In my current course of discrete dynamical systems, I prove that if $x_o \notin \{y_0,f(y_0), f^{2}(y_0),...\}=$ orbit$(y_0,f)=o(y_0,f)$ and $y_0 \notin \{x_0,f(x_0), f^{2}(x_0),...\}=$ orbit$(x_0,f)=o(x_0,f)$, where $f:\mathbb{R}\rightarrow \mathbb{R}$ is an injective function, then $o(x_0,f) \cap o(y_0,f)= \varnothing$, but my professor told me that it's a wrong result.
My proof by induction is the following:
The base case is proved using similar arguments to the following. Now, we suppose that $f^{p}(x_0) \notin o(y_0,f)$ and $f^{p}(y_0) \notin o(x_0,f)$, with $p \in \mathbb{N}$. Then
$ \left.\begin{matrix} f^{p}(x_0)\neq f^{k}(y_0), \forall k\geq 0 \\ f^{p}(y_0)\neq f^{m}(x_0), \forall m\geq 0 \end{matrix}\right\} \Rightarrow \left.\begin{matrix} f^{p+1}(x_0)\neq f^{k+1}(y_0), \forall k\geq 0 \\ f^{p+1}(y_0)\neq f^{m+1}(x_0), \forall m\geq 0 \end{matrix}\right\} \Rightarrow $
$ \left.\begin{matrix} f^{p+1}(x_0) \notin o(f(y_0),f) \\ f^{p+1}(y_0) \notin o(f(x_0),f) \end{matrix}\right\} $
Now, we want that $f^{p+1}(x_0) \neq y_0$ and $f^{p+1}(y_0) \neq x_0$ in order to obtain $o(x_0,f) \cap o(y_0,f)= \varnothing$. Is $f^{p+1}(x_0) = y_0$ or $f^{p+1}(y_0) = x_0$ possible? No, because if $f^{p+1}(x_0) = y_0$, then $y_0 \in o(x_0,f)$, which is a contradiction to the hypothesis $y_0 \notin o(x_0,f)$. Therefore, $o(x_0,f) \cap o(y_0,f)= \varnothing$
Anyway this seems to be wrong, could someone give me a counterexample? Thanks in advance.
