Question

It is asserted in A Course in Metric Geometry by Burago, Burago, Ivanov that

there can be no more than continuum of mutually nonisometric compact spaces

How is this proven?

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?

Answer 1

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.

Answer 2

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 $\mathbb{N}$.

Answer 3

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.