Compactness and Paracompactness

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

We now turn to the remaining result related to compactness: the Tychonoff theorem.

Tychonoff theorem

The arbitrary product of compact spaces is again compact. If the product is finite, this can be shown in a more intuitive manner, but for an infinite product we need the following lemma. It generalizes §Compactness and Convergence of Filters, ⁋Proposition 5 to the language of filters.

Lemma 1 A topological space \(X\) is compact if and only if every ultrafilter converges.

Proof

First, assume \(X\) is compact and let an arbitrary ultrafilter \(\mathcal{F}\) be given. Suppose for contradiction that \(\mathcal{F}\) has no limit point. Then for every \(x\in X\), there exists an open neighborhood \(U_x\) such that \(U_x\not\in \mathcal{F}\). By compactness of \(X\), there exists a finite subcover \(U_{x_1},\ldots, U_{x_n}\) of \(X\).

On the other hand, by [Set Theory] §Filters, Ideals, and Galois Connections, ⁋Proposition 5, \(\mathcal{F}\) is prime. That is, for any subset \(A\subseteq X\), exactly one of \(A\in \mathcal{F}\) or \(X\setminus A\in \mathcal{F}\) holds. Then for any \(A\in \mathcal{F}\),

\[A=A\cap X=(A\cap U_{x_1})\cup \cdots\cup (A\cap U_{x_n})\in \mathcal{F}\]

and since \(U_{x_i}\not\in \mathcal{F}\) by assumption, each \(A\cap U_{x_i}\) is also not in \(\mathcal{F}\); since \(\mathcal{F}\) is maximal, we must have \(X\setminus (A\cap U_{x_i})\in \mathcal{F}\). Then their finite intersection

\[X\setminus A=(X\setminus (A\cap U_{x_1}))\cap\cdots\cap (X\setminus (A\cap U_{x_n}))\]

must also belong to \(\mathcal{F}\), which contradicts the maximality of \(\mathcal{F}\).

Conversely, suppose that every ultrafilter \(\mathcal{F}\) has a limit point, and let \(\mathcal{A}\) be a family of closed subsets of \(X\) satisfying the finite intersection property. Then we can consider an ultrafilter \(\mathcal{F}\) containing the filter generated by \(\mathcal{A}\), and by assumption \(\mathcal{F}\) has a limit point \(x\). That is, \(\mathcal{N}(x)\subseteq \mathcal{F}\), and therefore for every \(F\in \mathcal{F}\) there exists a neighborhood \(U\) of \(x\) such that \(U\cap F\neq\emptyset\). In particular, for any \(A\in \mathcal{A}\) there exists a neighborhood \(U\) of \(x\) such that \(A\cap U\neq\emptyset\), and thus \(x\in \cl(A)=A\) always holds. From this we know that \(x\in\bigcap_{A\in \mathcal{A}}A\), and hence we obtain the desired result by §Compact Spaces, ⁋Proposition 11.

Then the following holds.

Theorem 2 (Tychonoff) The product \(X=\prod_{i\in I} X_i\) of compact spaces \((X_i)_{i\in I}\) is compact. Conversely, if the product space \(X\) is compact, then each \(X_i\) is compact.

Proof

If \(X\) is compact, then each \(X_i\) is compact by the continuity of \(\pr_i\) and §Compact Spaces, ⁋Proposition 8.

For the converse, given any ultrafilter \(\mathcal{F}\) on \(X\), we first verify that \(\pr_i(\mathcal{F})\) defines an ultrafilter base on \(X_i\); then from the assumption that \(X_i\) is compact and Lemma 1 we obtain a limit point \(x_i\) of this ultrafilter, and we can show that \(x=(x_i)_{i\in I}\) is a limit point of \(\mathcal{F}\). Thus the proof is complete by Lemma 1.

Locally Compact Spaces

Compactness is among the strongest properties a topological space can have, yet many spaces we actually deal with are not compact. Euclidean space \(\mathbb{R}^n\) is not compact because it is unbounded, and topological manifolds, which we will discuss later, also only resemble Euclidean space locally and need not be compact globally. Nevertheless, these spaces behave like compact spaces in a neighborhood of each point. We extract this local property so that we can revive compact-space arguments locally even when global compactness is absent. Furthermore, we will see that adding just a single point to such a space yields a compact space.

We formalize this local condition of compactness around each point as follows.

Definition 3 A topological space \(X\) is locally compact at a point \(x\in X\) if there exists a compact neighborhood of \(x\) in \(X\). When \(X\) is locally compact at every point, we call \(X\) a locally compact space. A space that is locally compact and Hausdorff is abbreviated as an LCH space.

Here, a neighborhood of \(x\) means a subset containing an open set that contains \(x\), so a compact neighborhood need not be open. For example, in \(\mathbb{R}\) the closed interval \([-1,1]\) is a compact neighborhood of \(0\) but is not open. Thus the definition requires only the weak form of the existence of a single compact neighborhood, but if the space is Hausdorff, this condition can be restated in a much more convenient form.

Proposition 4 For a point \(x\) in a Hausdorff space \(X\), the following two conditions are equivalent.

1. \(X\) is locally compact at \(x\).
2. For any open neighborhood \(V\) of \(x\), there exists an open set \(W\) such that \(x\in W\), \(\cl(W)\subseteq V\), and \(\cl(W)\) is compact.

Proof

If the second condition holds, taking \(V=X\) gives a compact set \(\cl(W)\) containing the open set \(W\) that contains \(x\), which forms a compact neighborhood of \(x\); thus the second condition implies the first.

Conversely, suppose \(X\) is locally compact at \(x\) and let an open neighborhood \(V\) of \(x\) be given. Take a compact neighborhood \(K\) of \(x\) and let \(U=\interior(K)\); then \(U\) is an open set containing \(x\) with \(U\subseteq K\). We may now replace \(V\) by \(V\cap U\) and assume from the outset that \(V\subseteq U\subseteq K\), since obtaining the conclusion for a smaller \(V\) implies it for the original \(V\).

\(K\) is a compact Hausdorff space and hence is regular. (§Compact Spaces, ⁋Lemma 6) Meanwhile \(V\subseteq K\) is an open set in \(X\) and thus also open in the subspace \(K\), so \(K\setminus V\) is a closed set in \(K\) not containing \(x\). Applying regularity of \(K\) to the point \(x\) and the closed set \(K\setminus V\), we obtain two disjoint open sets \(P\ni x\) and \(Q\supseteq K\setminus V\) in \(K\). Since \(P\subseteq K\setminus Q\subseteq V\) and \(K\setminus Q\) is closed in \(K\), we have \(\cl_K(P)\subseteq K\setminus Q\subseteq V\).

Now let \(W=P\cap U\). \(P\) is open in \(K\) and \(U\) is open in \(X\) with \(U\subseteq K\), so \(W\) is open in \(X\) and \(x\in W\). Since \(W\subseteq U\subseteq K\) and \(K\) is closed in \(X\) (§Compact Spaces, ⁋Corollary 5), we have \(\cl(W)\subseteq K\). Therefore \(\cl(W)=\cl(W)\cap K=\cl_K(W)\subseteq\cl_K(P)\subseteq V\). Finally, \(\cl(W)\) is a closed subset of the compact set \(K\) and hence is compact. (§Compact Spaces, ⁋Lemma 3) Thus the second condition holds.

