# Facts about the Urysohn Space – Some useful, some cool

(This is almost verbatim the talk I gave recently (Feb 23, 2012) at the Toronto Student Set Theory and Topology Seminar. I will be giving this talk again on April 5, 2012)

I have been working on a problem involving the Urysohn space recently, and I figured that I should fill people in with the basic facts and techniques involved in this space. I will give some useful facts, a key technique and 3 cool facts. First, the definition!

Definition: A metric space $U$ has the Urysohn property if

• $U$ is complete and separable
• $U$ contains every separable metric space as an isometric copy.
• $U$ is ultrahomogeneous in the sense that if $A,B$ are finite, isometric subspaces of $U$ then there is an automorphism of $U$ that takes $A$ to $B$.

You might already know a space that satisfies the first two properties – The Hilbert cube $[0,1]^\omega$ or $C[0,1]$ the continuous functions from $[0,1]$ to $[0,1]$. However, these spaces are not ultrahomogeneous. Should a Urysohn space even exist? It does, but the construction isn’t particularly illuminating so I will skip it.

Usually, ultrahomogeneity and universality are used in the following way:

Fact 1: A Urysohn space has the “1 point extension property”. That is, if $A$ is a finite subset of $U$ and $A\cup\{a\}$ is an arbitrary metric extension of $A$, then there is a point $z \in U$ such that $A\cup \{a\} \cong A \cup\{z\}$ with $A$ getting mapped to $A$ as the identity.

proof. Let $A$ be a finite subset of $U$ and let $A \cup \{c\}$ be a one point metric extension of $A$. Well, $A \cup \{c\}$ is a separable metric space, so find a copy of it in $U$, call it $A' \cup \{c'\}$. Now use ultrahomogeneity of $U$ to map $A'$ to $A$, by an isometry $\phi$. Then $A \cup \{\phi(c')\}$ is contained in $U$ and is isometric to $A \cup \{c\}$.[QED]

It turns out that the 1 point extension property is a very strong property. This is kind of expected because a space with the 1 point extension property contains a copy of every finite metric space. That is pretty big already!

Fact 2: For a complete, separable metric space $U$, TFAE:

• $U$ has the 1pt extension property
• $U$ is a Urysohn space

proof. We have already seen the converse, so let us assume that $U$ has the 1pt extension property and we will show that $U$ is universal with respect to separable metric spaces and that $U$ is ultrahomogeneous.

Universal: This is an easy induction. Let $Q = \{q_n : n<\omega\}$ be a dense subset of a metric space. Since $U$ is complete it will be enough to show that $Q \subseteq U$ (as an isometric copy of course!)

Any finite collection $\{q_i : i is contained (isometrically) in $U$ (by using the 1pt extension property $N$ times) and one more application gives that $\{q_i : i is contained (isometrically) in $U$.

Since at each stage we are extending, the union of all of these finite copies will be the desired copy of $Q$.

Ultrahomogeneous: For no cost we can actually show a more general fact that illustrates a helpful technique. (This is one of those cases where showing the more general fact more clearly indicates what properties are important.)

Ultrahomogeneity Theorem: Let $X,Y$ be separable, complete metric spaces with the 1pt extension property. Finite isomorphisms extend to the whole space. (i.e. If $A$ is a finite subset of $X$ and $B$ is a finite subset of $Y$ such that $A \cong B$ then there is an isomorphism $f: X \rightarrow Y$ that sends $A$ to $B$.)

This will be a back-and-forth argument. The “back” part will ensure that we have a surjection, and the “forward” will give us a function defined on all of $X$. The proof is by induction, but seeing the first step will be enough to give you the entire idea.

Let $\{x_n : n<\omega\} \subseteq X$ be countable, dense and let $\{y_n : n<\omega\}$ be countable dense. Since $X,Y$ are complete it will be enough to define a map $f$ on at least $\{x_n : n<\omega\}$ whose image contains $\{y_n : n<\omega\}$ as there is a unique function that extends $f$ to all of $X,Y$.

Let $f_{-1} : A\rightarrow B$ be an isomorphism.

We will define $f_0$ so that:

1. $f_0$ extends $f_{-1}$
2. $x_0 \in \textrm{dom}(f_0)$
3. $y_0 \in \textrm{ran}(f_0)$

(Sam’s comment: It might not be possible to map the countable dense subset of $X$ isometrically onto the countable dense subset of $Y$. For example, the rationals cannot be mapped isometrically onto the rationals with $\pi$.)

Now $A \cup \{x_0\} \subseteq X$ is a finite metric space, and by the 1pt extension property of $Y$ there is a $w_0 \in Y$ such that $B \cup \{w_0\}$ is isometric to $A \cup \{x_0\}$. Obviously there is in fact an isometry $f_{-0.5}: A \cup \{x_0\} \rightarrow B \cup \{w_0\}$ that extends $f_{-1}$.

