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 ${A \subset X}$ of a topological space ${X}$ has some given property, e.g. that it is open or closed. In many cases, however, the big space ${X}$ may not be easily understandable, but local pieces of it may be. For instance, ${X}$ might be a manifold, and we might not know what the global structure of ${X}$ is, but we do know that ${X}$ is locally homeomorphic (or diffeomorphic) to a ball in ${\mathbb{R}^n}$. So we need a way to go from local results to global results.

Proposition 1 Let ${\left\{U_\alpha\right\}}$ be an open cover of the topological space ${X}$. Suppose ${W \subset X}$ and ${W \cap U_{\alpha}}$ is open in ${U_{\alpha}}$ for each ${\alpha}$. Then ${W}$ is open in ${X}$.

So openness is a local property.

This is the easy result. Indeed, since ${U_{\alpha}}$ is an open set, each ${W \cap U_{\alpha} }$ is open in ${X}$. But

$\displaystyle W = \bigcup W \cap U_{\alpha}$

so that ${W}$ is a union of open sets, hence open.

Similarly, we can deduce the corresponding result for closed sets:

Corollary 2 Suppose ${U_{\alpha}}$ is an open cover of ${X}$. Let ${W \subset X}$. Suppose ${W \cap U_{\alpha}}$ is closed in ${ U_{\alpha}}$ for each ${U_{\alpha}}$. Then ${W}$ is closed in ${X}$.

Indeed, this follows from the previous result, with ${W}$ replaced with ${X - W}$.

However, the analogous result is no longer true if we look at closed covers. Consider for instance the closed cover of ${\mathbb{R}^2}$ by vertical lines. The set ${W}$, defined as the intersection of the graph ${y=x}$ 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 ${X}$ be a topological space. A collection of sets ${U_{\alpha} \subset X}$ (not necessarily open or closed) is said to be locally finite if to each ${x \in X}$, there is a neighborhood ${U}$ of ${x}$ that intersects only finitely many of the ${U_{\alpha}}$.

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 ${U_{\alpha}}$ is the same as saying it doesn’t intersect ${\overline{U_{\alpha}}}$.

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 ${F_{\alpha}}$ be a locally finite collection of closed subsets of a topological space ${X}$. Then ${F=\bigcup F_{\alpha}}$ 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 ${x \in X}$, we can find a neighborhood ${U}$ of ${x}$ with ${U \cap F}$ closed in ${U}$ (i.e. relatively), then ${F}$ is closed.

So for each ${x \in X}$, choose ${U}$ containing ${x}$ such that ${U}$ intersects only finitely many ${F_{\alpha}}$. Then ${U \cap F}$ is a finite union of the ${U \cap F_{\alpha}}$ by the assumption on ${U}$. Thus it is relatively closed in ${U}$.

1. Paracompactness

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 ${\left\{U_{\alpha}\right\} }$ be a cover of a space ${X}$. Then a cover ${\left\{V_{\beta}\right\}}$ is called a refinement if each ${V_{\beta}}$ sits inside some ${U_{\alpha}}$.

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 ${X}$ is locally compact. Then ${X}$ is paracompact if it is ${\sigma}$-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 ${\sigma}$-compactness implies the existence of a special kind of cover, one that increases rapidly like a sequence of shells.

Lemma 7 Suppose ${X}$ is locally compact and ${\sigma}$-compact. Then there is a covering collection of open sets ${U_i \subset X}$ with compact closure and ${\overline{U_i} \subset U_{i+1}}$.

To start with, we can write ${X = \bigcup_{i \geq 1} K_i}$ for each ${K_i}$ compact. We shall inductively define the increasing sequence ${U_i}$. First, ${U_0 = \emptyset}$. That was easy. Next, assuming ${U_{i-1}}$ is defined and has compact closure, we consider the compact set ${\overline{U_{i-1}} \cup K_i}$. We can find a neighborhood ${U_i}$ of this compact set which has compact closure in view of local compactness. Indeed, pick a small neighborhood of each point in ${\overline{U_{i-1}} \cup K_i}$ with compact closure, take a finite sub-cover of this open cover of ${\overline{U_{i-1}} \cup K_i}$, and take ${U_i}$ to be the union of the sets in this finite sub-cover. Then this will have compact closure and will contain ${K_i}$ and ${\overline{U_{i-1}}}$.

