I’ve been reviewing some basic general topology as of late. I will post some of this material here. Apologies to readers who prefer more advanced topics; my current focus is on foundational material.
Often, we’d like to prove that a given subset of a topological space has some given property, e.g. that it is open or closed. In many cases, however, the big space may not be easily understandable, but local pieces of it may be. For instance, might be a manifold, and we might not know what the global structure of is, but we do know that is locally homeomorphic (or diffeomorphic) to a ball in . So we need a way to go from local results to global results.
Proposition 1 Let be an open cover of the topological space . Suppose and is open in for each . Then is open in .
So openness is a local property.
This is the easy result. Indeed, since is an open set, each is open in . But
so that is a union of open sets, hence open.
Similarly, we can deduce the corresponding result for closed sets:
Corollary 2 Suppose is an open cover of . Let . Suppose is closed in for each . Then is closed in .
Indeed, this follows from the previous result, with replaced with .
However, the analogous result is no longer true if we look at closed covers. Consider for instance the closed cover of by vertical lines. The set , defined as the intersection of the graph with the upper right quadrant, is not closed, though its intersection with each vertical line is closed (in fact, is a point). So we need something more. The problem, as we will see, is that there are too many lines.
Definition 3 Let be a topological space. A collection of sets (not necessarily open or closed) is said to be locally finite if to each , there is a neighborhood of that intersects only finitely many of the .
If a collection of sets is locally finite, then their closures form a locally finite family. This is because to say that an open set does not intersect a set is the same as saying it doesn’t intersect .
Local finiteness is the condition that was missing in the previous counterexample: the vertical lines are far from locally finite; in fact, there are uncountably many through each neighborhood.
Before getting to the main results, we start with:
Proposition 4 Let be a locally finite collection of closed subsets of a topological space . Then is closed.
Normally, we cannot claim that an infinite union of closed sets is closed, only for finite unions. However, we have proved earlier that being “closed” is a local property. In other words, if for each , we can find a neighborhood of with closed in (i.e. relatively), then is closed.
So for each , choose containing such that intersects only finitely many . Then is a finite union of the by the assumption on . Thus it is relatively closed in .
Now that we have seen the usefulness of the local finiteness condition, we consider spaces where we can reduce to locally finite open coverings. This is extremely useful because it enables us to handle things like partitions of unity.
First, a bit of terminology. Let be a cover of a space . Then a cover is called a refinement if each sits inside some .
Definition 5 A Hausdorff space is paracompact if every open covering has a locally finite refinement.
Dugundji’s Topology goes into a lot of technical results about paracompactness. For now, I don’t think I need them, so I will skip over those.
So, trivially any compact space is paracompact, since a subcover is a refinement. However, many noncompact spaces are paracompact. Motivated by the case of manifolds, which are locally compact, we start with:
Theorem 6 (Dieudonné) Suppose is locally compact. Then is paracompact if it is -compact, i.e. the countable union of compact sets.
This is useful. In particular, it implies that a topological manifold with a countable base is paracompact.
We shall now prove this result. The first step is to show that -compactness implies the existence of a special kind of cover, one that increases rapidly like a sequence of shells.
Lemma 7 Suppose is locally compact and -compact. Then there is a covering collection of open sets with compact closure and .
To start with, we can write for each compact. We shall inductively define the increasing sequence . First, . That was easy. Next, assuming is defined and has compact closure, we consider the compact set . We can find a neighborhood of this compact set which has compact closure in view of local compactness. Indeed, pick a small neighborhood of each point in with compact closure, take a finite sub-cover of this open cover of , and take to be the union of the sets in this finite sub-cover. Then this will have compact closure and will contain and .
The are ascending by construction, and since they contain the covering , they themselves form an open cover of .
We shall now prove the first half of Dieudonné’s theorem. Suppose is locally compact and -compact; we shall prove that it is paracompact. Suppose is an open covering of ; we are to find a locally finite refinement.
We construct a nested covering as in the lemma. Then the sets form a covering of by compact sets. This corresponds to the area between the successive layers. This is good, but we need more. We look at the open covering of , which is in fact locally finite because of the nesting of the . With these two, we will make the construction.
