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Let $X$ be a separable Hausdorff topological space and $\phi \in C(X,X)$ be a topologically transitive map. Further, let $V$ be a fixed non-empty open subset of $X$. Then does there necessarily exist a countable open cover $\{U_i\}_{i \in \mathbb{N}}$ of $X$ and a sequence of natural numbers $\{N_i\}_{i \in \mathbb{N}}$ such that $$ \bigcup_{n \in \mathbb{N}} \phi^{N_i}(U_i) \subseteq V? $$

ie the entire space fits into $V$; eventually?

[$\phi$ is said to be topologically transitive iff for every two non-empty open subsets $U,V\subseteq X$ there exists some $N\in \mathbb{N}$ such that $\phi^N(U)\cap V \neq \emptyset$.]

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No.

Let $\phi$ be the left shift on the set $X = \{0,1\}^\mathbb{Z}$ of bi-infinite binary sequences with the prodiscrete topology, and let $V = \{ x \in X : x_0 = 0 \}$ be the set of sequences that have $0$ at the central coordinate. Then $X$ is Hausdorff (even metrizable and compact), $\phi$ is transitive (even mixing) and $V$ is nonempty and open (even clopen). The all-$1$ sequence $x = \ldots 1 1 1 \ldots$ satisfies $\phi^n(x) \notin V$ for all $n \in \mathbb{Z}$. Hence in any open cover $(U_i)_{i \in I}$, some $U_i$ contains $x$ and then $\phi^n(U_i)$ is not a subset of $V$ for any $n$.

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