Say $M$ is a manifold and $f: M \to M$ is a diffeomorphism. Assume also that, if we are given any nonempty open subsets $U$ and $V$, then there is $n \in \mathbb{Z}$ such that $f^n(U)$ intersects $V$.
Apparently it is then true that $f$ is transitive, i.e., $f$ has a dense orbit, i.e, there exists $p \in M$ such that $\{f^n(p) | n\in \mathbb{Z}\}$ is dense in $M$.
I can see that the converse is true (quite easily) but I'm not sure how to show that the statement above is true...how could I pinpoint this particular $p$ whose orbit is dense?
This is the classical proof: There exists a countable family of open subsets $(U_n)_{n\in\mathbb{N}}$ which is a basis of the topology. The orbit of $x$ is dense if for every $n$, there exists $m\in\mathbb{Z}$ such that $f^{m}(x)\in U_n$ or equivalently $x\in f^{-m}(U_n)$, this is equivalent to say that $x\in \cup_{m\in\mathbb{Z}}f^m(U_n)$ for every $n\in \mathbb{N}$ and this last condition is equivalent to say that $I=x\in\cap_{n\in\mathbb{N}}\cup_{m\in\mathbb{Z}}f^{-m}(U_n)$.
The set $\cup_{m\in\mathbb{Z}}f^{-m}(U_n)$ is dense since for every open subset $V$, there exists $m\in\mathbb{Z}$ such that $f^m(U_n)\cap V$ is not empty. The Baire category theorem implies that $I$ is not empty.