Cardinality of all compact metric spaces

I'm looking for cardinal number of all compact metric spaces. I know that:

1. Cardinal number of compact set is at most $\mathfrak{c}$ (it is a continous image of Cantor set)
2. Compact metric space is separable and complete, so we can look just at countable dense set. It bound our cardinal number to cardinality of $\mathbb{R}^\mathbb{N}$ which is $\mathfrak{c}$

How can I bound it from below?

• Every finite metric space is compact. Are you perhaps interested only in the cardinality of infinite compact metric spaces? – Daniel Fischer Jun 2 '14 at 12:47
• There are of course countably infinite metric spaces too, for instance the one-point closure of the naturals. – Dan Rust Jun 2 '14 at 13:09
• Are you asking for the possible cardinalities of compact metric spaces? Obviously they can be finite, countable, or have cardinality that of $\mathbb{R}$. I guess you are saying they can't have higher cardinality. I didn't understand your argument. Maybe someone else can explain? – Seth Jun 2 '14 at 13:10
• Im looking for cardinality of set of all compact metric spaces, with homeomorphism as equivalence relation, not a cardinality of just compact metric spaces. I think that i found the upper limit of it: continuum. Am i wrong with conlcusions? Now im trying to find if continuum is really a number of all compact metric spaces. Im sorry for bad english. I`m doing my best. – Luke Jun 2 '14 at 14:06
• See math.stackexchange.com/questions/1602978/… for some ways of constructing $\mathfrak{c}$ non-homeomorphic compact metric spaces (specifically, subsets of $\mathbb{R}$). – Eric Wofsey Oct 31 '16 at 7:10

• Well, that's a nontrivial proof. Every Polish space is a $G_\delta$ subspace of the Hilbert cube, and every compact metric space is Polish; but a compact subspace of the Hilbert cube is closed so it suffices to talk about closed subspaces (in the other direction, note that every closed subspace of the Hilbert cube is a compact metric space). – Asaf Karagila Jun 3 '14 at 4:10