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Suppose $S$ is a set of n points, that is $|S| =n$ seen as a discrete smooth manifold. Then is the cartesian product of manifolds $\mathbb{R} \times S \simeq \mathbb{R}^n$? If not what is it?

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    Is it possible to have a *discrete smooth manifold*? If it is discrete, how can it be smooth?2012-03-11

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$\mathbb R\times S$ is homeomorphic to a disjoint union of $|S|$ copies of $\mathbb R$. This is not the same as $\mathbb R^n$, unless $n=1$. The space $\mathbb R\times S$ is a $1$-dimensional manifold, whereas $\mathbb R^n$ is $n$-dimensional. Also $\mathbb R\times S$ is disconnected, for $|S|>1$, whereas $\mathbb R^n$ is connected.

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    @MarkNeuhas: No. If you have two disjoint lines, there is no path that leads continuously from one to the other. So this space is not "path-connected." One line is, on the other hand, path-connected. Diffeomorphisms (and actually more generally homeomorphisms) preserve the property of path-connectivity.2012-03-10