# Spaces of (complete) separable metric spaces

First, I want to make sure that I've got something right. By my understanding:

• There are separable (complete) metric spaces which are universal, i.e. contain isometrically embedded copies of every separable metric space. For example, Urysohn's space.
• Any such universal space, being a separable metric space, is second-countable and hence has at most continuum-many $$G_\delta$$ subsets.
• The Mazurkiewicz theorem says that, given a complete metric space, any subspace which is complete with respect to the induced metric must be a $$G_\delta$$ set.

Conclusion: up to isometry, there are only continuum-many complete separable metric spaces!

Is this correct?

A Course in Metric Geometry by Burago, Burago, and Ivanov defines a Gromov–Hausdorff distance on the class $$\mathcal{M}$$ of all metric spaces; the distance is possibly infinite, but we can partition $$\mathcal{M}$$ to obtain equivalence classes on which the distance is finite. I wish to restrict to those $${M \in \mathcal{M}}$$ which are separable, so that these equivalence classes are sets and hence "true" metric spaces.

If we restrict further e.g. to $$M$$ compact, then the resulting "space of spaces" is itself complete and separable. What happens if we weaken this condition?

(1a) If we require only that the $$M$$ be separable, are the resulting spaces of spaces complete with respect to the Gromov–Hausdorff metric? If not, how should we visualize points in their formal completion?

(1b) What if we require that the $$M$$ be separable and also complete?

(1c) Is the space of complete separable spaces itself separable?

This question originally included a section about a weaker notion of Gromov–Hausdorff convergence for pointed metric spaces. That section is now a separate question here: Spaces of separable metric spaces II: pointed spaces

there are only continuum-many complete separable metric spaces

Correct. It's not necessary to appeal to those theorems. A countable metric space $C$ is determined up to isometry by its infinite distance matrix $D_{ij} = d(c_i, c_j)$, $i, j\in\mathbb{N}$. There are continuum-many such matrices. The map sending each countable metric space to its completion is a surjection onto the space of all complete separable spaces (up to isometry).

(1a) The Gromov–Hausdorff limit $X$ of a sequence of separable metric spaces $X_n$ is separable. Indeed, for each $\epsilon$ we can pick $n$ such that $d_{GH}(X, X_n)<\epsilon$, and a countable $\epsilon$-net in $X_n$. Throw this net over to $X$ (pick some neighbor for each point), and you'll get a countable $(3\epsilon)$-net for $X$. Repeat for $\epsilon = 1/k$, take the union.

(1b) Completeness should be required of all spaces, otherwise the Gromov–Hausdorff distance is not a metric. Indeed, the distance between a space and its completion is always zero. So we may as well quotient out all non-complete spaces, replacing them by their completion.

For a concrete example, $X_n = [0, 1-1/n]$ converge to $X=[0, 1)$ (all spaces being considered with Euclidean metric). Of course $X$ is not complete. Then again, the sequence also converges to $[0, 1]$. There is no reason to consider $[0, 1)$ in this context.

(1c) No, the space of complete separable spaces contains an uncountable uniformly separated subset. One way to construct such a family of spaces is to define, for each subset $A\subset \mathbb{N}$, $$X_A = \{0\} \cup A$$ Then $X_A$ and $X_B$ are not isometric unless $A=B$. Indeed, $0$ is distinguished in $X_A$ as the only point $x$ that is not "between two others", i.e., $d(x, y)+d(x, z)=d(y, z)$ never holds unless $x=y=z$. So an isometry must send $0$ to $0$. But then distances from $0$ determine where everything else goes.

Recall that $d_{GH}(X, Y) = \frac12\inf_R \operatorname{dis}(R)$ where the infimum is taken over all left-right-total relations $\sim$ between $X$ and $Y$, and $$\operatorname{dis}(R) = \sup \{|d_X(x, x') - d_Y(y, y')| : x\sim y, x'\sim y'\}$$ A proof is in Burago-Burago-Ivanov and also here. In our case, all distances in $X_A$ and $X_B$ are integers. So, the distortion of a relation cannot be less than $1$ unless the relation is an isometry. In conclusion, $$d_{GH}(X_A, X_B)\ge \frac12,\quad \text{ whenever }\ A\ne B$$ The distance may be infinite in general, but there are also uncountably many sets $A$ within finite distance of each other: for example, the sets that contain all odd positive integers.

• Thank you for this comprehensive answer! I will split the question about pointed spaces as you advised. Mar 14, 2018 at 16:14