Let $f\colon X\rightarrow X$ be a topological dynamical system with at least two orbits. ($f$ is homeomorphism, the orbit $\mathcal{O}_f(x):=\{f^n(x)|n\in \mathbb{Z}\}$.)
Show that if $f$ has an attracting fixed point, then it is not topologically transitive. ($X$ is locally compact, metrizable, and second countable.)
Attracting fixed point $p$ means that the fixed point $p$ has a neighborhood $U$ such that $\bar{U}$ is compact, $f(\bar{U})\subset U$, and $\cap_{n\geq 0}f^{n}(U)=\{p\}$.
Topologically transitive: there is a point $x\in X$ whose forward orbit is dense in $X$.
I don't know what to do with this condition $\cap_{n\geq 0}f^{n}(U)=\{p\}$.
Suppose that $f$ is a homeomorphism (as otherwise counterexamples exist). Let $x$ be a point with dense orbit. Let $m$ be such that $f^m(x) \in U$, which exists by the density of the orbit of $x$. Well, then $V:=f^{-m}(U)$ also satisfies $V$ is open and contains $p$, $\bar{V}$ is compact, $f(\bar{V}) \subset V$ and $\cap_{n \geq 0} f^n(V) = \{p\}$ (check these properties). But note that $V$ contains $x$.
Can you see how this must imply that $f^n(x)$ converges to $p$? Why does this contradict the fact that $x$ supposedly has a dense orbit?