The ${U_i}$ are ascending by construction, and since they contain the covering ${K_i}$, they themselves form an open cover of ${X}$.

We shall now prove the first half of Dieudonné’s theorem. Suppose ${X}$ is locally compact and ${\sigma}$-compact; we shall prove that it is paracompact. Suppose ${V_{\alpha}}$ is an open covering of ${X}$; we are to find a locally finite refinement.

We construct a nested covering ${\left\{U_i\right\}}$ as in the lemma. Then the sets ${\overline{U_i} - U_{i-1}}$ form a covering of ${X}$ 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 ${U_{i+1} - \overline{U_{i-2}}}$ of ${X}$, which is in fact locally finite because of the nesting of the ${U_i}$. With these two, we will make the construction.

So, for each ${i}$, we know that the ${V_{\alpha}}$ cover ${\overline{U_i} - U_{i-1}}$. We can construct a finite refinement ${\mathfrak{A }_i}$ of this cover ${\left\{V_{\alpha}\right\}}$ which covers the compact set ${\overline{U_i} - U_{i-1}}$ and which is contained in the open set ${U_{i+1} - \overline{U_{i-2}}}$. I claim that the union ${\mathfrak{A} = \bigcup_i \mathfrak{A }_i}$ satisfies the conditions.

Well, first of all, there are only finitely many of the sets in the cover ${\mathfrak{A }}$ that intersect each set ${U_{i+1} - \overline{U_{i-2}}}$ because for large ${j}$ the sets in ${\mathfrak{A }_j}$ lie in ${U_{i+1}^c}$. Since the ${U_{i+1} - \overline{U_{i-2}}}$ cover ${X}$, we get local finiteness. Moreover, we know that the union actually covers ${X}$. Furthermore, each ${\mathfrak{A}_i}$ is a refinement of ${\left\{V_{\alpha}\right\}}$, hence their union is too. This completes the proof of one direction.

The converse is not quite true, but it is true if ${X}$ is connected. I refer the interested reader to Bourbaki for the proof.

2. Normality

We shall now prove that a paracompact space is normal. Recall that this means that if ${A, B \subset X}$ are disjoint closed sets, then they are separated by disjoint open sets ${U \supset A, V \supset B}$. One of the dandy consequences of this is that if ${K \subset V}$ is an inclusion of a closed set in an open set, then there is a continuous function ${f: X \rightarrow [0,1]}$ which is equal to 1 on ${K}$ and to 0 outside ${V}$; this is Urysohn’s lemma.

Before establishing the full force of rormality, we start with the weaker result of regularity.

Lemma 8 Let ${X}$ be a paracompact space, ${ A \subset X}$ a closed subspace, and ${x \in X - A}$. Then there are open neighborhoods ${U,V}$ of ${x, A}$, respectively, which do not intersect.

This is a straightforward exercise in the properties of local finiteness and closedness. We shall construct ${V}$ first and show that its closure does not contain ${x}$.

Indeed, for each ${a \in A}$, we can choose a neighborhood ${V_{\alpha}}$ of ${a}$ and a neighborhood ${U_{a}}$ of ${x}$ which do not intersect by Hausdorff-ness. In particular, we have ${x \notin \overline{V_{a}}}$. Now ${A}$, as a closed subspace, is paracompact (easy exercise). So we can choose a locally finite refinement ${\left\{W_{\beta}\right\}}$ of the ${\left\{V_a\right\}}$. Then the ${\overline{W_{\beta}}}$ do not contain ${x}$ because the ${V_a}$‘s are separated from ${x}$. Let ${W = \bigcup W_{\beta}}$. We have ${\overline{W} = \bigcup \overline{W_{\beta}}}$ by local finiteness, so ${x \notin W}$.

