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paracompact Hausdorff spaces equivalently admit subordinate partitions of unity

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Topology

topology (point-set topology, point-free topology)

see also differential topology, algebraic topology, functional analysis and topological homotopy theory

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Contents

Statement

Theorem

Assuming the axiom of choice then:

Let (X,τ)(X,\tau) be a topological space that is T 1T_1 (meaning singletons are closed subsets; see separation property). Then the following are equivalent:

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

  2. Every open cover of (X,τ)(X,\tau) admits a subordinate partition of unity (def. 4).

We give the proof below.

Background

Definition

(locally finite cover)

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

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

Definition

(refinement of open covers)

Let (X,τ)(X,\tau) be a topological space, and let {U iX} iI\{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 jX} jJ\{V_j \subset X\}_{j \in J} which is still an open cover in itself and such that for each jJj \in J there exists an iIi \in I with V jU iV_j \subset U_i.

Definition

(paracompact topological space)

A topological space (X,τ)(X,\tau) is called paracompact if every open cover of XX has a refinement (def. 2) by a locally finite open cover (def. 1).

Definition

(partition of unity)

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

such that with

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

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

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

  2. {Supp(f i)X} iI\left\{ Supp(f_i) \subset X \right\}_{i \in I} is a locally finite cover (def. 1);

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

Remark

Due to the second clause in def. 4, 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,τ)(X,\tau) is a paracompact topological space, then for every open cover {U iX} iI\{U_i \subset X\}_{i \in I} there is a subordinate partition of unity (def. 4).

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,τ)(X,\tau) be a topological space, let {U iX} iI\{U_i \subset X\}_{i \in I} be an open cover, and let (ϕ:JI),{V jX} jJ)(\phi \colon J \to I), \{V_j \subset X\}_{j \in J}), be a refinement to a locally finite cover.

Then {W iX} iI\left\{ W_i \subset X \right\}_{i \in I} with

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

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

Proof

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

Hence consider xXx \in X. By the assumption that {V jX} jJ\{V_j \subset X\}_{j \in J} is locally finite, it follows that there exists an open neighbourhood U x{x}U_x \supset \{x\} and a finitee subset KJK \subset J such that

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

Hence by construction

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

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

Lemma

(shrinking lemma)

Let XX be a topological space which is normal and let {U iX} iI\{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 iX} iI\{V_i \subset X\}_{i \in I} such that the topological closure Cl(V i)Cl(V_i) of its elements is contained in the original patches:

iI(V iCl(V i)U i). \underset{i \in I}{\forall} \left( V_i \subset Cl(V_i) \subset U_i \right) \,.
Proof

of prop. 1

By paracmpactness of XX, for every open cover there exists a locally finite refinement {U iX} iI\{U_i \subset X\}_{i \in I}, and by lemma 1 we may assume that this has same index set. It is now sufficient to show that this locally finite cover {U iX} iI\{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 2 to the given locally finite open cover {U iX} i\{U_i \subset X\}_i, to obtain a smaller locally finite open cover {V iX} iI\{V_i \subset X\}_{i \in I}, and then apply the lemma once more to that result to get a yet small open cover {W iX} iI\{W_i \subset X\}_{i \in I}, so that now

iI(W iCl(W i)V iCl(V i)U i). \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 iIi \in I we have two disjoint closed subsets, namely the topological closure Cl(W i)Cl(W_i) and the complement X\V iX \backslash V_i

Cl(W i)X\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:X[0,1] h_i \;\colon\; X \longrightarrow [0,1]

with the property that

h i(Cl(W i))={1},AAAh i(X\V i)={0}. 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])V ih_i^{-1}((0,1]) \subset V_i and hence that

Supp(h i)=Cl(h i 1((0,1]))Cl(V i)U i. Supp(h_i) = Cl(h_i^{-1}((0,1])) \subset Cl(V_i) \subset U_i \,.

By construction, the set of function {h i} iI\{h_i\}_{i \in I} already satisfies conditions 1) and 2) of the three conditions on a partition of unity subordinate to {U iX} iI\{U_i \subset X\}_{i \in I} from def. 4. It just remains to normalize these functions so that they indeed sum to unity. To that end, consider the continuous function

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

defined on xXx \in X by

h(x)iIh i(x). 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

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

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

Hence it makes sense to define

f ih i/h. f_i \;\coloneqq\; h_i/h \,.

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

{f i} iI \left\{ f_i \right\}_{i \in I}

is a partition of unity as required.

Proof

of theorem 1

In one direction, assume that every open cover {U iX} iI\{U_i \subset X\}_{i \in I} admits a subordinate partition of unity {f i} iI\{f_i\}_{i \in I}. Then by definition (def. 4) {Int(Supp(f) i)X} iI\{ Int(Supp(f)_i) \subset X\}_{i \in I} is a locally finite open cover refining the original one, so XX is paracompact. Moreover, since XX is T 1T_1, for any distinct points x,yXx, y \in X there is a cover consisting of the open sets ¬{x},¬{y}\neg \{x\}, \neg \{y\} (the complements of the singletons). If f x,f yf_x, f_y is a subordinate partition of unity (with f xf_x supported on ¬{x}\neg \{x\}, and similarly for f yf_y), then f x(x)=0f_x(x) = 0 and f y(y)=0f_y(y) = 0. Then from f x+f y1f_x + f_y \equiv 1 we derive f x(y)=1f_x(y) = 1, whereupon {zX:f x(z)<1/2}\{z \in X: f_x(z) \lt 1/2\} and {zX:f x(z)>1/2}\{z \in X: f_x(z) \gt 1/2\} are disjoint open neighborhoods containing xx and yy respectively. Hence XX is Hausdorff.

The other direction is the statement of prop. 1.

Revised on June 6, 2017 05:08:49 by Urs Schreiber (88.77.226.246)