nLab paracompact Hausdorff spaces equivalently admit subordinate partitions of unity

Statement

Theorem

Assuming the axiom of choice then:

Let $(X,\tau)$ be a topological space that is $T_1$ (meaning singletons are closed subsets; see separation property). Then the following are equivalent:

1. $(X,\tau)$ is paracompact and Hausdorff.

2. Every open cover of $(X,\tau)$ admits a subordinate partition of unity (def. ).

We give the proof below.

Background

Definition

(locally finite cover)

Let $(X,\tau)$ be a topological space.

An open cover $\{U_i \subset X\}_{i \in I}$ of $X$ is called locally finite if for all point $x \in X$, there exists a neighbourhood $U_x \supset \{x\}$ such that it intersects only finitely many elements of the cover, hence such that $U_x \cap U_i \neq \emptyset$ for only a finite number of $i \in I$.

Definition

(refinement of open covers)

Let $(X,\tau)$ be a topological space, and let $\{U_i \subset X\}_{i \in I}$ be a open cover.

Then a refinement of this open cover is a set of open subsets $\{V_j \subset X\}_{j \in J}$ which is still an open cover in itself and such that for each $j \in J$ there exists an $i \in I$ with $V_j \subset U_i$.

Definition

(paracompact topological space)

A topological space $(X,\tau)$ is called paracompact if every open cover of $X$ has a refinement (def. ) by a locally finite open cover (def. ).

Definition

(partition of unity)

Let $(X,\tau)$ be a topological space, and let $\{U_i \subset X\}_{i \in I}$ be an open cover. Then a partition of unity subordinate to the cover is

• a set $\{f_i\}_{i \in I}$ of continuous functions

  $$f_i \;\colon\; X \longrightarrow [0,1]$$

  (where $[0,1] \subset \mathbb{R}$ is equipped with the subspace topology of the real numbers $\mathbb{R}$ regarded as the 1-dimensional Euclidean space equipped with its metric topology);

such that with

$$Supp(f_i) \coloneqq Cl\left( f_i^{-1}( (0,1] ) \right)$$

denoting the support of $f_i$ (the topological closure of the subset of points on which it does not vanish) then

1. $\underset{i \in I}{\forall} \left( Supp(f_i) \subset U_i \right)$;

2. $\left\{ Supp(f_i) \subset X \right\}_{i \in I}$ is a locally finite cover (def. );

3. $\underset{x \in X}{\forall} \left( \underset{i \in I}{\sum} f_i(x) = 1 \right)$.

Remark

Due to the second clause in def. , the sum in the third clause involves only a finite number of elements not equal to zero, and therefore is well defined.

Proof

The non-trivial direction to be shown is:

Proposition

If $(X,\tau)$ is a paracompact topological space, then for every open cover $\{U_i \subset X\}_{i \in I}$ there is a subordinate partition of unity (def. ).

We give the proof below.

The following says that if there exists a locally finite refinement of a cover, then in fact there exists one with the same index set as the original cover.

Lemma

Let $(X,\tau)$ be a topological space, let $\{U_i \subset X\}_{i \in I}$ be an open cover, and let $(\phi \colon J \to I), \{V_j \subset X\}_{j \in J})$, be a refinement to a locally finite cover.

Then $\left\{ W_i \subset X \right\}_{i \in I}$ with

$$W_i \;\coloneqq\; \left\{ \underset{j \in \phi^{-1}(\{i\})}{\cup} V_j \right\}$$

is still a refinement of $\{U_i \subset X\}_{i \in I}$ to a locally finite cover.

Proof

It is clear by construction that $W_i \subset U_i$, hence that we have a refinement. We need to show local finiteness.

Hence consider $x \in X$. By the assumption that $\{V_j \subset X\}_{j \in J}$ is locally finite, it follows that there exists an open neighbourhood $U_x \supset \{x\}$ and a finite subset $K \subset J$ such that

$$\underset{j \in J\backslash K}{\forall} \left( U_x \cap V_j = \emptyset \right) \,.$$

Hence by construction

$$\underset{i \in I\backslash \phi(K)}{\forall} \left( U_x \cap W_i = \emptyset \right) \,.$$

Since the image $\phi(K) \subset I$ is still a finite set, this shows that $\{W_i \subset X\}_{i \in I}$ is locally finite.

