I was wondering today, that if $\mathscr{M}$ is the collection of all sets that admit a metric generating a compact topology, then

  1. Is $\mathscr{M}$ a set in ZFC?
  2. 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.

  • 3
    $\begingroup$ The collection of all one-element sets is not a set. $\endgroup$ – Hurkyl Mar 11 '13 at 15:38
  • $\begingroup$ @Hurkyl. How would you proceed to prove it? $\endgroup$ – T. Eskin Mar 11 '13 at 15:42
  • 2
    $\begingroup$ The union of its elements is the collection of all sets. $\endgroup$ – Hurkyl Mar 11 '13 at 15:43
  • 2
    $\begingroup$ A related concept: Urysohn's universal space is a separable metric space containing an isometric copy of every separable metric space. $\endgroup$ – Martin Mar 11 '13 at 16:13
  • 2
    $\begingroup$ @ThomasE. Let $A$ be the collection of one-element sets. Assume $A$ is a set. Then $\{A\}$ is a set by Pairing Axiom and is one-element, hence $\{A\}\in A$. Again by Pairing, $B:=\{A,\{A\}\}$ is a set. By Axiom of Foundation, there exists $C\in B$ with $C\cap B=\emptyset$. Clearly, $C\ne A$ because then $\{A\}\in C\cap B$. Also, $C\ne\{A\}$ because then $A\in C\cap B$. Contradiction! Therefore $A$ is not a set. $\endgroup$ – Hagen von Eitzen Mar 11 '13 at 16:32

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!).

  • $\begingroup$ Thanks Brian. Since you also addressed the isometric part, I have one more question. I figured that the quotient space under the relation of being isometric should have more equivalence classes than the homeomorphism relation, since not all homeomorphic spaces are isometric. What would be the cardinality of this quotient space? (notice the edit in this comment) $\endgroup$ – T. Eskin Mar 11 '13 at 15:57
  • $\begingroup$ @Thomas: Let $X$ be a fixed set of cardinality $\kappa\le 2^\omega$. The set of metrics on $X$ is a subset of the set of functions from $X\times X$ to $\Bbb R$, so we can form the set $$\big\{\langle X,d\rangle:d\text{ is a metric on }X\big\}\;.$$ Do this with an $X$ for each $\kappa\le 2^\omega$, and you have your $\mathscr{M}_0$. $\endgroup$ – Brian M. Scott Mar 11 '13 at 16:02
  • $\begingroup$ @Thomas: I’ve incorporated a few more specifics in my answer, including an answer to your question about the relationship between the homeomorphism and isometry classes. $\endgroup$ – Brian M. Scott Mar 11 '13 at 18:15
  • $\begingroup$ Thank you Brian, this is an excellent answer. Thanks for your effort. A further question for your added part: you probably use the notation $\mathscr{M}$ in a different way as before, where you noted that $\mathscr{M}$ was not a set. Also, is it easy to prove that for a compact metric space $(X,d)$ of cardinality $\kappa$ there exists an isometry $\phi:\kappa\to X$ for a suitable metric on $\kappa$? Do we just take the bijection $\phi:\kappa\to X$ given by the cardinality of $X$, and pull back the metric $d$ to $\kappa$ by defining $e(\alpha,\beta)=d(\phi(\alpha),\phi(\beta))$ ? $\endgroup$ – T. Eskin Mar 11 '13 at 21:40
  • $\begingroup$ @Thomas: You’re welcome. Yes, I used $\mathscr{M}$ in two different ways. And yes, just use the bijection to transfer the metric; it really is that simple. (Nice when that happens.) $\endgroup$ – Brian M. Scott Mar 11 '13 at 21:42

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.

  • $\begingroup$ Thanks Asaf for this clarification. $\endgroup$ – T. Eskin Mar 11 '13 at 15:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.