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



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



How to show the cardinality of nonisometric compact metric spaces is the continuum
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.