The second condition of Proposition 4 can be read as saying that in an LCH space, each point has a neighborhood basis consisting of open sets with compact closure. This will be used repeatedly whenever we unfold local arguments, and it also plays a key role in determining the Hausdorff property of the one-point compactification. Moreover, from this property it follows immediately that any LCH space is regular: given a point \(x\) and a closed set \(C\) not containing \(x\), applying the second condition to \(V=X\setminus C\) yields \(W\) and \(X\setminus\cl(W)\) separating \(x\) and \(C\).

The most basic example is that Euclidean space \(\mathbb{R}^n\) is LCH. For any point \(x\), the closed ball \(\{y:\lVert y-x\rVert\leq 1\}\) is compact by the Heine–Borel theorem and contains the open ball, so it forms a compact neighborhood of \(x\), and we already know that \(\mathbb{R}^n\) is Hausdorff. Similarly, any discrete space is also LCH, since for each point \(x\) the singleton \(\{x\}\) is an open finite set and hence a compact neighborhood, and a discrete space is Hausdorff. A slightly less obvious example is a topological manifold.

Example 5 A Hausdorff space in which each point has an open neighborhood homeomorphic to an open subset of \(\mathbb{R}^n\) is an LCH space. A topological manifold, which we will define later, is precisely such a space. Indeed, for each point \(x\) of such a space \(M\), take an open neighborhood \(U\) homeomorphic to an open subset of \(\mathbb{R}^n\); then since the point corresponding to \(x\) under this homeomorphism has a compact neighborhood in \(\mathbb{R}^n\) (because \(\mathbb{R}^n\) is LCH), pulling this back to \(U\) gives a compact neighborhood of \(x\).

The condition of being locally compact may appear very weak at first glance, but it is by no means automatic. The following is a representative example of a space that is not locally compact.

Example 6 The rational number space \(\mathbb{Q}\) is not locally compact at any point as a subspace of \(\mathbb{R}\). By symmetry, it suffices to show this at \(0\). Suppose for contradiction that there exists a compact neighborhood \(K\subseteq\mathbb{Q}\) of \(0\). Then \(K\) contains an open set containing \(0\), so for some \(\delta>0\) we have \(\mathbb{Q}\cap(-\delta,\delta)\subseteq K\), and fixing \(0<r<\delta\) we have \(\mathbb{Q}\cap[-r,r]\subseteq K\).

Compactness is an intrinsic property independent of the ambient space in which a subspace sits, so \(K\) is also compact as a subspace of \(\mathbb{R}\), and since \(\mathbb{R}\) is Hausdorff, \(K\) is closed in \(\mathbb{R}\). (§Compact Spaces, ⁋Corollary 5) However, the closure of \(\mathbb{Q}\cap[-r,r]\) in \(\mathbb{R}\) is the entire interval \([-r,r]\), so since \(K\) is closed and contains \(\mathbb{Q}\cap[-r,r]\), we obtain \([-r,r]\subseteq K\). This contradicts the fact that \(K\subseteq\mathbb{Q}\) while \([-r,r]\) contains irrational numbers.

Local compactness is inherited by suitable subspaces, but not by arbitrary subspaces—only by those that are open or closed.

Proposition 7 Open subspaces and closed subspaces of an LCH space are both LCH spaces.

Proof

Since subspaces of a Hausdorff space are again Hausdorff, we only need to verify local compactness.

First, let \(A\subseteq X\) be an open set and let \(x\in A\) be given. Since \(A\) is an open neighborhood of \(x\) in \(X\), by the second condition of Proposition 4 there exists an open set \(W\) in \(X\) such that \(x\in W\) and \(\cl_X(W)\subseteq A\) with \(\cl_X(W)\) compact. Since \(\cl_X(W)\subseteq A\), the closure \(\cl_A(W)\) in \(A\) equals \(\cl_X(W)\), which is compact and contains the open set \(W\) in \(A\); thus it is a compact neighborhood of \(x\) in \(A\).

Now let \(A\subseteq X\) be a closed set and let \(x\in A\) be given. Take a compact neighborhood \(K\) of \(x\) in \(X\) and choose an open set \(U\subseteq K\) in \(X\) containing \(x\). Then \(K\cap A\) is a closed subset of the compact set \(K\) and hence is compact (§Compact Spaces, ⁋Lemma 3), and \(U\cap A\) is an open set in \(A\) containing \(x\) and contained in \(K\cap A\); thus \(K\cap A\) is a compact neighborhood of \(x\) in \(A\).

Construction of the One-Point Compactification

The most economical way to make a non-compact space compact is to fill in the missing part with a single point. Intuitively, this is like gathering all points escaping to both ends of \(\mathbb{R}\) into a single point at infinity to form a circle. We explicitly formalize this compactification for an arbitrary topological space, show that it yields a compact space, and then determine when this construction yields a Hausdorff space.

Definition 8 Let a topological space \(X\) be given. Write a new point not belonging to \(X\) as \(\infty\) and consider the set \(X^+=X\cup\{\infty\}\). Among the subsets of \(X^+\), we declare the following two kinds to be open: first, open sets \(U\) of \(X\); second, for subsets \(C\) of \(X\) that are compact and closed, the sets \(X^+\setminus C\). The topological space \(X^+\) obtained in this way is called the one-point compactification of \(X\), or the Alexandroff compactification.

We must verify that this declaration actually satisfies the axioms of a topology from §Open Sets, ⁋Definition 1. The empty set is an open set of \(X\), so it belongs to the first kind, and the empty set is compact and closed, so \(X^+=X^+\setminus\emptyset\) belongs to the second kind. For the intersection of two open sets, we distinguish three cases. The intersection of two sets of the first kind is an open set of \(X\). The intersection of two sets of the second kind is \((X^+\setminus C)\cap(X^+\setminus D)=X^+\setminus(C\cup D)\), and since \(C\cup D\) is a compact closed set, it again belongs to the second kind. The intersection of different kinds is \(U\cap(X^+\setminus C)=U\cap(X\setminus C)\), and since \(C\) is closed, \(X\setminus C\) is open, so this is an open set of \(X\), that is, of the first kind. For arbitrary unions, similarly, the union of sets of the first kind is open, the union of sets of the second kind \(\bigcup_\alpha(X^+\setminus C_\alpha)=X^+\setminus\bigcap_\alpha C_\alpha\) is of the second kind since \(\bigcap_\alpha C_\alpha\) is a closed subset of some \(C_\alpha\) and hence compact (§Compact Spaces, ⁋Lemma 3), and a mixed union is \(U\cup(X^+\setminus C)=X^+\setminus(C\cap(X\setminus U))\) where \(C\cap(X\setminus U)\) is a compact closed set, so it is of the second kind.

We first organize how \(X\) sits inside \(X^+\) under this topology.

Proposition 9 The inclusion map \(X\hookrightarrow X^+\) is an open embedding, that is, a homeomorphism placing \(X\) as an open subspace of \(X^+\). Moreover, \(X\) is dense in \(X^+\) if and only if \(X\) is not compact.

Proof

\(X\) is an open set of \(X\), so it is an open set of \(X^+\) by the first kind. Intersecting an open set of \(X^+\) with \(X\), for a set \(U\) of the first kind we have \(U\cap X=U\), and for a set \(X^+\setminus C\) of the second kind we have \((X^+\setminus C)\cap X=X\setminus C\); both are open sets of \(X\), and conversely any open set of \(X\) is an open set of \(X^+\) as a set of the first kind. Therefore the subspace topology on \(X\) coincides with the original topology, and the inclusion map is a homeomorphism onto an open subspace.

