Let $X\subset \mathbb R$ be an interval and define the continuous map $f:X\longrightarrow X.$ If for every pair $I,J \subset X$ of closed intervals ($\neq$ single set ) exists $n\in \mathbb N:J\subset f^n(I),$ prove that the topological dynamical system $f$ is Topologically Transitive.
My work so far is that for every open set $U\subset X$ exists a closed interval $I\subset X:U\subset I\subset f^n(J)$ for some integer $n\ge 1.$
Im not sure if that leads anywhere, any hints?
Note: If $(X,f)$ is a dynamical system. Then $f$ is said to be topologically transitive if for every pair of non-empty open sets $U$ and $V$ in $X$ there exists $n \geq 1$ such that $f^n(U) \cap V\neq \emptyset.$
Also, $\exists$ x $\in X: \overline{\,\mathcal O_{f}^+\,}(x)=X \implies f$ is topologically transitive, where $\mathcal O_f^+(x) $ is the semi-orbit of $f(x).$
Let $U$ and $V$ be two non empty open sets and $I \subset U$, $J \subset V$ two included non-empty intervals. Then there is $n$ such that $J \subset f^n(I)$ and so $J \cap f^n(I) \ne \emptyset$. That is there is $x$ in $I$ such that $f^n(x) \in J$.
But as $J \subset V$, we also have $f^n(x) \in V$ and as $I \subset U$, we have $x \in U$ and so $x \in f^n(U) \cap V \ne \emptyset$. This is the definition you gave of topologically mixing.