On Urysohn’s universal separable metric space

Martin Ziegler

May 15, 2003

The theorems in the paper are not new. I will rewrite the paper soon and give the references.

1 Introduction

In [2] Urysohn constructed a separable metric space U with the following two properties

• (universality) every separable metric space is embeddable in U,

• (homogeneity) every isomorphism between two finite subspaces extends to an automorphism of U.

and asked if U is the only separable metric space with these properties. We give a negative answer to his question in section 4. We also show

Theorem 1.1. Urysohn’s space is characterised by each of the following conditions:

1. U is separable, homogeneous, complete and embeds every finite metric space.

2. U is separable, universal and strongly homogeneous , i.e. every isomorphism between totally bounded subspaces extends to an automorphism.

In section 2 we develop the tools needed for the proof of the theorem, which is proved in section 3. We construct the counterexample in 4.

2 Urysohn’s space

It is easy to see that metric spaces can be amalgamated:

Lemma 2.1. If a metric space A is (isometrically) embedded in B_{1} and B_{2} by f_{1} and f_{2}, there exist a metric space C and embedings g_{i} : B_{1} → C such that g_{1}f_{1} = g_{2}f_{2}.

Proof. We may assume that the f_{i} are inclusions and that A = B_{1} ∩ B_{2}. Define a pseudometric on B_{1} ∪ B_{2} by

δ(b_{1}, b_{2}) = inf_{a∈A} δ(b_{1}, a) + δ(a, b_{2})  (1)

if b_{i} ∈ B_{i}. We obtain C and the f_{i} by factoring B_{1} ∪B_{2} through the equivalence relation δ(x, y) = 0.

We call a metric space rational if all distances are rational numbers. The C constructed in the last proof is rational if the B_{i} are rational and A is finite. The class of all finite rational spaces is therefore closed under amalgamation, and since there are only countably many isomorphism types of finite rational spaces, this yields^{1}

Lemma 2.2. There is a (unique) countable rational space U_{0}, which is homogeneous and embeds every finite rational space.

Of course U_{0} embeds not only all finite but all countable rational spaces.

Let A be a metric space. A type over A is a family p = (p_{a})_{a∈A} of reals which is realized by an element x of an extension of A, in the sense that p_{a} = δ(x, a). In this case we call p = tp(x/A) the type of x over A. We denote by S(A) the set of all types over A. A type p is rational if all p_{a} are rational.

The following can be proved by standard arguments.

Lemma 2.3. A countable rational space X_{0} is isomorphic to U_{0} iff every rational type over every finite subset of X_{0} is realized in X_{0}.

Definition. Urysohn’s space U is the completion of U_{0}.

The next proposition extends a result of C. Joiner [1], who proved that isomorphism between Cauchy sequences extend to automorphisms of U.

Proposition 2.4. U realizes every type over every totally bounded subset.

We will give the proof at the end of this section.

Let A and B be two non-empty subsets of a metric space. The distance δ(A, B) maximum of sup_{a∈A} inf_{b∈B} δ(a, b) and sup_{b∈B} inf_{a∈A} δ(a, b). It is easy to check, that this defines a pseudometric. In order to define a pseudometric on the class of types over non-empty sets of finite diameter, we attach to every p ∈ S(A) the set p̄ = {(p_{a}, a) | a ∈ A} ⊂ ℝ × A. We give ℝ × A the maximum metric

1Every metric space embeds the one-point space. Thus, by amalgamation, every two metric spaces have a “joint embedding”.

δ((α, a), (α′, a′)) = max (|α − α′|, δ(a, a′)).

p̄ has then finite diameter, and we define δ(p, q) as the distance between p̄ and q̄.

One proves easily

Lemma 2.5. Let p = tp(x/A) and q = tp(y/B). Then

δ(A, B) ≤ δ(p, q) ≤ δ(A, B) + δ(x, y).

Note that, by Lemma 2.1, we can assume that x and y lie in the same space.