So we have extended the domain by one point. Now to extend the range. (The next paragraph is almost a copy of the one above.)

Notice that $B \cup \{w_0\} \cup \{y_0\}$ is a finite metric space, and by the 1pt extension property of $X$ there is a $z_0 \in X$ such that $A \cup \{x_0\}\cup \{z_0\}$ is isometric to $B \cup \{w_0\}\cup \{y_0\}$. Obviously there is in fact an isometry $f_{0}: A \cup \{x_0\}\cup \{z_0\} \rightarrow B \cup \{w_0\} \cup \{y_0\}$ that extends $f_{-0.5}$.

So by induction there is a family of isometries $\{f_n : n<\omega\}$ such that

1. $f_n$ extends $f_{n-1}$
2. $\{x_i : 0\leq i \leq n\} \in \textrm{dom}(f_n)$
3. $\{y_i : 0\leq i \leq n\} \in \textrm{ran}(f_n)$

Thus $f := \bigcup_{n<\omega} f_n$ is an isometry from $A \cup \{x_n : n<\omega\}\cup \{z_n : n<\omega\}$ onto $B\cup \{w_n : n<\omega\}\cup \{y_n : n<\omega\}$. [QED]

(Sam and Ivan’s comment: It is true a posteriori that $\{w_n : n<\omega\}$ is dense in $Y$ as isometries map dense sets to dense sets. In fact continuous surjections map dense sets to dense sets.)

By letting $X=Y=U$ and $A = B = \emptyset$ in the above theorem we get:

FACT 3: Any two Urysohn spaces are isometric.

And letting $X=Y=U$ we get what some people would call “ultrahomogeneity”:

FACT 4: In the Urysohn space $U$, any partial isometry $\phi: A \rightarrow B$, with $A,B \subseteq U$ finite, then $\phi$ extends to all of $U$.

### More on the 1pt extension property

In practise the 1 point extension property isn’t really the right way to describe 1pt extensions. (Wait, what?) It turns out that instead of describing abstract one point metric extensions (which requires using an element outside of the metric space in question) we instead define the distances from a metric space to an outside point. (Stay with me, this will make sense soon.)

Definition: Let $(X,d)$ be a metric space. A function $f: X \rightarrow \mathbb{R}$ is a Katetov function if
$\displaystyle \forall x,y \in X, |f(x)-f(y)| \leq d(x,y) \leq f(x) + f(y)$
We let $E(X)$ be the set of Katetov functions on $X$.

At first these seem like weird constraints to put on a function. The first inequality is just saying that $f$ is 1-Lipschitz, the second inequality is a type of triangle inequality (which in particular guarantees that $f$ is non-negative). Intuitively, these functions are meant to represent distances to a point.

Example 1: Let $(X,d)$ be your favourite metric space with $\heartsuit \in X$. Consider $f: X \rightarrow \mathbb{R}$ given by $f(x)=d(x, \heartsuit)$. Notice that $\forall x,y \in X$:

 $|f(x)-f(y)|$ $=|d(x,\heartsuit) - d(y,\heartsuit)|$ $\leq |d(x,y)+d(y,\heartsuit) - d(y,\heartsuit)|$ $\leq d(x,y)$ $\leq d(x, \heartsuit)+d(y, \heartsuit)$ $= f(x)+f(y)$

Example 2: If $f \in E(X), c \geq 0$, then $g(x) = f(x)+c$ is also a katetov function.

Example 3: If $f \in E(X)$ we can get a 1pt extension of $(X, d)$ to $(X \cup \{z\}, d^\prime)$ by letting $d^\prime (x,y)=d(x,y)$ if $x,y \in X$ and $d^\prime (x,z)=f(x)$ if $x \in X$. Notice now that being a katetov function ensures that d’ satisfies the triangle inequality.

FACT 5: If $X \subseteq U$ is finite, and $f \in E(X)$, then there is a $z \in U$ such that $f(x)=d(x,z)$ for all $x\in X$.

This says that $f$ being a katetov function means that it describes distances to an (abstract) point. (i.e. $f$ describes a 1pt extension). Since $U$ has the 1pt extension property, there is a point $z\in U$ that witnesses this.

### Cool Stuff

Enough with all of that abstract stuff, let’s show you some cool stuff.

You are probably wondering about the topological properties of $U$. Well you already know that $U$ is a complete, separable metric space (which is really nice), but what about connectedness?

FACT 6: $U$ has a strong form of path-connectedness, that is, $U$ has geodesics.

