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A dynamical system $(X,\phi)$ is said to be topologically transitive if for any non-empty open sets $U,V\subset X$, there exists a positive integer $n$ such that $\phi^n(U)\cap V \neq \emptyset$.

Birkhoff's Transitivity theorem asserts tat if $X$ is a second countable, complete metric space, then topological transitivity implies that there is a dense set of points in $X$ with dense orbit.

In theorem 2.5 here, the proof goes that $\cup_{m=1}^\infty \phi^{-m}(U_k)$ is dense, by topological transitivity. I cannot see why this claim is true. Can anyone help?

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  • $\begingroup$ What happend if that set is not dense? $\endgroup$ – Anubhav Mukherjee Mar 8 '18 at 20:27
  • $\begingroup$ Then there exists an open $V$ such that $V \cap \cup_{m=1}^\infty(U_k) = \emptyset$. And so $\phi^{-m}(U_k) \cap V = \emptyset$ for all $m$. Then I'm not sure how to proceed. $\endgroup$ – Zhanfeng Lim Mar 8 '18 at 20:46
  • $\begingroup$ Hint: show that $\phi$ is topologically transitive if and only if for any nonempty open sets $U,V\subset X$, there exists a positive integer $n$ such that $\phi^{-n}(U)\cap V \neq \emptyset$. $\endgroup$ – John B Mar 9 '18 at 0:03

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