I know that $\mathbb{R}^n$ has the structure of CW complex by considering lattices. What about $\mathbb{R}^\infty$?
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1What do you mean by $\Bbb{R}{\infty}$? If you mean $\bigcup_n\Bbb{R}^n$, the answer is yes. If you mean an infinite product of copies of $\Bbb{R}$, I think the answer is probably no.. – Rob Arthan Sep 14 '17 at 15:31
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I mean $\cup_n \mathbb{R}^n$. – baby bunny Sep 14 '17 at 15:37
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View $\Bbb{R}^n$ as the set of sequences $(x_1, \ldots, x_n, 0, 0, \ldots)$, and make each $\Bbb{R}^n$ into a CW complex in such a way that $\Bbb{R}^n$ is a subcomplex of $\Bbb{R}^{n+1}$. (The natural lattice construction will have this property.) The union of the cells in all the $\Bbb{R}^n$ gives a CW structure on $\bigcup_n\Bbb{R}^n$.
Rob Arthan
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