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?
Cartesian Product of the Real line with a discrete sets
0
$\begingroup$
general-topology
differential-geometry
-
0Is it possible to have a *discrete smooth manifold*? If it is discrete, how can it be smooth? – 2012-03-11
1 Answers
5
$\mathbb R\times S$ 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.
-
0@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