Cardinality of the collection of all compact metric spaces
I was wondering today, that if $\mathscr{M}$ is the collection of all sets that admit a metric generating a compact topology, then
- Is $\mathscr{M}$ a set in ZFC?
- If it is, what is the cardinality of $\mathscr{M}$?
The reason why I found $1.$ interesting is because in general, the collection of all sets is not a set in ZFC, and each set admits a metric space structure via discrete metric and discrete topology. However, only those sets that are finite are compact with this metric. So could compactness assumption restrict this collection to become a set?
And about $2.$, I figured that atleast under the equivalence relation of homeomorphism the quotient space should be a set with cardinality less or equal than $2^{\mathfrak{c}}$, since every compact metric space admits an embedding to $[0,1]^{\mathbb{N}}$, which is of the size $\mathfrak{c}$. So for each equivalence class we could pick a representative from the power-set of $[0,1]^{\mathbb{N}}$. But how about if we consider the relation of being isometric instead of homeomorphic? And finally, if we drop the quotient spaces entirely?
My motivation for these questions arose when I was reading about the Gromov-Hausdorff distance for compact metric spaces. Maybe there is a trivial answer that $\mathscr{M}$ is not a set.
Solution 1:
Generally collections of "all possible" are classes, simply because there is a proper class of sets of cardinality $1$, all of which are compact metric spaces.
But indeed every compact metric space can be embedded into $[0,1]^\Bbb N$, therefore we only need to find out how many closed sets this space has, and since it is a second countable metric space it has exactly $\frak c$ many closed subsets, so there are at most $\frak c$ compact metric spaces up to homeomorphism.
Solution 2:
For any $x$, $\{x\}$ with the metric $d(x,x)=0$ is a compact metric space, so it’s clear that $\mathscr{M}$ cannot be a set. However, it is true that $|X|\le 2^\omega$ for every $X\in\mathscr{M}$, so there is a set $\mathscr{M}_0$ of compact metric spaces such that every compact metric space is homeomorphic (indeed isometric) to one in $\mathscr{M}_0$.
Added: For each cardinal $\kappa\le 2^\omega$ let $$\mathscr{M}_\kappa=\big\{\langle\kappa,d\rangle:d\text{ is a metric on }\kappa\big\}\subseteq{}^{\kappa\times\kappa}\Bbb R\;;$$ this is clearly a set, as is $\mathscr{M}=\bigcup\{\mathscr{M}_\kappa:\kappa\le 2^\omega\text{ is a cardinal}\}$, and every compact metric space is isometric to some space in $\mathscr{M}$. (For $\kappa>1$ $\mathscr{M}_\kappa$ contains $2^\kappa$ isometric copies of each space, corresponding to permutations of $\kappa$, so you might want to choose representatives of the isometry classes.)
To answer your last question in the comments, a compact metric space $X$ is separable, and every compatible metric is continuous on $X\times X$, so it has at most $2^\omega$ compatible metrics. On the other hand, it $d$ is a compatible metric, then so is $\alpha d$ for every $\alpha>0$, so $X$ has $2^\omega$ compatible metrics (unless $X$ is the one-point space!).