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The Hilbert cube $H$ in $\ell^2=\ell^2(\mathbb{R})$ is the subset given by $$H=\lbrace(x_n)=(x_1,x_2,\ldots)\in\ell^2:|x_n|\le2^{-n} \text{ for }n=1,2,\ldots\rbrace.$$ I've heard that $\overline{\mathrm{Sp}{(H)}}=H$ and core$(H)=\varnothing$, where $\overline{\mathrm{Sp}{(H)}}$ denotes the closed linear span of $H$. An element $x\in H$ is in core$(H)$ whenever $$\ell^2=\bigcup_{t>0}t\cdot(H\setminus\{x\})$$ where $t\cdot(H\setminus\{x\})=\{t\cdot y: y\in (H\setminus\{x\})\}$.

Is there a book where I can find a proof of there statements?

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I don't understand the definition of $core(H)$. With this definition, we should that that $\ell^2=\bigcup_{t>0}tH$, which is not possible, considering $x_k=2^{-k/2}$. – Davide Giraudo Jul 18 '12 at 11:20

Let $$D:=\bigcup_{n\geq 1}\{(x_k),x_k=0,k\geq n\}.$$ $D$ is a dense subset of $\ell^2$ and is contained in the vector subspace generated by the elements of $H$.

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