Definition: A metric space $X$ has geodesics if for any two points $a,b \in X$ there is a closed interval $I = [x,y]$ and an isometry $f$ from $I$ into $X$ such that $f(x)=a$ and $f(y)=b$. So there is “a line between $a$ and $b$“.

proof. The picture says everything, but for those of you so inclined, here are the details.

Take $a \neq b$, both in $U$. Let $I = [0, d(a,b)]$. Hey, $I$ is a separable metric space, so find an isometric copy $J$ in $U$. Now by ultrahomogeneity of $U$, there is an isometry $f$ of $U$ that maps the end-points of $J$ to $\{a,b\}$. Thus $f$ and $J$ fulfill the requirements of the definition of geodesics. [QED]

(Dominic’s question: Does $U$ have unique geodesics?)

Fact 7: $U$ has uncountably many geodesics between any two points.

proof. Now we get to use a standard proof technique in Urysohn spaces- We construct katetov functions that code different geodesics.

Fix a single geodesic $I$ between two points $a \neq b$, and let $m$ be the midpoint of $I$. For simplicity, assume that $I$ has length 8.

Consider $f_\delta: \{a,b,m\} \rightarrow \mathbb{R}$ defined by $f_\delta (a)=f_\delta (b)= 4$ and $f_\delta (m)=\delta$, where $0<\delta < 4$.

If we can show that $f_\delta$ is a katetov function, then it will produce a point $z_\delta \in U$ that is distance $4$ to both $x$ and $y$ (so by the triangle inquality it lies on a geodesic from $x$ to $y$), and each $z_\delta$ is distinct as $d(z_\delta, z_\gamma) \neq 0$ if $\delta \neq \gamma$.

Claim: $f_\delta$ is a katetov function.

(I’ll supress the subscript for readability)
$\displaystyle |f(a)-f(b)|=|4-4|\leq d(a,b) =8 \leq 4+4 = f(a)+f(b)$
and
$\displaystyle |f(a)-f(m)|= |4-\delta| \leq 4 \leq d(a,m) \leq 4 + \delta = f(a)+f(m)$
[QED]

### Cool Isometric copies of $U$

Now we look at an interesting copy of $U$. For the record, $0 \in U$, which is clear from the construction of $U$ (which I have omitted).

FACT 7: There are isometric copies of $U$ in $U$ with empty interior.

proof. For a finite set $F \subseteq U$ we define $m(F) = \{x \in U : d(x,a)=d(x,b), \forall a,b \in F\}$.

For example, in $\mathbb{R}^2$, if $F= \{(0,0),(2,0)\}$ then $m(F)= \{(1,y): y \in \mathbb{R} \}$.

Claim: For any nonempty finite set $F \subseteq U$, $m(F) \cong U$.

It is pretty easy to swallow that $m(F)$ is a closed subset of $U$, a complete, separable metric space. Thus $m(F)$ is itself a complete separable metric space and by the ultrahomogeneity theorem it is enough to show that $m(F)$ has the 1pt extension property.

Let $A \subseteq m(F)$ be finite and let $f: A \rightarrow \mathbb{R}$ be a katetov function.

Let $g: A\cup F \rightarrow \mathbb{R}$ be defined by $g(a)=f(a)$ for $a\in A$ and $g(x)=\inf_{a\in A}\{f(a) + d(a,x)\}$.
[This is called the katetov extension of $f$ to $A \cup F$].

We will show later that this is a katetov function, but if we accept that $g$ is a katetov function then there is a $z \in U$ such that $g(x)=d(z,x)$ for all $x \in A \cup F$.

Since $g(a)=f(a)$ for all $a \in A$, it only remains to see that $z \in m(F)$. This follows since $d(a,x) = d(b,x)$ for all $a,b \in A$, so $d(x,z)=g(x)=g(y)=d(y,z)$ for all $x,y \in F$; the infimum does not depend on $x$ or $y$.
[QED]

This next fact stands in direct opposition of Fact 7:

FACT 8: There are (proper) isometric copies of $U$ in $U$ with non-empty interior.

This will be shown using the following proposition:

Proposition: For $X$ a compact subset of $U$, $M \in \mathbb{R}$, we have $U\setminus \{z \in U : d(z,X )< M\} \cong U$. Thus $U$ is isometric to $U\setminus B(0,1)$.

proof. Note that $\{z \in U : d(z,X )< M\}$ is an open subset of $U$, so $U\setminus \{z \in U : d(z,X )< M\}$ is a complete separable metric space. Thus we only need to show that it has the 1pt extension property.

Let $A \subseteq U\setminus \{z \in U : d(z,X )< M\}$ be finite, and let $f: A \rightarrow \mathbb{R}$ be a katetov function. Without loss of generality we may assume that $\epsilon := \inf_{a \in A} f(a) >0$.

