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I have the next unitary representation, $\pi : G\rightarrow \mathcal{U}(H)$, where G is a closed subgroup of $S_{\infty}$ (the group of bijective functions from $\mathbb{N}\rightarrow \mathbb{N}$), and $\mathcal{U}(H)$ is the group of unitary operators on a separable hilbert space $H$, and I want to show that $\bigcup\{H_V : V \text{ clopen subgroup of } G\}$ is dense in $H$, where $H_V=\{ x \in H : \pi(v)x=x, \forall v \in V \}$.

I think it's easy but I am not sure, I take a sequence $x_n \in H_V$, and assume that $x_n \rightarrow x$, where $x \in H$, then because $\pi$ is a unitary operator, it's bounded and thus continuous as well, so we know that $\pi(v)x_n \rightarrow \pi(v)x$, and because $\pi(v)x_n=x_n$, we get from uniqueness of the limit that $\pi(v)x=x$, so $x\in H_V$.

Where did I get it wrong?

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up vote 3 down vote accepted

Your proof is perfectly valid, but it doesn't prove that $$W = \bigcup \{ H_{V} \,|\, V \text{ clopen subgroup of G} \}$$ is dense in $H$, it proves that each $H_{V}$ is closed in $H$. To prove that $W$ is dense, you could pick an arbitrary element $x \in H$ and show that there exists some sequence $(x_{n})_{n = 1}^{\infty} \subseteq W$ such that $x_{n} \to x$.

Edit: If $G$ is discrete (which it is probably not), then the result is completely trivial, since $\{e\}$ (the trivial subgroup containing only the identity) would be a closed and open subgroup of $G$, and $H_{\{e\}} = H$. However, I doubt it is that easy...

Edit 2: I found a proof of the result in an article I had lying around. See lemma 3.1 in

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Do you have some hint as to how to show this? Thanks. – MathematicalPhysicist Feb 7 '12 at 6:12
@MathematicalPhysicist: see second edit. – Martin Wanvik Feb 7 '12 at 10:58
Great, thanks!! – MathematicalPhysicist Feb 7 '12 at 14:56

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