most recent 30 from http://mathoverflow.net 2013-05-24T02:29:16Z http://mathoverflow.net/feeds/question/48951 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/48951/how-to-show-the-cardinality-of-nonisometric-compact-metric-spaces-is-the-continuu Otis Chodosh 2010-12-10T18:12:30Z 2010-12-10T18:31:47Z

<p>It is asserted in <em>A Course in Metric Geometry</em> by Burago, Burago, Ivanov that</p> <blockquote> <p>there can be no more than continuum of mutually nonisometric compact spaces</p> </blockquote> <p>How is this proven? </p> <p>Its clear that there must be at least a continuum of mutually nonisometric compact spaces, i.e. $([0,\alpha], d_{\mathbb{R}})$ for $\alpha>0$ are a family of nonisometric metric spaces, but I don't know enough set theory to have any ideas how to bound the cardinality from above. A first guess was that the fact that compact metric spaces are totally bounded should be useful?</p>

http://mathoverflow.net/questions/48951/how-to-show-the-cardinality-of-nonisometric-compact-metric-spaces-is-the-continuu/48954#48954 Pietro Majer 2010-12-10T18:27:33Z 2010-12-10T18:27:33Z

<p>Since a compact metric space is in particular separable, its type of isometry is determined by a dense countable subspace. There are continuum many distances on , say $\mathbb{N}$.</p>

http://mathoverflow.net/questions/48951/how-to-show-the-cardinality-of-nonisometric-compact-metric-spaces-is-the-continuu/48955#48955 Mark Meckes 2010-12-10T18:29:31Z 2010-12-10T18:29:31Z

<p>I think "compact" can be even weakened here to "separable and complete" (and regarding your first guess, total boundedness is essentially used to prove that compact implies separable). Here's a sketch: any such space is determined, up to isometry, by the restriction of the metric to a countable dense subset. Thus the number of such isometry classes is bounded above by the cardinality of $\mathbb{R}^{\mathbb{N}\times \mathbb{N}}$, which is the same as the cardinality of $\mathbb{R}$, which is the continuum.</p>

http://mathoverflow.net/questions/48951/how-to-show-the-cardinality-of-nonisometric-compact-metric-spaces-is-the-continuu/48957#48957 Gerald Edgar 2010-12-10T18:31:47Z 2010-12-10T18:31:47Z

<p>Extend Qiaochu's deleted comment. Compact metric spaces are separable, can be isometrically embedded as closed subsets of the separable metric space $C[0,1]$. This space has continuum many closed subsets.</p>