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?

  • 2
    $\begingroup$ Every finite metric space is compact. Are you perhaps interested only in the cardinality of infinite compact metric spaces? $\endgroup$ – Daniel Fischer Jun 2 '14 at 12:47
  • $\begingroup$ There are of course countably infinite metric spaces too, for instance the one-point closure of the naturals. $\endgroup$ – Dan Rust Jun 2 '14 at 13:09
  • $\begingroup$ 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? $\endgroup$ – Seth Jun 2 '14 at 13:10
  • $\begingroup$ 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. $\endgroup$ – Luke Jun 2 '14 at 14:06
  • $\begingroup$ See math.stackexchange.com/questions/1602978/… for some ways of constructing $\mathfrak{c}$ non-homeomorphic compact metric spaces (specifically, subsets of $\mathbb{R}$). $\endgroup$ – Eric Wofsey Oct 31 '16 at 7:10


There is no set of all compact metric spaces, because every singleton is a compact metric space, and the collection of all singletons is not a set.

However, note that every compact metric space is homeomorphic to a closed subspace of the Hilbert cube. So if you are only interested in equivalence classes of compact metric spaces, it suffices to consider subspaces of the Hilbert cube.

  • $\begingroup$ Why are compact metric spaces homeomorphic to subspaces of the Hilbert cube? $\endgroup$ – Seth Jun 3 '14 at 0:16
  • $\begingroup$ 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). $\endgroup$ – Asaf Karagila Jun 3 '14 at 4:10

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