \Bbb N$ such that
$$h^n[\theta[U]]\cap \theta[V]\neq \emptyset$$
but applying $(1)$ we have
$$\emptyset \neq \theta[f^n[U]] \cap \theta[V] =\theta[f^n[U] \cap V]$$
so $f^n[U] \cap V \neq \emptyset$.
The other direction ($f$ top. mixing then $h$ too) follows by lifting $U,V \subseteq Y$ to $X$ by using the mixing property of $f$ on the open sets $\theta^{-1}[U]$ and $\theta^{-1}[V] \subseteq X$ plus $(2)$, quite in the same way.