\(\{\infty\}=X^+\setminus X\) is the complement of the open set \(X\), so it is closed. That \(X\) is dense means \(\infty\in\cl(X)\), which is equivalent to every open neighborhood of \(\infty\) meeting \(X\). An open set containing \(\infty\) must be of the second kind \(X^+\setminus C\), and this meets \(X\) if and only if \(X\setminus C\neq\emptyset\), that is, \(C\neq X\). Therefore the necessary and sufficient condition for \(X\) not to be dense is that \(C=X\) for some compact closed set \(C\), that is, that \(X\) itself is compact.

When \(X\) is compact, \(X\) itself is compact and closed, so \(\{\infty\}=X^+\setminus X\) becomes an open set and \(\infty\) is an isolated point. In this case \(X^+\) is nothing more than \(X\) with an isolated point attached, which is not interesting. The one-point compactification plays its intended role when \(X\) is not compact; in this case \(\infty\) serves as the limit point of all directions escaping outside \(X\).

Theorem 10 For any topological space \(X\), \(X^+\) is compact.

Proof

Let an arbitrary open covering \((O_i)_{i\in I}\) of \(X^+\) be given. There exists at least one open set \(O_j\) covering \(\infty\), which must be of the second kind, so we can write \(O_j=X^+\setminus C\) for a compact closed subset \(C\) of \(X\). The remaining \((O_i)_{i\neq j}\) must cover \(C\subseteq X^+\setminus O_j\), and each \(O_i\cap X\) is an open set of \(X\), so \((O_i\cap X)_{i\neq j}\) is an open covering of \(C\) in \(X\). Since \(C\) is compact, we can choose a finite \(J\subseteq I\setminus\{j\}\) such that \(C\subseteq\bigcup_{i\in J}(O_i\cap X)\subseteq\bigcup_{i\in J}O_i\). (§Compact Spaces, ⁋Proposition 2) Then \((O_i)_{i\in J\cup\{j\}}\) is a finite subcover of \(X^+\).

Hausdorff Criterion and Universality

The one-point compactification yields a compact space for any space, but whether the result is again Hausdorff is a separate problem. For example, \(\mathbb{Q}^+\) is compact but not Hausdorff. The following theorem reveals that the condition for \(X^+\) to be Hausdorff is exactly the local compactness defined above.

Theorem 11 For a topological space \(X\), \(X^+\) is a Hausdorff space if and only if \(X\) is an LCH space.

Proof

First, suppose \(X^+\) is Hausdorff. The subspace \(X\) is Hausdorff since it is a subspace of a Hausdorff space. To show local compactness, fix \(x\in X\); by the Hausdorff property of \(X^+\) there exist disjoint open sets \(U\ni x\) and \(W\ni\infty\) separating \(x\) and \(\infty\). Since \(W\) contains \(\infty\), it is of the second kind, and thus \(W=X^+\setminus C\) for a compact closed subset \(C\) of \(X\). From \(U\cap W=\emptyset\) we have \(U\subseteq C\), and \(U\) does not contain \(\infty\) so it is an open set contained in \(X\). Therefore \(C\) is a compact set containing the open set \(U\) that contains \(x\), that is, a compact neighborhood of \(x\), and \(X\) is locally compact at \(x\).

Conversely, suppose \(X\) is LCH. We must separate two distinct points of \(X^+\). If both points lie in \(X\), then since \(X\) is Hausdorff we obtain disjoint open sets separating them in \(X\), and these are also open in \(X^+\) as sets of the first kind. The remaining case is when one point is \(x\in X\) and the other is \(\infty\). Since \(X\) is Hausdorff, by Proposition 4 there exists an open neighborhood \(U\) of \(x\) such that \(K=\cl(U)\) is compact. Since \(X\) is Hausdorff, \(K\) is closed (§Compact Spaces, ⁋Corollary 5), and therefore \(X^+\setminus K\) is an open set of the second kind containing \(\infty\). Since \(U\subseteq K\), the sets \(U\) and \(X^+\setminus K\) are disjoint and separate \(x\) and \(\infty\).

Combining Theorem 10 and Theorem 11, when \(X\) is an LCH space, \(X^+\) becomes a compact Hausdorff space and by Proposition 9 \(X\) is embedded in it as an open subspace. In particular, if \(X\) is not compact, this embedding is dense. This is the existence part of Alexandroff’s theorem: any LCH space can be embedded as a dense open subspace of some compact Hausdorff space. What remains is that such a compactification is essentially unique, which is formalized by the following universality property.

Theorem 12 Let an LCH space \(X\) be given. If a compact Hausdorff space \(Y\), a point \(p\in Y\), and a homeomorphism \(\varphi:X\rightarrow Y\setminus\{p\}\) are given, then there exists a unique homeomorphism \(h:X^+\rightarrow Y\) such that \(h=\varphi\) on \(X\) and \(h(\infty)=p\).

Proof

Identifying \(X\) and \(Y\setminus\{p\}\) via \(\varphi\), we regard \(X\) as a subset of \(Y\). Since \(Y\) is Hausdorff, \(\{p\}\) is a closed set of \(Y\), and therefore \(X=Y\setminus\{p\}\) is an open subspace of \(Y\). That is, the topology of \(X\) coincides with the subspace topology induced from \(Y\).

We now show that the open sets of \(Y\) correspond exactly to the open sets of \(X^+\). If an open set \(O\) of \(Y\) does not contain \(p\), then \(O\subseteq X\) and since \(X\) is an open subspace of \(Y\), \(O\) is an open set of \(X\), that is, of the first kind. Conversely, an open set of \(X\) is open in \(Y\) since \(X\) is open in \(Y\). On the other hand, if an open set \(O\) of \(Y\) contains \(p\), then \(Y\setminus O\) is a closed subset of the compact space \(Y\) and hence is compact (§Compact Spaces, ⁋Lemma 3), and \(Y\setminus O\subseteq X\) and is closed in \(Y\) so also closed in \(X\). Therefore \(O=X^+\setminus(Y\setminus O)\) is the complement of a compact closed subset of \(X\), that is, of the second kind. Conversely, for a compact closed subset \(C\) of \(X\), since \(C\) is also compact in \(Y\) and \(Y\) is Hausdorff, it is closed (§Compact Spaces, ⁋Corollary 5), so \(Y\setminus C\) is an open set of \(Y\) containing \(p\).

Therefore the correspondence \(h:X^+\rightarrow Y\) identifying \(p\) with \(\infty\) sends open sets to open sets and vice versa, bijectively, and hence is a homeomorphism. Since \(h\) must agree with \(\varphi\) on \(X\) and send \(\infty\) to \(p\), it is unique.

Theorem 12 says that for an LCH space \(X\), there is only one way to add a point to make a compact Hausdorff space, up to homeomorphism. This is the uniqueness part of Alexandroff’s theorem expressed in the language of universality. Thanks to this uniqueness, we can treat \(X^+\) independently of its concrete construction, and in actual calculations we may choose any convenient compact Hausdorff model and identify it with \(X^+\).