So, for each , we know that the cover . We can construct a finite refinement of this cover which covers the compact set and which is contained in the open set . I claim that the union satisfies the conditions.
Well, first of all, there are only finitely many of the sets in the cover that intersect each set because for large the sets in lie in . Since the cover , we get local finiteness. Moreover, we know that the union actually covers . Furthermore, each is a refinement of , hence their union is too. This completes the proof of one direction.
The converse is not quite true, but it is true if is connected. I refer the interested reader to Bourbaki for the proof.
We shall now prove that a paracompact space is normal. Recall that this means that if are disjoint closed sets, then they are separated by disjoint open sets . One of the dandy consequences of this is that if is an inclusion of a closed set in an open set, then there is a continuous function which is equal to 1 on and to 0 outside ; this is Urysohn’s lemma.
Before establishing the full force of rormality, we start with the weaker result of regularity.
Lemma 8 Let be a paracompact space, a closed subspace, and . Then there are open neighborhoods of , respectively, which do not intersect.
This is a straightforward exercise in the properties of local finiteness and closedness. We shall construct first and show that its closure does not contain .
Indeed, for each , we can choose a neighborhood of and a neighborhood of which do not intersect by Hausdorff-ness. In particular, we have . Now , as a closed subspace, is paracompact (easy exercise). So we can choose a locally finite refinement of the . Then the do not contain because the ‘s are separated from . Let . We have by local finiteness, so .
Now is an open neighborhood of , whose closure does not contain . We can thus use and as the two open sets in the statement of the lemma.
We are now ready for:
Theorem 9 A paracompact space is normal.
Suppose are two disjoint closed subsets and is paracompact. For each , we can find containing and containing such that . In particular, we have
The cover , so we can find a locally finite refinement of this cover. Call this locally finite refinement . The union of the closures is the same thing as the closure of the union by local finiteness (see the proof of the lemma), so this union satisfies
and this, as before, enables us to construct via . This proves normality.
3. Partitions of unity
One of the basic reasons we care about paracompactness is that it enables us to get partitions of unity. So, suppose is a topological space and a cover.
Definition 10 A partition of unity subordinate to the cover is a collection of continuous functions such that each is supported in and . Moreover, we assume that the supports of the are locally finite (which ensures that the sum is always well-defined).
Partitions of unity are fairly ubiquitous. The point is that, by decomposing any as , we can reduce problems about the whole space into problems about the constituent parts , which are generally much simpler. For instance, the Mayer-Vietoris sequence for de Rham cohomology is obtained using a partition of unity. Partitions of unity are also used to prove the general Stokes theorem by reducing to the case of a half-space.
This is why the following is so important:
Theorem 11 Suppose is an open cover of the paracompact space . Then there is a partition of unity subordinate to .
We shall start with a lemma.
Lemma 12 Let be a paracompact space and an open covering. Then there is a locally finite refinement with, for each ,
The point of this lemma is twofold. One, we can make a substantial shrinking of the cover (because we take closures on the left-hand-side). Two, we can make the covering locally finite while keeping the same indexing, which is often convenient.
We now prove the lemma. First, for each , we choose an open neighborhood whose closure is contained in some ; we can do this by normality of a paracompact space. The form an open covering of , so we have a locally finite refinement, call it . For each , define
In other words, we are collecting together the ‘s to make the ‘s. Clearly all the ‘s are collected together in this manner, so that the form an open cover of . Local finiteness of the implies that of the , which are just bundles of the former; moreover, in the same way we deduce
We have now proved the lemma.
We now proceed to the proof of the theorem. By replacing the by a locally finite refinement as in the lemma with the same indexing, we can assume that the are themselves locally finite.
Now take a refinement as in the lemma. The key property we now want is that . From this, we can use Urysohn’s lemma to choose a continuous nonnegative which is equal to one on but vanishes outside .
Then the sum is well-defined, and we can take
these are supported in the , and clearly add to one everywhere. The proof is complete.