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Let $S$ be the space of all real sequences with direct product topology (i.e. $S=\prod_{n=0}^\infty \mathbb{R}$) and let $U$ be an open set in $\mathbb{R}^n$. Is there a way to define smooth function in such a way that $\mathcal{C}^\infty(U,\prod_{n=0}^\infty \mathbb{R}) = \prod_{n=0}^\infty \mathcal{C}^\infty(U,\mathbb{R})$?

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What kind of differentiability are you providing to functions F : U -> S? I mean, differentiability in a space of sequences can be defined in a myriad of ways. Most of them not as easy as the ones in R^n. –  Henrique Tyrrell Mar 27 '12 at 17:32
Is a toplogoy a non-Jewish topology? –  Gerry Myerson Mar 27 '12 at 22:59

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