Remark 13 The one-point compactification is characterized as the smallest among Hausdorff compactifications. A Hausdorff compactification of a non-compact LCH space \(X\) is a compact Hausdorff space containing \(X\) as a dense subspace; from any such compactification, collapsing all points outside \(X\) into a single point yields a unique continuous surjection onto \(X^+\). The proof of this fact requires the observation that if an LCH space is densely embedded in a Hausdorff space, it is always an open subspace; the detailed argument follows standard literature. [Mun] At the opposite extreme is the Stone–Čech compactification, the largest Hausdorff compactification that a completely regular space can have, but this requires a separate construction so we only mention its name here.

Examples of One-Point Compactifications

The most familiar example is that the one-point compactification of Euclidean space becomes a sphere.

Example 14 Consider the north pole \(N=(0,\ldots,0,1)\) of the \(n\)-sphere \(S^n=\{x\in\mathbb{R}^{n+1}:\lVert x\rVert=1\}\). The stereographic projection

\[\sigma:S^n\setminus\{N\}\rightarrow\mathbb{R}^n,\qquad \sigma(x_1,\ldots,x_{n+1})=\frac{1}{1-x_{n+1}}(x_1,\ldots,x_n)\]

is well known to be a homeomorphism between \(S^n\setminus\{N\}\) and \(\mathbb{R}^n\). Since \(S^n\) is a closed bounded subset of \(\mathbb{R}^{n+1}\), it is compact by the Heine–Borel theorem, and it is Hausdorff as a subspace. Therefore \(S^n\) is a compact Hausdorff space, and removing the single point \(N\) yields a space homeomorphic to \(\mathbb{R}^n\); thus by Theorem 12 there is a unique homeomorphism

\[(\mathbb{R}^n)^+\cong S^n\]

and the point at infinity \(\infty\) corresponds to the north pole \(N\). In particular, \((\mathbb{R})^+\) is the circle \(S^1\).

The one-point compactification of a discrete space gives a very concrete picture of a convergent sequence.

Example 15 Endow the set of natural numbers \(\mathbb{N}=\{1,2,3,\ldots\}\) with the discrete topology. This is LCH as a discrete space, so by Theorem 11 \(\mathbb{N}^+\) is a compact Hausdorff space. In a discrete space, the only compact subsets are finite sets, so an open neighborhood of \(\infty\) in \(\mathbb{N}^+\) is the complement of a finite set, that is, a cofinite set containing \(\infty\). This means precisely that a sequence in \(\mathbb{N}\) converges to \(\infty\) if and only if it eventually leaves any finite set.

This space is realized by a familiar subset of the real numbers. Consider the function

\[f:\mathbb{N}^+\rightarrow\mathbb{R},\qquad f(n)=\frac1n\quad(n\in\mathbb{N}),\qquad f(\infty)=0\]

Then \(f\) is a bijection between \(\mathbb{N}^+\) and \(\{0\}\cup\{1/n\mid n\geq 1\}\). Each \(n\in\mathbb{N}\) is an isolated point in \(\mathbb{N}^+\) and its image \(1/n\) is also an isolated point in \(\{0\}\cup\{1/n\}\), and since the cofinite neighborhoods of \(\infty\) map to neighborhoods of \(0\), \(f\) is continuous. Since the domain is compact and the codomain is Hausdorff, by §Compact Spaces, ⁋Proposition 9 \(f\) is a homeomorphism. That is, \(\mathbb{N}^+\) is homeomorphic to a convergent sequence with one limit point.

Complete Regularity

The one-point compactification is not merely an existential tool; it is also used to characterize intrinsic properties of LCH spaces. A compact Hausdorff space is normal, so (§Compact Spaces, ⁋Proposition 7) it has abundant continuous functions via Urysohn’s lemma, and we show that this property is inherited by LCH spaces through open subspaces.

Corollary 16 Any LCH space is completely regular. Therefore any LCH space is a Tychonoff space. (§Hausdorff Spaces, ⁋Definition 3)

Proof

We first show that any compact Hausdorff space \(Y\) is completely regular. Let a point \(p\in Y\) and a closed set \(C\subseteq Y\) not containing \(p\) be given. Since \(Y\) is Hausdorff, it is \(T_1\) and thus \(\{p\}\) is a closed set; \(\{p\}\) and \(C\) are two disjoint closed sets. Since \(Y\) is normal, (§Compact Spaces, ⁋Proposition 7) by Urysohn’s lemma there exists a continuous function \(g:Y\rightarrow[0,1]\) taking the value \(0\) on \(\{p\}\) and \(1\) on \(C\). (§Urysohn’s Lemma and Tietze Extension Theorem, ⁋Theorem 2) This means precisely that \(p\) and \(C\) can be separated by a continuous function, so \(Y\) is completely regular.

Now let \(X\) be LCH. By Theorem 10 and Theorem 11, \(X^+\) is a compact Hausdorff space and hence completely regular as shown above. By Proposition 9, \(X\) is a subspace of \(X^+\). We show that complete regularity is inherited by subspaces. Let \(x\in X\) and a closed set \(C\) of \(X\) not containing \(x\) be given; by the property of closed sets in a subspace, there exists a closed set \(C'\) of \(X^+\) such that \(C=C'\cap X\). Since \(x\in X\) and \(x\notin C\), we have \(x\notin C'\), and since \(X^+\) is completely regular, there exists a continuous function \(g:X^+\rightarrow[0,1]\) such that \(g(x)=0\) and \(g\) takes the value \(1\) on \(C'\). The restriction \(g\vert_X\) to \(X\) is a continuous function separating \(x\) and \(C\subseteq C'\), so \(X\) is completely regular. Finally, \(X\) is Hausdorff and hence \(T_0\), so \(X\) is a Tychonoff space.

Corollary 16 guarantees that on an LCH space, we can always obtain a continuous function separating two distinct points or a point and a closed set. This is a fundamental fact indicating how well-behaved locally compact spaces are as analytic objects, and it serves as the starting point for various constructions that patch together locally defined data into global ones using continuous functions.

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References

[Mun] J. R. Munkres, Topology, 2nd ed., Prentice Hall, 2000.

[Wil] S. Willard, General Topology, Addison-Wesley, 1970.

[Kel] J. L. Kelley, General Topology, Springer, 1975.

Bridging Locality and Globality

The objects we wish to handle on a topological space are usually given locally first: continuous functions defined in a neighborhood of each point, sections obtained locally, constructions placed on each coordinate patch. To patch such local data into a single global object, we need a device that smoothly vanishes each piece within its own domain of definition while distributing values so that they do not overlap across the whole space. The standard tool serving this role is the partition of unity, and the natural stage on which it exists is the paracompact Hausdorff space we discuss in this article.

Compactness demands that from any open cover we can extract a finite subcover (§Compact Spaces, ⁋Definition 1), but most spaces we actually deal with are not compact. Paracompactness weakens compactness appropriately by replacing finiteness with local finiteness, allowing us to revive many arguments that held for compact spaces in a local manner. We introduce this concept, show that paracompact Hausdorff spaces are normal, and then use this as a foundation to prove that a partition of unity subordinate to any open cover can always be constructed.

Paracompact Spaces

We first name the operation of slicing an open cover into finer pieces so that each piece fits entirely inside some original piece.

Definition 17 Let two covers \((U_i)_{i\in I}\) and \((V_j)_{j\in J}\) of a topological space \(X\) be given. The latter is called a refinement of the former if for every \(j\in J\) there exists \(i\in I\) such that \(V_j\subseteq U_i\). When all elements of a refinement \((V_j)_{j\in J}\) are open sets, we call it an open refinement.

