(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 has the Urysohn property if
- is complete and separable
- contains every separable metric space as an isometric copy.
- is ultrahomogeneous in the sense that if are finite, isometric subspaces of then there is an automorphism of that takes to .
You might already know a space that satisfies the first two properties – The Hilbert cube or the continuous functions from to . 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 is a finite subset of and is an arbitrary metric extension of , then there is a point such that with getting mapped to as the identity.
proof. Let be a finite subset of and let be a one point metric extension of . Well, is a separable metric space, so find a copy of it in , call it . Now use ultrahomogeneity of to map to , by an isometry . Then is contained in and is isometric to .[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 , TFAE:
- has the 1pt extension property
- is a Urysohn space
proof. We have already seen the converse, so let us assume that has the 1pt extension property and we will show that is universal with respect to separable metric spaces and that is ultrahomogeneous.
Universal: This is an easy induction. Let be a dense subset of a metric space. Since is complete it will be enough to show that (as an isometric copy of course!)
Any finite collection is contained (isometrically) in (by using the 1pt extension property times) and one more application gives that is contained (isometrically) in .
Since at each stage we are extending, the union of all of these finite copies will be the desired copy of .
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 be separable, complete metric spaces with the 1pt extension property. Finite isomorphisms extend to the whole space. (i.e. If is a finite subset of and is a finite subset of such that then there is an isomorphism that sends to .)
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 . The proof is by induction, but seeing the first step will be enough to give you the entire idea.
Let be countable, dense and let be countable dense. Since are complete it will be enough to define a map on at least whose image contains as there is a unique function that extends to all of .
Let be an isomorphism.
We will define so that:
(Sam’s comment: It might not be possible to map the countable dense subset of isometrically onto the countable dense subset of . For example, the rationals cannot be mapped isometrically onto the rationals with .)
Now is a finite metric space, and by the 1pt extension property of there is a such that is isometric to . Obviously there is in fact an isometry that extends .
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 is a finite metric space, and by the 1pt extension property of there is a such that is isometric to . Obviously there is in fact an isometry that extends .
So by induction there is a family of isometries such that
Thus is an isometry from onto . [QED]
(Sam and Ivan’s comment: It is true a posteriori that is dense in as isometries map dense sets to dense sets. In fact continuous surjections map dense sets to dense sets.)
By letting and in the above theorem we get:
FACT 3: Any two Urysohn spaces are isometric.
And letting we get what some people would call “ultrahomogeneity”:
FACT 4: In the Urysohn space , any partial isometry , with finite, then extends to all of .
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 be a metric space. A function is a Katetov function if
We let be the set of Katetov functions on .
At first these seem like weird constraints to put on a function. The first inequality is just saying that is 1-Lipschitz, the second inequality is a type of triangle inequality (which in particular guarantees that is non-negative). Intuitively, these functions are meant to represent distances to a point.
Example 1: Let be your favourite metric space with . Consider given by . Notice that :
Example 2: If , then is also a katetov function.
Example 3: If we can get a 1pt extension of to by letting if and if . Notice now that being a katetov function ensures that d’ satisfies the triangle inequality.
FACT 5: If is finite, and , then there is a such that for all .
This says that being a katetov function means that it describes distances to an (abstract) point. (i.e. describes a 1pt extension). Since has the 1pt extension property, there is a point that witnesses this.
Enough with all of that abstract stuff, let’s show you some cool stuff.
You are probably wondering about the topological properties of . Well you already know that is a complete, separable metric space (which is really nice), but what about connectedness?
FACT 6: has a strong form of path-connectedness, that is, has geodesics.
Definition: A metric space has geodesics if for any two points there is a closed interval and an isometry from into such that and . So there is “a line between and “.
proof. The picture says everything, but for those of you so inclined, here are the details.
Take , both in . Let . Hey, is a separable metric space, so find an isometric copy in . Now by ultrahomogeneity of , there is an isometry of that maps the end-points of to . Thus and fulfill the requirements of the definition of geodesics. [QED]
(Dominic’s question: Does have unique geodesics?)
Fact 7: 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 between two points , and let be the midpoint of . For simplicity, assume that has length 8.
Consider defined by and , where .
If we can show that is a katetov function, then it will produce a point that is distance to both and (so by the triangle inquality it lies on a geodesic from to ), and each is distinct as if .
Claim: is a katetov function.
(I’ll supress the subscript for readability)
Cool Isometric copies of
Now we look at an interesting copy of . For the record, , which is clear from the construction of (which I have omitted).
FACT 7: There are isometric copies of in with empty interior.
proof. For a finite set we define .
For example, in , if then .
Claim: For any nonempty finite set , .
It is pretty easy to swallow that is a closed subset of , a complete, separable metric space. Thus is itself a complete separable metric space and by the ultrahomogeneity theorem it is enough to show that has the 1pt extension property.
Let be finite and let be a katetov function.
Let be defined by for and .
[This is called the katetov extension of to ].
We will show later that this is a katetov function, but if we accept that is a katetov function then there is a such that for all .
Since for all , it only remains to see that . This follows since for all , so for all ; the infimum does not depend on or .
This next fact stands in direct opposition of Fact 7:
FACT 8: There are (proper) isometric copies of in with non-empty interior.
This will be shown using the following proposition:
Proposition: For a compact subset of , , we have . Thus is isometric to .
proof. Note that is an open subset of , so is a complete separable metric space. Thus we only need to show that it has the 1pt extension property.
Let be finite, and let be a katetov function. Without loss of generality we may assume that .
Let be an -net. (Every point of is within of some point of )
Now consider be defined by . Let us accept for now that this is a katetov function.
Then there is a such that for all . For we just get . For we get
And so it is clear that for all , not just those in .
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 be defined by really are katetov functions (where is katetov).
FACT 9: Amalgamation is possible.
proof. Let and be metric spaces both having an isometric copy of , a compact metric space. Let , but identify the copies of . Now we need to define a metric on that agrees with and .
So let if . Similarly, let if .
Now we only need to deal with the case where the points are in different spaces.
Let . So it is the shortest way from to while going through . This makes sense because is compact, we need only show that is actually a metric. The only thing that isn’t clear is the triangle inequality, so here goes.
FACT 10: If , with compact, and is a katetov function, then is a katetov function that (obviously) extends , where
proof. If , the result is clear (although we didn’t really have to point out this case). Let .
[Note that above, was chosen to be .]
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).