Embedding Riemannian manifolds
"...the isometric embedding is understood in the global sense, i.e it must preserve all distances, not only infinitesimally small ones; for example no compact Riemannian manifold of negative sectional curvature admits such an embedding into Euclidean space."
It is not clear to me what the distinction is between the isometric embedding being discussed here and the one implied by the Nash Embedding Theorem when the manifolds are viewed as metric spaces. Yasmar (talk) 15:20, 16 August 2010 (UTC)
- Here the path-distance and the ambient distance must coincide. A moment's reflection will convince you that you can't even imbed the circle in Euclidean space in this sense. The appeal to manifolds of negative curvature is superfluous here. Tkuvho (talk) 15:24, 16 August 2010 (UTC)
Set or class?
Is "the set of all isometry classes of compact metric spaces" really a set?--22.214.171.124 10:43, 10 May 2007 (UTC)
Yes, actually there aren't that many. Every compact metric space is the completion of a countable dense subset. Therefore the cardinality of the set of all compact metric spaces is the same as the reals. 126.96.36.199 (talk) 16:21, 20 April 2009 (UTC)
- How do you prove it is a set? By its definition it is not a set (and therefore can not have a cardinality as cardinality is a property of sets). 188.8.131.52 (talk) 17:20, 22 November 2009 (UTC)
- As he said before; the cardinality of a compact metric space is at most the continuum, so up to isometry, we may assume the metric space is a subset of the reals. Clearly there are only set-many such spaces, and there are only set-many metrics possible on any given set. So it's a set, regardless of the fact that the class of compact metric spaces (without considering isometry classes) is obviously a proper class. 184.108.40.206 (talk)