§ Urhyson's lemma
We don't know any continuous functions on compact Haussdorf spaces; Let be a topological space. What functions
are continuous? We only have the constant functions!
If is a metric space, can we get more continuous functions? We can probably do something
like . But the compact Haussdorf spaces are very nice, we should know
something about them!
we often asumme we have an embedding (a homeomorphism) of a compact Haussdorff space into ,
I then get so many continuous functions! I can take the polynomials on and restrict to .
Polynomials are dense in the compact-open topology! (Stone-Weirstrass)
§ Normal space (T4)
A space where points are closed, and two disjoint closed subsets can be separated by open neighbourhoods. It's
stronger than haussdorf, because we can separate subsets, not just points.
§ Urhyson lemma
If is normal and are closed disjoint subsets of , there exists a continuous
function such that and . This gives us
"interesting continuous function" on an arbitrary topological space.
We ask for normal, because compact Hausfdorff spaces are normal. So "good smooth manifolds"
for example, are normal.
§ Lemma: Re-characterization of normality
If is normal and with closed, open, then there
exists a set that is open such that .
(In fact, this is iff!) See that we "reverse the direction"; We started with closed-open, we end with
- Consider and . These are two closed sets. Since is contained in , does not meet (, we have ).
- By normality, we have two opens such that , , and .
- So we have , and . This gives us .
- We have as and is closed, and thus contains all of its limit points.
- This together gives .
§ Proof: Intuition
- Suppse we succeeded. Then the only thing we know is that the space is normal, so it isrich in open sets. We're going to convert the existence of a continuous function into propertiesof open pre-images. We will then show that we have "enough opens" in to build the continuousfunction using the pre-image characterization.
- Suppose we succeeded. Then is open in the subspace topology.
- Define . Each of these are included in one another as we make larger.
- In fact, we have
- We have that for .
- If we now think of the original sets, we needed , . So we must have that for all .
- Similarly, we have that for , as till reache , we cannot get to .
- This gives us .
- This is the only properties we will use to reconstruct !
- Really, I only need a dense subset of . So let's say I pick .
- I can reconstruct by first thinking of . There are sets that reach towards .Consider the closest such
- So take . Because is dense in , this works out and we get
- But if , this means since covers . So we write this as:. Because is dense in , this works out and reconstructs for us us (see that we did not use in the definition of ).
§ Claim 1
Claim: If is dense, and is a collection of open subsets of
indexed by , such that:
Then is continuous and , .
See that we don't even need normality! (Time: 29:30 in video)
- Proof: It's clearly well defined based on . If it's continuous, then it obeys the propertiesbased on the containment assumptions (1) and (2).
- For , We claim . If ,then is in the set of points we take a over. Hence, we have that since is the .
- The contrapositive is that .
- To show is continuous, it suffices to show that preimages of open sets are open for a basis. We know a basis consisting of intervals .We need to show that is open for with .[The cases where or are easy modifications].
- Choose a . Since is dense, we can find such that.
- We claim that . is open as it is the complement of a closed set. Hence, we have shown that is in this open nbhd.
- Since , we must have by our contrapositive.
- We claim that . Proof by contradiction; suppose ,hence .Then, for any such that , we would have (by 2).We claim that this is A CONTRADICTION.
- Since , and we have something in the that is bigger than .Let this thing be such that . But this is a contradiction to .
- TODO: there is more to proof!
With there exists a collection as we need.
We will prove this by induction. Choose an bijection .
Let . To define choose an open set such
Such a set exists by our characterization of normality.
Now suppose have been constructed such that
satisfies the claims (i), (ii). To construct . Recall that the
can be in arbitrary order, since we choose an arbitrary bijection. So let
be the index such that comes right before , and comes
right after . This gives us ,
. Then and .
Thus by our characterization of normality, there exists an open such that:
§ Tietze Extension theorem
If is normal, is closed, given a continuous bounded function
then there exists and continuous and bounded function such that .
§ Urhyson's Metrization theorem
A normal space with a countable basis is metrizable. We know that metrizable is normal. This says
that normal is not so far away from metrizable.
§ Embedding of topological manifolds
If is a compact topological -manifold, then there exists an embedding into some .
That's saying that there are "m" very interesting continuous functions, the coordinate functions! So it
makes sense Urhyson's is involved.