A refinement is a much more flexible notion than a subcover. While a subcover merely selects some of the original pieces as they are, a refinement allows freely cutting each piece into smaller ones as long as each resulting piece remains contained in some original piece. Combining this with local finiteness yields a new finiteness condition that can replace compactness. Recall that a family \((A_i)_{i\in I}\) is locally finite if every point has a neighborhood meeting only finitely many \(A_i\). (§Interior, Closure, and Boundary, ⁋Definition 3)

Definition 18 A topological space \(X\) is paracompact if every open cover of \(X\) has a locally finite open refinement.

The definition demands that for any open cover, there exists a locally finite open cover refining it. Compactness demands that any open cover still covers the whole space after discarding all but finitely many sets; a cover consisting of finitely many open sets is itself locally finite, so paracompactness can be read as relaxing the finiteness required by compactness to finiteness in a neighborhood of each point.

Proposition 19 Any compact space is paracompact.

Proof

Let \(X\) be a compact space and let an open cover \((U_i)_{i\in I}\) be given. By compactness there exists a finite \(J\subseteq I\) such that \((U_j)_{j\in J}\) still covers \(X\). (§Compact Spaces, ⁋Definition 1) This finite subcover is an open refinement of the original cover, and a finite family is always locally finite, so (§Interior, Closure, and Boundary, ⁋Definition 3) it is a locally finite open refinement of \((U_i)_{i\in I}\). Therefore \(X\) is paracompact.

That paracompactness is substantially broader than compactness must be verified on non-compact spaces. The following is a representative case, showing that even in Euclidean space, which has no finiteness at all, we can explicitly construct a locally finite refinement.

Example 20 Euclidean space \(\mathbb{R}^n\) is paracompact. \(\mathbb{R}^n\) is not compact because it is unbounded, but we can exhaust the space by open balls centered at the origin with increasing radii and obtain a locally finite refinement.

Let an arbitrary open cover \(\mathcal{U}=(U_i)_{i\in I}\) be given. For each integer \(k\geq 1\), let \(B_k=\{x:\lVert x\rVert<k\}\) be the open ball of radius \(k\), and agree that \(B_0=B_{-1}=\emptyset\). Then the shells

\[A_k=\cl(B_k)\setminus B_{k-1}=\{x: k-1\leq\lVert x\rVert\leq k\}\]

are closed bounded subsets of \(\mathbb{R}^n\) and hence compact by the Heine–Borel theorem, and their union is all of \(\mathbb{R}^n\). Meanwhile,

\[O_k=B_{k+1}\setminus\cl(B_{k-2})\]

is an open set satisfying \(A_k\subseteq O_k\). Here for \(k\geq 3\) we have \(O_k=\{x: k-2<\lVert x\rVert<k+1\}\), and for \(k=1,2\) since \(B_{k-2}=\emptyset\) we have \(\cl(B_{k-2})=\emptyset\) so \(O_k=B_{k+1}\).

We now treat each compact set \(A_k\). Each \(x\in A_k\) belongs to some \(U_i\), so \(x\in U_i\cap O_k\), and such open sets cover \(A_k\). Since \(A_k\) is compact, we can choose finitely many indices, that is, a finite set \(F_k\subseteq I\), such that \((U_i\cap O_k)_{i\in F_k}\) covers \(A_k\). (§Compact Spaces, ⁋Proposition 2) Consider the family gathered over all \(k\geq 1\):

\[\mathcal{V}=(U_i\cap O_k)_{k\geq 1,i\in F_k}\]

Each element is an open set contained in \(U_i\), so \(\mathcal{V}\) is an open refinement of \(\mathcal{U}\); and since the \(A_k\) cover \(\mathbb{R}^n\), \(\mathcal{V}\) also covers \(\mathbb{R}^n\). Finally, we verify that \(\mathcal{V}\) is locally finite. For a point \(x\), letting \(r=\lVert x\rVert\), the neighborhood \(B_{r+1}=\{y:\lVert y\rVert<r+1\}\) meets \(O_k\) only when \(k-2<r+1\), that is, \(k<r+3\), so it meets only finitely many \(O_k\). For each \(k\), the elements of \(\mathcal{V}\) are only finitely many (\(\lvert F_k\rvert\) many), so \(B_{r+1}\) meets only finitely many elements of \(\mathcal{V}\). Therefore \(\mathcal{V}\) is a locally finite open refinement and \(\mathbb{R}^n\) is paracompact.

The argument of this example relies not on any special property of \(\mathbb{R}^n\) but only on two properties: local compactness and exhaustion by countably many compact sets. Indeed, the same method shows that any second countable LCH space, and more generally any \(\sigma\)-compact LCH space, is paracompact. This forms the basis for the paracompactness of topological manifolds, which we will discuss later.

Normality of Paracompact Hausdorff Spaces

The power of paracompactness is fully revealed when combined with the Hausdorff condition. Just as a compact Hausdorff space is normal (§Compact Spaces, ⁋Proposition 7), we aim to show that a paracompact Hausdorff space is also normal. Once normality is secured, we can obtain abundant continuous functions via Urysohn’s lemma, and this becomes the key ingredient for constructing partitions of unity. (§Urysohn’s Lemma and Tietze Extension Theorem, ⁋Theorem 2)

The proof repeatedly relies on the following property of locally finite families: in a locally finite family, the closure of a union equals the union of the closures, so the closure operation freely commutes with infinite unions.

Lemma 21 Let \((A_i)_{i\in I}\) be a locally finite family of subsets of a topological space \(X\). Then

\[\cl\Bigl(\bigcup_{i\in I} A_i\Bigr)=\bigcup_{i\in I}\cl(A_i)\]

holds.

Proof

For each \(i\), since \(A_i\subseteq\bigcup_j A_j\), we have \(\cl(A_i)\subseteq\cl(\bigcup_j A_j)\), and therefore \(\bigcup_i\cl(A_i)\subseteq\cl(\bigcup_j A_j)\).

For the reverse inclusion, it suffices to show that \(\bigcup_i\cl(A_i)\) is a closed set, for then the closure of \(\bigcup_i A_i\subseteq\bigcup_i\cl(A_i)\) is also contained in \(\bigcup_i\cl(A_i)\). We first observe that the family \((\cl(A_i))_{i\in I}\) is also locally finite. We can choose a neighborhood \(V\) of a point \(x\) such that \(V\) meets only finitely many \(A_i\); if we choose \(V\) to be open, then whenever \(V\cap\cl(A_i)\neq\emptyset\), since \(V\) is a neighborhood of a point of \(\cl(A_i)\) we have \(V\cap A_i\neq\emptyset\). Therefore \(V\) meets only finitely many \(\cl(A_i)\), and \((\cl(A_i))_{i\in I}\) is also locally finite. This is a locally finite family of closed sets, so their union \(\bigcup_i\cl(A_i)\) is closed. (§Interior, Closure, and Boundary, ⁋Proposition 4)

We first show that a paracompact Hausdorff space is regular. The skeleton of the argument is as follows: to separate a point and a closed set, we use the Hausdorff property to obtain, for each point of the closed set, an open set whose closure avoids the point in question; then we form an open cover from these together with the complement of the closed set, refine it to be locally finite using paracompactness, and control the closure using Lemma 21.

