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Given a separable Hilbert space $H$, $U$ is a unitary operator. A cyclic subspace, denoted as $Z(x)$ for some $x\in H$, is defined as the closure of linear span of $U^nx$, where $n\in \Bbb Z$ is any integer number.

Now we have a sequence of cyclic subspaces, namely, $Z(x_1)\subset Z(x_2)\subset \cdots$. Then the closure of its union, $\overline{\bigcup_iZ(x_i)}$, is also a cyclic subspace.

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    The problem is that we don't know what this $x$ could be. We can assume that $||x_j||=1$, maybe taking $x\in\bigcap_n C_n$, where $C_n$ is the closed convex hull of $\{x_k,k\geq n\}$, will work.2011-12-30

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Adding some details to @DavideGiraudo's comment: in the context of Hilbert spaces, one might not think of using (an easy case of) the Banach-Alaoglu theorem, namely, that closed unit balls are compact in the weak dual topology. Thus, the closure of the convex hull of $C_n=\{x_k:k\ge n\}$ is compact in the weak dual topology, and their intersection is non-empty.

Given the existence of a (non-zero) vector in that intersection, it is not hard to prove it's cyclic.