Lemma 2.6. Let p ∈ S(A) and q ∈ S(B). Then there are realizations x and y of p and q such that

δ(x, y) ≤ 2(δ(p, q) + δ(A, B)).

If A and B are non-empty finite subsets of U_{0}, and p and q are rational, we find x, y in U_{0}.

Proof. Assume first that A = B. Choose realizations x, y of p and q in two different extensions of A. We amalgamate the two spaces A ∪ {x} and A ∪ {y} by setting δ(x, y) = μ where

μ = sup_{a∈A} |d(x, a) − d(y, a)| = sup_{a∈A} |p_{a} − q_{a}|.

The inequality

|p_{a} − q_{a}| ≤ |p_{a} − q_{b}| + |q_{b} − q_{a}| ≤ |p_{a} − q_{b}| + δ(a, b) ≤ 2 max(|p_{a} − q_{b}|, δ(a, b))

shows that μ ≤ 2δ(p, q) and we are finished.

For the general case, choose first an arbitrary realization x of p and set p′ = tp(x/B). The first case gives us a realization y of q such that δ(x, y) ≤ 2δ(p′, q). The result follows now from δ(p′, q) ≤ δ(p, q) + δ(p, p′) = δ(p, q) + δ(A, B).

The last part of the lemma follows from an inspection of the above proof.

Lemma 2.7. Let p a type over the non-empty finite set A and ϵ > 0. Then there is a rational type q over A such that δ(p, q) < ϵ.

Proof. Left to the reader^{2}.

2Note that A may not be rational.

Proof of the proposition. Let B be a totally bounded subset of U and q ∈ S(B). We may assume that B is non-empty. Realize q in an extension of U by an element y (using Lemma 2.1) and fix a convergent series ∑_{i=0}^{∞} ϵ_{i} with positive summands.

Since B is totally bounded and U_{0} is dense in U, we find for every i a finite A_{i} ⊂ U_{0} with

δ(A_{i}, B) ≤ ϵ_{i}.

Also, for each i, we choose a rational p_{i} ∈ S(A_{i}) with δ(p_{i}, tp(y/A_{i})) ≤ ϵ_{i}. We have then

δ(p_{i}, p) ≤ ϵ_{i} + δ(tp(y/A_{i}), p) = ϵ_{i} + δ(A_{i}, B) ≤ 2ϵ_{i}.

We define recursively a sequence of realizations x_{i} ∈ U_{0} of the p_{i}. x_{0} is an arbitrary realization of p_{0}. If x_{i} defined, we apply Lemma 2.6 to get a realization x_{i+1} ∈ U_{0} of p_{i+1} with δ(x_{i}, x_{i+1}) ≤ 2 (δ(p_{i}, p_{i+1}) + δ(A_{i}, A_{i+1})). From δ(p_{i}, p_{i+1}) ≤ 2ϵ_{i} + 2ϵ_{i+1} and δ(A_{i}, A_{i+1}) ≤ ϵ_{i} + ϵ_{i+1} we get

δ(x_{i}, x_{i+1}) ≤ 6ϵ_{i} + 6ϵ_{i+1}

which shows that (x_{i}) is a Cauchy sequence, which has a limit z in U. z realizes p, since for all i

δ(tp(z/B), p) ≤ (δ(tp(z/B), tp(z/A_{i})) + δ(tp(z/A_{i}), p_{i}) + δ(p_{i}, p)
≤ δ(A_{i}, B) + δ(z, x_{i}) + δ(p_{i}, p)

This shows that δ(tp(z/B), p) = 0 and therefore, by 2.6, tp(z/B) = p.

3 Proof of Theorem 1.1

Theorem 1.1 follows Proposition 2.4 and the next three lemmas.

Lemma 3.1. A separable space which realizes all types over totally bounded sets is universal and strongly homogeneous.