Proposition 22 Any paracompact Hausdorff space is a regular space. (§Hausdorff Spaces, ⁋Definition 3)

Proof

Let \(X\) be a paracompact Hausdorff space, and let a point \(a\in X\) and a closed set \(B\subseteq X\) not containing \(a\) be given. For each \(b\in B\), since \(a\neq b\), by the Hausdorff property of \(X\) there exist disjoint open sets \(P_b\ni a\) and \(U_b\ni b\). Since \(U_b\subseteq X\setminus P_b\) and \(X\setminus P_b\) is closed, we have \(\cl(U_b)\subseteq X\setminus P_b\), and in particular \(a\notin\cl(U_b)\).

Adding the open set \(X\setminus B\) to the family \((U_b)_{b\in B}\) yields an open cover of \(X\). Since \(X\) is paracompact, choose a locally finite open refinement \(\mathcal{C}\) of this cover. Among the elements of \(\mathcal{C}\), let \(\mathcal{D}=\{C\in\mathcal{C}\mid C\cap B\neq\emptyset\}\) be those meeting \(B\). Each element \(C\) of \(\mathcal{D}\) is contained in \(X\setminus B\) or some \(U_b\) by the refinement property; since \(C\) meets \(B\), it cannot be contained in \(X\setminus B\), so it is contained in some \(U_b\) and thus \(a\notin\cl(C)\). Also, each point of \(B\) belongs to some element of \(\mathcal{C}\) containing it, and that element meets \(B\) so it belongs to \(\mathcal{D}\). That is, \(\mathcal{D}\) covers \(B\).

Let \(V=\bigcup\mathcal{D}\); then \(V\) is an open set containing \(B\). Since \(\mathcal{D}\) is a subfamily of the locally finite family \(\mathcal{C}\), it is locally finite, and by Lemma 21 we have \(\cl(V)=\bigcup_{C\in\mathcal{D}}\cl(C)\). Since each \(\cl(C)\) does not contain \(a\), we have \(a\notin\cl(V)\). Then \(W=X\setminus\cl(V)\) is an open set containing \(a\) and disjoint from \(V\supseteq B\). Therefore \(a\) and \(B\) are separated by neighborhoods and \(X\) is regular.

Expanding the same argument with the point \(a\) replaced by a closed set \(A\) yields normality. Here we use the regularity just proved instead of the Hausdorff property, to ensure that the closures of the open sets covering each point of \(B\) avoid \(A\).

Theorem 23 Any paracompact Hausdorff space is a normal space. (§Hausdorff Spaces, ⁋Definition 3)

Proof

Let \(X\) be a paracompact Hausdorff space and let two disjoint closed sets \(A,B\subseteq X\) be given. By Proposition 22, \(X\) is regular. For each \(b\in B\), since \(b\notin A\), applying regularity to the point \(b\) and the closed set \(A\) yields disjoint open sets \(U_b\ni b\) and \(Q_b\supseteq A\). Since \(U_b\subseteq X\setminus Q_b\) and \(X\setminus Q_b\) is closed, we have \(\cl(U_b)\subseteq X\setminus Q_b\subseteq X\setminus A\), that is, \(\cl(U_b)\cap A=\emptyset\).

Add the open set \(X\setminus B\) to the family \((U_b)_{b\in B}\) to form an open cover, and by paracompactness choose a locally finite open refinement \(\mathcal{C}\); let \(\mathcal{D}=\{C\in\mathcal{C}\mid C\cap B\neq\emptyset\}\). As in the proof of Proposition 22, each element of \(\mathcal{D}\) is contained in some \(U_b\) and thus satisfies \(\cl(C)\cap A=\emptyset\), and \(\mathcal{D}\) covers \(B\).

\(V=\bigcup\mathcal{D}\) is an open set containing \(B\), and since \(\mathcal{D}\) is locally finite, by Lemma 21 we have \(\cl(V)=\bigcup_{C\in\mathcal{D}}\cl(C)\), which is disjoint from \(A\). Therefore \(W=X\setminus\cl(V)\) is an open set containing \(A\) and disjoint from \(V\supseteq B\). Then \(V\) and \(W\) are disjoint open sets containing \(B\) and \(A\) respectively, so \(X\) is normal.

Theorem 23 is the true generalization of the fact that a compact Hausdorff space is normal. Indeed, by Proposition 19 a compact space is paracompact, so the result that a compact Hausdorff space is normal is recovered as a special case of Theorem 23. With normality secured, we can now always obtain a continuous function separating two disjoint closed sets, and this makes the construction of partitions of unity in the next section possible.

Among the most abundant sources of paracompact spaces are metric spaces. We already know that every metric space is normal (§Urysohn’s Lemma and Tietze Extension Theorem, ⁋Proposition 4), but in fact they are always paracompact as well. This is known as A. H. Stone’s theorem.

Theorem 24 (Stone) Any metric space is paracompact.

Proof

We only sketch the key idea of the proof and leave the details to standard literature. [Mun] Let a space \(X\) with metric \(d\) and an open cover \((U_\alpha)_{\alpha\in J}\) be given. First, using the axiom of choice we impose a well-ordering on the index set \(J\). For each integer \(n\geq 1\) and each \(\alpha\), from the points of \(U_\alpha\) we retain only those that are at least \(2^{-n}\) away from the boundary and are not already in the corresponding set for some earlier \(U_\beta\) in the order, and then take the union of open balls of radius \(2^{-n-1}\) centered at these remaining points to define the set \(V_{n,\alpha}\). Here the well-ordering on \(\alpha\) and the geometrically shrinking radii interact so that the family \((V_{n,\alpha})\) becomes an open cover refining \((U_\alpha)\) and simultaneously locally finite: for each point \(x\), looking at the stage \(n\) where \(x\) is first covered, a neighborhood of radius about \(2^{-n-1}\) meets only finitely many \(V_{n',\alpha}\). Therefore \(X\) is paracompact. This construction is widely known in the form recorded by M. E. Rudin.

Existence of Partitions of Unity

We now formalize the partition of unity, the tool that patches locally defined data into a global object. The support of a continuous function \(\phi:X\rightarrow[0,1]\) is defined as \(\supp\phi=\cl(\{x\in X\mid\phi(x)\neq 0\})\), which is the smallest closed set containing the region where \(\phi\) takes non-zero values.

Definition 25 A family \((\phi_i)_{i\in I}\) of continuous functions on a topological space \(X\) is called a partition of unity if each \(\phi_i:X\rightarrow[0,1]\) satisfies the following two conditions.

1. The family \((\supp\phi_i)_{i\in I}\) is locally finite.
2. For every \(x\in X\), \(\sum_{i\in I}\phi_i(x)=1\) holds.

Furthermore, given an open cover \((U_i)_{i\in I}\) of \(X\), a partition of unity \((\phi_i)_{i\in I}\) with the same index set is called subordinate to \((U_i)\) if \(\supp\phi_i\subseteq U_i\) for all \(i\).

The local finiteness in the first condition ensures that the sum \(\sum_i\phi_i(x)\) in the second condition reduces to a finite sum in some neighborhood of each point, so it actually makes sense. The subordination condition \(\supp\phi_i\subseteq U_i\) means that each \(\phi_i\) vanishes not only outside \(U_i\) but even near the boundary of \(U_i\), so multiplying locally defined data on \(U_i\) by \(\phi_i\) allows us to extend the product continuously to all of \(X\) by filling in \(0\) outside \(U_i\). This is the principle by which a partition of unity globalizes local constructions.

