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A linear automorphism $T: X \rightarrow X$, where $X$ is Banach, is said to be topologically transitive if for all $U$ and $V$ open, there exists $n \in \mathbb{Z}$ such that $T^n(U)\cap V \neq \varnothing.$

When $X$ is separable, transitive is equivalent to $T$ having a dense orbit, by a well known theorem of Birkhoff.

Every separable Banach space supports a transitive operator. Concrete examples include certain weighted shifts on $l^p$, $p \in (1, \infty)$.

Clearly $T$ cannot have a dense orbit in $l^{\infty}$, since it is not separable.

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