Lemma

(shrinking lemma)

Let $X$ be a topological space which is normal and let $\{U_i \subset X\}_{i \in I}$ be a locally finite open cover.

Assuming the axiom of choice then:

There exists another open cover $\{V_i \subset X\}_{i \in I}$ such that the topological closure $Cl(V_i)$ of its elements is contained in the original patches:

$$\underset{i \in I}{\forall} \left( V_i \subset Cl(V_i) \subset U_i \right) \,.$$

Proof

of prop.

By paracompactness of $X$, for every open cover there exists a locally finite refinement $\{U_i \subset X\}_{i \in I}$, and by lemma we may assume that this has same index set. It is now sufficient to show that this locally finite cover $\{U_i \subset X\}_{i \in I}$ admits a subordinate partition of unity, since this will then also be subordinate to the original cover.

Since paracompact Hausdorff spaces are normal we may apply lemma to the given locally finite open cover $\{U_i \subset X\}_i$, to obtain a smaller locally finite open cover $\{V_i \subset X\}_{i \in I}$, and then apply the lemma once more to that result to get a yet smaller open cover $\{W_i \subset X\}_{i \in I}$, so that now

$$\underset{i \in I}{\forall} \left( W_i \subset Cl(W_i) \subset V_i \subset Cl(V_i) \subset U_i \right) \,.$$

It follows that for each $i \in I$ we have two disjoint closed subsets, namely the topological closure $Cl(W_i)$ and the complement $X \backslash V_i$

$$Cl(W_i) \cap X\backslash V_i = \emptyset \,.$$

Now since paracompact Hausdorff spaces are normal, Urysohn's lemma says that there exist continuous functions

$$h_i \;\colon\; X \longrightarrow [0,1]$$

with the property that

$$h_i( Cl(W_i) ) = \{1\} \,, \phantom{AAA} h_i( X \backslash V_i ) = \{0\} \,.$$

This means in particular that $h_i^{-1}((0,1]) \subset V_i$ and hence that

$$Supp(h_i) = Cl(h_i^{-1}((0,1])) \subset Cl(V_i) \subset U_i \,.$$

By construction, the set of function $\{h_i\}_{i \in I}$ already satisfies conditions 1) and 2) of the three conditions on a partition of unity subordinate to $\{U_i \subset X\}_{i \in I}$ from def. . It just remains to normalize these functions so that they indeed sum to unity. To that end, consider the continuous function

$$h \;\colon\; X \longrightarrow [0,1]$$

defined on $x \in X$ by

$$h(x) \coloneqq \underset{i \in I}{\sum} h_i(x) \,.$$

Notice that the sum on the right has only a finite number of non-zero summands, due to the local finiteness of the cover, so that this is well-defined.

Moreover, notice that

$$\underset{x \in X}{\forall} \left( h(x) \neq 0 \right)$$

because $\{Cl(W_i) \subset X\}_{i \in I}$ is a cover so that there is $i_x \in I$ with $x \in Cl(W_{i_x})$, and since $h_i(Cl(W_{i_x})) = \{1\}$, by the above.

Hence it makes sense to define

$$f_i \;\coloneqq\; h_i/h \,.$$

This is now manifestly such that $\underset{i \in I}{\sum} f_i = 1$, and so

$$\left\{ f_i \right\}_{i \in I}$$

is a partition of unity as required.

Proof

of theorem

In one direction, assume that every open cover $\{U_i \subset X\}_{i \in I}$ admits a subordinate partition of unity $\{f_i\}_{i \in I}$. Then by definition (def. ) $\{ Int(Supp(f)_i) \subset X\}_{i \in I}$ is a locally finite open cover refining the original one, so $X$ is paracompact. Moreover, since $X$ is $T_1$, for any distinct points $x, y \in X$ there is a cover consisting of the open sets $\neg \{x\}, \neg \{y\}$ (the complements of the singletons). If $f_x, f_y$ is a subordinate partition of unity (with $f_x$ supported on $\neg \{x\}$, and similarly for $f_y$), then $f_x(x) = 0$ and $f_y(y) = 0$. Then from $f_x + f_y \equiv 1$ we derive $f_x(y) = 1$, whereupon $\{z \in X: f_x(z) \lt 1/2\}$ and $\{z \in X: f_x(z) \gt 1/2\}$ are disjoint open neighborhoods containing $x$ and $y$ respectively. Hence $X$ is Hausdorff.

The other direction is the statement of prop. .

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