The key to the existence proof is that normality alone is not sufficient; we need to “shrink” the cover twice so that closed sets still cover the whole space. We first prepare a lemma for this purpose. A family \((U_i)_{i\in I}\) is point-finite if each point \(x\in X\) belongs to only finitely many \(U_i\); a locally finite family is always point-finite.

Lemma 26 (Shrinking lemma) Let \((U_\alpha)_{\alpha\in J}\) be a point-finite open cover of a normal space \(X\). Then there exists an open cover \((V_\alpha)_{\alpha\in J}\) such that \(\cl(V_\alpha)\subseteq U_\alpha\) for all \(\alpha\).

Proof

Using the axiom of choice, we impose a well-ordering on the index set \(J\). We define open sets \(V_\alpha\) for each \(\alpha\in J\) by transfinite induction, maintaining the following invariant at every stage:

\[(\ast_\alpha)\qquad \{V_\beta\mid\beta<\alpha\}\cup\{U_\beta\mid\beta\geq\alpha\}\ \text{covers}\ X.\]

We first verify that \((\ast_\alpha)\) holds for arbitrary \(\alpha\) from point-finiteness. Suppose that for each \(\beta<\alpha\), \(V_\beta\) has already been defined satisfying \(\cl(V_\beta)\subseteq U_\beta\). Given a point \(x\in X\), it belongs to only finitely many \(U_\gamma\). If among these there is one with \(\gamma\geq\alpha\), then \(x\) is covered by \(U_\gamma\). Otherwise, all \(\gamma\) with \(x\in U_\gamma\) are less than \(\alpha\); let \(\gamma_0\) be the largest among them. By the already valid \((\ast_{\gamma_0})\), \(x\) is covered by either some \(V_\beta\) with \(\beta\leq\gamma_0\) or some \(U_\beta\) with \(\beta>\gamma_0\); but by maximality of \(\gamma_0\), if \(\beta>\gamma_0\) then \(x\notin U_\beta\), so \(x\) is covered by some \(V_\beta\) with \(\beta\leq\gamma_0<\alpha\). In any case, \(x\) is covered by the family in \((\ast_\alpha)\), so \((\ast_\alpha)\) holds.

Now, assuming \(V_\beta\) has been defined for \(\beta<\alpha\), we define \(V_\alpha\). The set

\[C_\alpha=X\setminus\Bigl(\bigcup_{\beta<\alpha}V_\beta\cup\bigcup_{\beta>\alpha}U_\beta\Bigr)\]

is closed. By \((\ast_\alpha)\), points outside this complement, that is, points of \(C_\alpha\), do not belong to any \(V_\beta\) with \(\beta<\alpha\) or any \(U_\beta\) with \(\beta>\alpha\), so they must belong to \(U_\alpha\). That is, \(C_\alpha\subseteq U_\alpha\). Since \(X\) is normal, for the closed set \(C_\alpha\) and the open set \(U_\alpha\) containing it, there exists an open set \(V_\alpha\) such that \(C_\alpha\subseteq V_\alpha\subseteq\cl(V_\alpha)\subseteq U_\alpha\). (§Urysohn’s Lemma and Tietze Extension Theorem, ⁋Lemma 1) Then since \(C_\alpha\subseteq V_\alpha\), the family \(\{V_\beta\mid\beta\leq\alpha\}\cup\{U_\beta\mid\beta>\alpha\}\) covers \(X\), continuing the invariant to the next stage.

Finally, we show that the \((V_\alpha)_{\alpha\in J}\) obtained in this way covers \(X\). Since the \(U_\gamma\) containing a point \(x\) are only finitely many, let \(\gamma_0\) be the largest index among them; then by \((\ast_{\gamma_0})\) and the maximality of \(\gamma_0\), as before, \(x\) is covered by some \(V_\beta\) with \(\beta\leq\gamma_0\). Therefore \((V_\alpha)_{\alpha\in J}\) is an open cover of \(X\) satisfying \(\cl(V_\alpha)\subseteq U_\alpha\) for each \(\alpha\).

Note that point-finiteness was used crucially to maintain the covering property at the limit stages and the final stage of the transfinite induction. With only a well-ordering, there would be a risk that some point is not covered when infinitely many \(U_\beta\) are replaced by \(V_\beta\) all at once, but the fact that each point belongs to only finitely many pieces pins down the stage at which it is covered to a finite one. We are now ready to prove the main theorem.

Theorem 27 For any open cover \((U_\alpha)_{\alpha\in J}\) of a paracompact Hausdorff space \(X\), there exists a partition of unity subordinate to \((U_\alpha)\).

Proof

The proof consists of four steps. First, we refine the cover to a locally finite one with the same index set; then we shrink this twice to obtain a closed cover; next we construct bump functions for each piece using Urysohn’s lemma; and finally we normalize them.

(1) Precise locally finite refinement. Since \(X\) is paracompact, there exists a locally finite open refinement \((W_\beta)_{\beta\in K}\) of \((U_\alpha)\). By the definition of refinement, for each \(\beta\) we can choose \(\alpha(\beta)\in J\) such that \(W_\beta\subseteq U_{\alpha(\beta)}\). Now for each \(\alpha\in J\) define

\[V_\alpha=\bigcup\{W_\beta\mid\alpha(\beta)=\alpha\}\]

(if there is no such \(\beta\), set \(V_\alpha=\emptyset\)). Then \(V_\alpha\) is an open set with \(V_\alpha\subseteq U_\alpha\), and since the \(W_\beta\) cover \(X\), \((V_\alpha)_{\alpha\in J}\) also covers \(X\). Moreover, if a neighborhood of a point \(x\) meets only finitely many \(W_\beta\), then it meets only finitely many \(V_\alpha\) (since a neighborhood meeting \(V_\alpha\) must meet some \(W_\beta\) with \(\alpha(\beta)=\alpha\)), so \((V_\alpha)_{\alpha\in J}\) is locally finite. Thus we have obtained a locally finite open cover with the same index set \(J\) satisfying \(V_\alpha\subseteq U_\alpha\).

(2) Two shrinkings. \(X\) is normal by Theorem 23, and the locally finite \((V_\alpha)\) is point-finite. Apply Lemma 26 to \((V_\alpha)\) to obtain an open cover \((P_\alpha)_{\alpha\in J}\) with \(\cl(P_\alpha)\subseteq V_\alpha\) for all \(\alpha\). Since \(P_\alpha\subseteq V_\alpha\), \((P_\alpha)\) is also point-finite, and applying Lemma 26 again to \((P_\alpha)\) yields an open cover \((Q_\alpha)_{\alpha\in J}\) with \(\cl(Q_\alpha)\subseteq P_\alpha\) for all \(\alpha\). In summary, both covers \((P_\alpha)\) and \((Q_\alpha)\) cover \(X\) and satisfy

\[\cl(Q_\alpha)\subseteq P_\alpha\subseteq\cl(P_\alpha)\subseteq V_\alpha\subseteq U_\alpha\]

for all \(\alpha\).