Proof. Let X satisfy the hypothesis. Since X realizes types over finite sets, a standard argument shows that X embeds every countable space. Since a Cauchy sequence (x_{i}) in X is totally bounded, the type of its limit of (x_{i}) is realized in X, which proves that X is complete. Now let Y be a separable space with countable dense subspace Y_{0}. Since X is complete any embedding of Y_{0} in X extends to an embedding of Y. This proves that X is universal.

Now let us show that X is strongly homogeneous. We start with an isomorphism f : A → B between two totally bounded subset of X. If we increase A by finitely many elements, still every type over the new set is realized in X, the same holds for B. This shows that f can be extended to an isomorphism g_{0} : Y_{0} → Z_{0} between two subsets of X which contain any given countable set. So, if we choose choose Y_{0} and Z_{0} to be dense in X, g_{0} extends uniquely to an automorphism of X.

Lemma 3.2. A space which is universal and strongly homogeneous is complete.

Proof. Let X be universal and and strongly homogeneous and (x_{i})_{i∈ℕ} be a Cauchy sequence in X. Since X is universal, it contains a space {y_{i}}_{i∈ℕ} ∪ {y}, where f : y_{i} ↦ x_{i} is an isomorphism and y is the limit of (y_{i}). Extend f to an automorphism g of X. Then (x_{i}) converges to g(y).

Lemma 3.3. Two separable spaces which are homogeneous, complete and embed every finite space are isomorphic.

Proof. Let X and Y as in the hypothesis. Consider a type p over a finite subset A of X. Choose a realization b of p in an extension of A. X contains an isomorphic copy A′ ∪ {b′} of A ∪ {b}. The isomorphism between A′ and A extends to an automorphism of X. This automorphism maps b′ to a realization of p which lies in X. We have thus proved that X (and Y) realizes every type over any finite subset.

The argument in the second part of the proof of Lemma 3.1 now shows that X and Y are isomorphic.

4 The counterexample

A filtered space is a metric space X with an increasing sequence X^{0} ⊂ X^{1} ⊂ ⋯ of non–empty closed subsets, with union X. A filtered space determines an f–space, this is X together with the distance functions

δ_{i}(x) = δ(x, X^{i}).

A subset of an f–space, with its induced metric and distance functions, is called an f–set.

Remark 4.1. Let A be metric space with a family (δ_{i}) of real valued functions. (A, δ_{i})_{i∈ℕ} is an f–set iff the following holds

• δ_{i}(x) ≥ 0

• δ_{i+1}(x) ≤ δ_{i}(x)

• δ_{i}(x) ≤ δ(x, y) + δ_{i}(y)

• For all x almost all δ_{i}(x) are zero.

Proof. Fix a_{0} ∈ A and i_{0} ∈ ℕ. Set B = A ∪ {b} for a new b and define δ(a, b) = δ(a, a_{0}) + δ_{i_{0}}(a_{0}) and

δ_{i}(b) = \begin{cases}0 & if i ≥ i_{0} \\ δ_{i_{0}}(a_{0}) + δ_{i}(a_{0}) & if i < i_{0} \end{cases}

B is an extension of A and the δ_{i} still have the four properties. It follows that A has an extension X such that For all x ∈ X and i ∈ ℕ there exists a d such that δ_{i}(d) = 0 and δ(x, d) = δ_{i}(x). We set X^{i} = {x | δ_{i}(x) = 0} and obtain a filtered space with distances functions δ_{i}.

The proof shows that every finite f–set can be extended to a finite filtered space.

An f–set (A, δ_{i})_{i∈ℕ} is rational if A is rational and if all δ_{i} have only rational values on A. (A, δ_{i})_{i∈ℕ} is finite if almost all δ_{i} are zero.

Lemma 4.2. The class of (finite rational) f–sets has the amalgamation property.