Let $Y \subseteq X$ be an $\epsilon$-net. (Every point of $X$ is within $\epsilon$ of some point of $Y$)

Now consider $g: Y \cup A \rightarrow \mathbb{R}$ be defined by $g(x) = \inf_{a \in A} \{f(a)+d(x,a)\}$. Let us accept for now that this is a katetov function.

Then there is a $z \in U$ such that $d(z,x) = g(x)$ for all $x\in Y \cup A$. For $a\in A$ we just get $d(z,a) = f(a)$. For $y \in Y$ we get
$\displaystyle d(z,y) = \inf_{a \in A} \{f(a)+d(x,a)\} > \epsilon + M$

And so it is clear that $d(z,x)> M$ for all $x \in X$, not just those in $Y$.
[QED]

### Katetov Extensions

The only thing we have left to justify is why the functions we defined in Fact 1 and 2 are katetov functions. This comes from a general notion of the “amalgamation of two metric spaces along a compact subspace”, which is mumbo jumbo for gluing two space together.

The idea is that if you have two unrelated spaces that both have an (isometric) copy of the unit circle (or your favourite compact metric space) in them, then you can define a bigger space that contains your original spaces, glued together at the circle.

Showing that there is such a space will prove that the functions $g: Y \cup A \rightarrow \mathbb{R}$ be defined by $g(x) = \inf_{a \in A} \{f(a)+d(x,a)\}$ really are katetov functions (where $f: A \rightarrow \mathbb{R}$ is katetov).

FACT 9: Amalgamation is possible.

proof. Let $(A, d_A)$ and $(B, d_B)$ be metric spaces both having an isometric copy of $X$, a compact metric space. Let $Z = A \cup B$, but identify the copies of $X$. Now we need to define a metric $d$ on $Z$ that agrees with $A$ and $B$.

So let $d(a_1, a_2)= d_A(a_1, a_2)$ if $a_1, a_2 \in A$. Similarly, let $d(b_1, b_2)= d_B (b_1, b_2)$ if $b_1, b_2 \in B$.

Now we only need to deal with the case where the points are in different spaces.

Let $d(a,b) = \inf_{x\in X} (d_A (a,x)+d_B (x,b))$. So it is the shortest way from $a$ to $b$ while going through $X$. This makes sense because $X$ is compact, we need only show that $d$ is actually a metric. The only thing that isn’t clear is the triangle inequality, so here goes.

FACT 10: If $A \subseteq X$, with $A,X$ compact, and $f: A \rightarrow \mathbb{R}$ is a katetov function, then $\hat{f}: X \rightarrow \mathbb{R}$ is a katetov function that (obviously) extends $f$, where
$\displaystyle \hat{f}(x) := \inf_{a \in A} \{f(a)+d(a,x)\}$

proof. If $a,b \in A$, the result is clear (although we didn’t really have to point out this case). Let $x,y \in X$.

 $|\hat{f}(x)-\hat{f}(y)|$ $=| \inf_{a \in A} \{f(a)+d(a,x)\} - \inf_{b \in A} \{f(b)+d(b,y)\}|$ $\leq | \inf_{a \in A} \{f(a)+d(a,x)\} - (f(b_0)+d(b_0,y))|$ $\leq | f(b_0)+d(b_0,x) - f(b_0)-d(b_0,y)|$ $\leq |d(b_0,x) - d(b_0,y)|$ $\leq d(x,y)$ $\leq d(x,a) + d(a,y) + f(a) + f(a), (\forall a \in A)$

Thus $d(x,y) \leq \inf_{a \in A} \{f(a)+d(a,x)\} + \inf_{a \in A} \{f(a)+d(a,y)\} = \hat{f}(x)+\hat{f}(y)$.

[Note that above, $b_0$ was chosen to be $\sup_{b\in A} \{f(b)+d(b,y)\}$.]

[QED]

### Conclusion

There is the end of the easily digestible facts about the Urysohn space (that I know). It is interesting to me that I was able to get all of this information across to our seminar in a mere 90 minutes! As presented here the material seems so dense, but it really isn’t. I hope the pictures help you grok this stuff, because I really think that the ideas here are simple and the techniques are useful.

Most of this material was taken from the wonderful article “On the Geometry of Urysohn’s Universal Metric Space” by Julien Melleray (2007), and chapter 5 of “Dynamics of Infinite Dimensional Groups” by Vladimir Pestov (2006). Melleray’s paper contains very useful exercises, some of which I have included in this talk (as facts).

## 2 thoughts on “Facts about the Urysohn Space – Some useful, some cool”

1. m. says:

nice, but there is a typo in the def of Katetov function. (Fixed. -Mike)

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2. Aristotelis says:

Thank you very much for sharing this useful notes. Just for an unimportant betterment in the definition of ultrahomogeneous : automorphism of U instead of isomorphism of U. Again thanks

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