(3) Bump functions. For each \(\alpha\), the sets \(\cl(Q_\alpha)\) and \(X\setminus P_\alpha\) are two disjoint closed sets. Since \(X\) is normal, by Urysohn’s lemma there exists a continuous function \(\psi_\alpha:X\rightarrow[0,1]\) taking the value \(1\) on \(\cl(Q_\alpha)\) and \(0\) on \(X\setminus P_\alpha\). (§Urysohn’s Lemma and Tietze Extension Theorem, ⁋Theorem 2) Then \(\{x\mid\psi_\alpha(x)\neq 0\}\subseteq P_\alpha\), so

\[\supp\psi_\alpha=\cl(\{x\mid\psi_\alpha(x)\neq 0\})\subseteq\cl(P_\alpha)\subseteq V_\alpha\subseteq U_\alpha\]

In particular, \((\supp\psi_\alpha)_{\alpha\in J}\) is locally finite since \(\supp\psi_\alpha\subseteq V_\alpha\) and \((V_\alpha)\) is locally finite.

(4) Normalization. Since the family \((\psi_\alpha)\) has locally finite support, for each point \(x\) there is a neighborhood where only finitely many \(\psi_\alpha\) are non-zero. On that neighborhood the sum \(\psi=\sum_{\alpha\in J}\psi_\alpha\) is a finite sum, and a finite sum of continuous functions is continuous, so \(\psi\) is continuous on that neighborhood. Since continuity is a local property, \(\psi:X\rightarrow\mathbb{R}\) is a continuous function. Also, since \((Q_\alpha)\) covers \(X\), any \(x\) belongs to some \(Q_\alpha\), and then \(\psi_\alpha(x)=1\) on \(\cl(Q_\alpha)\), so \(\psi(x)\geq 1>0\). Therefore, defining

\[\phi_\alpha=\frac{\psi_\alpha}{\psi}\]

each \(\phi_\alpha:X\rightarrow[0,1]\) is a continuous function (since \(\psi\) is nowhere zero). Since \(\psi>0\), we have \(\{x\mid\phi_\alpha(x)\neq 0\}=\{x\mid\psi_\alpha(x)\neq 0\}\), and thus \(\supp\phi_\alpha=\supp\psi_\alpha\subseteq U_\alpha\) and \((\supp\phi_\alpha)\) is locally finite. Finally, for each \(x\),

\[\sum_{\alpha\in J}\phi_\alpha(x)=\frac{1}{\psi(x)}\sum_{\alpha\in J}\psi_\alpha(x)=\frac{\psi(x)}{\psi(x)}=1\]

Therefore \((\phi_\alpha)_{\alpha\in J}\) is a partition of unity subordinate to \((U_\alpha)\).

Remark 28 The converse of Theorem 27 also holds. Let a space \(X\) be given such that for any open cover \((U_i)_{i\in I}\) there exists a subordinate partition of unity \((\phi_i)_{i\in I}\). Then the open sets \(G_i=\{x\mid\phi_i(x)>0\}\) satisfy \(G_i\subseteq\supp\phi_i\subseteq U_i\), so they form an open refinement of \((U_i)\); since \((\supp\phi_i)\) is locally finite, \((G_i)\) is also locally finite; and from \(\sum_i\phi_i(x)=1\), for each \(x\) there exists \(i\) with \(\phi_i(x)>0\), so \((G_i)\) covers \(X\). Therefore \(X\) is paracompact. Together with the Hausdorff condition, this yields the fact that a topological space is paracompact Hausdorff if and only if every open cover admits a subordinate partition of unity.

Globalizing Local Constructions

Once a partition of unity is available, the standard procedure for patching locally defined data into a global object opens up. In this section we illustrate the principle with the case of continuous functions.

Example 29 Let an open cover \((U_\alpha)_{\alpha\in J}\) of a paracompact Hausdorff space \(X\) be given, and let continuous functions \(f_\alpha:U_\alpha\rightarrow\mathbb{R}\) defined only on each \(U_\alpha\) be given. By Theorem 27, choose a partition of unity \((\phi_\alpha)_{\alpha\in J}\) subordinate to \((U_\alpha)\). For each \(\alpha\), the product \(\phi_\alpha f_\alpha\) is continuous on \(U_\alpha\); since \(\supp\phi_\alpha\subseteq U_\alpha\) is a closed set of \(X\), we can extend this product continuously to all of \(X\) by setting it to \(0\) outside \(U_\alpha\). Writing this extension again as \(\phi_\alpha f_\alpha\), since the family \((\supp\phi_\alpha)\) is locally finite,

\[f=\sum_{\alpha\in J}\phi_\alpha f_\alpha\]

reduces to a finite sum in a neighborhood of each point and defines a continuous function on all of \(X\). Here \(f\) is the global function obtained by averaging the local data \((f_\alpha)\) with the weights of the partition of unity. Conversely, given any continuous function \(g\) on \(X\), we have \(g=\sum_\alpha\phi_\alpha g\), so \(g\) is decomposed into pieces \(\phi_\alpha g\) on \(U_\alpha\).

The stage where this construction is most essentially used is topological manifolds. In this section we define topological manifolds and verify that they always admit the partitions of unity from the previous section. The model for a topological manifold is the familiar Euclidean space \(\mathbb{R}^m\), and a topological manifold is a space that locally resembles this model.

Topological Manifolds

Definition 30 A topological space \(M\) is locally Euclidean of dimension \(m\) if for every \(x\in M\) there exists an open neighborhood \(U\) of \(x\) such that \(U\) is homeomorphic to an open subset of \(\mathbb{R}^m\).

The locally Euclidean condition captures the local essence of a topological manifold, but by itself it is too weak; we add two global conditions.

Definition 31 A space that is second countable, Hausdorff, and locally Euclidean of dimension \(m\) is called a topological manifold of dimension \(m\).

The reasons for requiring Hausdorff and second countability are revealed in the following theorem. This theorem tells us that in a locally Euclidean space, second countability is essentially the same condition as paracompactness, thereby confirming that a topological manifold sits exactly on the stage of partitions of unity prepared in the previous section.

Theorem 32 For a Hausdorff and locally Euclidean topological space \(M\), the following are equivalent: \(M\) is second countable; and \(M\) is paracompact and has countably many connected components. (§Connected Spaces, ⁋Definition 7)

Proof

We only sketch the key idea and follow [Lee] for the details. From the locally Euclidean condition, each point has a neighborhood homeomorphic to an open subset of \(\mathbb{R}^m\), so the space behaves like a locally compact space at each point. If it is second countable, we can generalize the exhaustion argument of Example 20 to cover \(M\) by countably many relatively compact open sets, and from this we obtain a locally finite refinement in the same manner as in Proposition 19, yielding that \(M\) is paracompact and has countably many components. Conversely, if \(M\) is paracompact and has countably many components, then each component is Lindelöf and hence has a countable base, so \(M\) is second countable.

As a direct consequence of this theorem, any topological manifold \(M\) is a paracompact Hausdorff space, and therefore by Theorem 27 there exists a partition of unity subordinate to any coordinate cover of \(M\). Thanks to this fact, objects defined in the language of Euclidean space on each coordinate patch can be patched together to the whole manifold, and this is the reason why partitions of unity are an essential tool in manifold theory and bundle theory.

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References

[Mun] J. R. Munkres, Topology, 2nd ed., Prentice Hall, 2000.

[Wil] S. Willard, General Topology, Addison-Wesley, 1970.

[Lee] J. M. Lee, Introduction to Topological Manifolds, 2nd ed., Springer, 2011.