Proof. Let (X, X^{i})_{i∈ℕ} and (Y, Y^{i})_{i∈ℕ} be two filtered spaces and A a common subset, on which X and Y induce the same metric and distance functions. Define the metric on Z = X ∪ Y as in the proof of 2.1 and Z^{i} = X^{i} ∪ Y^{i}. It is easy to check, that Z induces on X and Y the old distance functions.

We call a filtered space homogeneous if every isomorphism between finite f–subsets can be extended to an automorphism.

Lemma 4.3. There is a (unique) countable rational filtered space V_{0}, which is homogeneous and embeds every finite rational f–space.

Proof. By Lemma 4.2. Note that there are only countably many finite rational f–spaces.

Let V^{i} be the completion of V_{0}^{i}. Our counterexample V is the filtered space which is the union of the V^{i}.

Let A be an f–space. An f–type over A is a family p = (p_{ξ})_{aξ∈A∪ℕ} of reals which is realized by an element x of an extension of A, in the sense that p_{a} = δ(x, a) and p_{i} = δ_{i}(x). In this case we call p = tpf(x/A) the f–type of x over A. We denote by S_{f}(A) the set of all types over A. An f-type p is rational if all p_{ξ} are rational.

Let p and q be f–types over two non-empty f-sets A and B of finite diameter. We define δ(p, q) as the maximum of the distance between (p_{a})_{a∈A} and (q_{b})_{b∈B} and of the differences |p_{i} − q_{i}|.

One proves again

Lemma 4.4. Let p = tpf(x/A) and q = tpf(y/B). Then

δ(A, B) ≤ δ(p, q) ≤ δ(A, B) + δ(x, y).

Lemma 4.5. Let p ∈ S_{f}(A) and q ∈ S_{f}(B). Then there are realizations x and y of p and q such that

δ(x, y) ≤ 2(δ(p, q) + δ(A, B)).

If A and B are non-empty finite subsets of V_{0}, and p and q are rational, we find x, y in V_{0}.

Proof. Assume first that A = B. Choose realizations x, y of p and q in two different extensions of A. We amalgamate the two spaces A ∪ {x} and A ∪ {y} by setting δ(x, y) = μ where

μ = sup_{ξ∈A∪ℕ} |p_{ξ} − q_{ξ}|.

It is easy to check that this defines an f–space. Also by the proof of Lemma 2.6 μ ≤ 2δ(p, q), and we are finished.

The rest of the proof is as in Lemma 2.6.

Lemma 4.6. Let p an f–type over the non-empty finite set A and ϵ > 0. Then there is a rational f–type q over A such that δ(p, q) < ϵ.

Proof. Left to the reader

The same proof as for Proposition 2.4 yields

Proposition 4.7. V realizes every f–type over every totally bounded subset.

Lemma 4.8. V is not complete

Proof. Let X = {x_{i}}_{i∈ℕ} a rational space and (x_{i}) a Cauchy sequence. Define X^{i} = {0, 1, …, i}. (X, X^{i})_{i∈ℕ} is a rational filtered space, which embeds in the filtered space V_{0}. (x_{i}) cannot converge to any element which lies in some V^{i}. Therefore V is not complete.

Lemma 4.9. V is universal

Proof. V^{0} is universal.

Lemma 4.10. Every automorphism between two totally bounded subspaces of some V^{n} extends to an automorphism over V.

Proof. Let f : A → B be an isomorphism between two totally bounded subsets of V^{n}. Consider the filtered space \overline{V} = (V, V^{n+i})_{i∈ℕ}. It is easy to see that \overline{V} satisfies still the claim of Proposition 4.7. And f is an isomorphism of f– subspaces of \overline{V}. It follows that f extends to an automorphism of the filtered space \overline{V}.

References

[1] Charles Joiner. On Urysohn’s universal separable metric space. Fundamenta Mathematicae, LXXIII:51–57, 1971.

[2] P. Urysohn. Sur un espace métrique universel. Bull. Sci. Math., 51:43–64, 74–90, 1927.

The bibliography is incomplete.