Now ${W}$ is an open neighborhood of ${A}$, whose closure does not contain ${x}$. We can thus use ${V = W}$ and ${U = X - \overline{W}}$ as the two open sets in the statement of the lemma.

We are now ready for:

Theorem 9 A paracompact space is normal.

Suppose ${A, B \subset X}$ are two disjoint closed subsets and ${X}$ is paracompact. For each ${a \in A}$, we can find ${U_a}$ containing ${a}$ and ${V_a}$ containing ${B}$ such that ${U_a \cap V_a = \emptyset}$. In particular, we have

$\displaystyle \overline{U_a} \cap B = \emptyset.$

The ${\left\{U_a\right\}}$ cover ${A}$, so we can find a locally finite refinement of this cover. Call this locally finite refinement ${\left\{U'_\alpha\right\}}$. 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 ${U}$ satisfies

$\displaystyle \overline{U} \cap B = \emptyset$

and this, as before, enables us to construct ${V}$ via ${V = X - \overline{U}}$. 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 ${X}$ is a topological space and ${U_{\alpha}}$ a cover.

Definition 10 A partition of unity subordinate to the cover ${U_\alpha}$ is a collection of continuous functions ${\phi_\alpha: X \rightarrow [0, 1]}$ such that each ${\phi_\alpha}$ is supported in ${U_\alpha}$ and ${\sum \phi_\alpha \equiv 1}$. Moreover, we assume that the supports of the ${\phi_\alpha}$ 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 ${f: X \rightarrow \mathbb{R}}$ as ${f = \sum \phi_\alpha f}$, we can reduce problems about the whole space ${X }$ into problems about the constituent parts ${U_\alpha}$, 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 ${\left\{U_\alpha\right\}}$ is an open cover of the paracompact space ${X}$. Then there is a partition of unity subordinate to ${\left\{U_\alpha\right\}}$.

We shall start with a lemma.

Lemma 12 Let ${X}$ be a paracompact space and ${\left\{U_\alpha\right\}}$ an open covering. Then there is a locally finite refinement ${\left\{V_\alpha\right\}}$ with, for each ${a}$,$\displaystyle \overline{V_\alpha} \subset U_\alpha.$

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 ${x \in X}$, we choose an open neighborhood ${U_x}$ whose closure is contained in some ${U_\alpha}$; we can do this by normality of a paracompact space. The ${U_x}$ form an open covering of ${X}$, so we have a locally finite refinement, call it ${\left\{W_\beta\right\}}$. For each ${\alpha}$, define

$\displaystyle V_\alpha = \bigcup_{\overline{W_\beta} \subset U_\alpha} W_\beta.$

In other words, we are collecting together the ${W}$‘s to make the ${V}$‘s. Clearly all the ${W}$‘s are collected together in this manner, so that the ${V_\alpha}$ form an open cover of ${X}$. Local finiteness of the ${W_\beta}$ implies that of the ${V_\alpha}$, which are just bundles of the former; moreover, in the same way we deduce

$\displaystyle \overline{V_\alpha} \subset U_\alpha.$

We have now proved the lemma.

We now proceed to the proof of the theorem. By replacing the ${U_{\alpha}}$ by a locally finite refinement as in the lemma with the same indexing, we can assume that the ${U_{\alpha}}$ are themselves locally finite.

Now take a refinement ${V_\alpha}$ as in the lemma. The key property we now want is that ${\overline{V_\alpha} \subset U_\alpha}$. From this, we can use Urysohn’s lemma to choose a continuous nonnegative ${\psi_\alpha}$ which is equal to one on ${\overline{V_\alpha}}$ but vanishes outside ${U_\alpha}$.

Then the sum ${\psi = \sum \psi_\alpha}$ is well-defined, and we can take

$\displaystyle \phi_\alpha = \frac{\psi_\alpha}{\psi};$

these are supported in the ${U_\alpha}$, and clearly add to one everywhere. The proof is complete.

About these ads