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locally compact and sigma-compact spaces are paracompact

Context

Topology

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

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

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Universal constructions

Extra stuff, structure, properties

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Theorems

Analysis Theorems

topological homotopy theory

Contents

Statement

Lemma

Let XX be a topological space which is

  1. locally compact

    in the sense that every open neighbourhood contains the topological closure of a smaller open neighbourhood;

  2. sigma-compact.

Then there exists a countable open cover {U iX} i\{U_i \subset X\}_{i \in \mathbb{N}} of XX such that for each iIi \in I

  1. the topological closure Cl(U i)Cl(U_i) is a compact subspace

  2. Cl(U i)U i+1Cl(U_i) \subset U_{i +1}.

Proof

By sigma-compactness of XX there exists a countable cover {K iX} i\{K_i \subset X\}_{i \in \mathbb{N}} of compact subspaces. We use these to construct the required cover by induction.

For i=0i = 0 set

U 0. U_0 \coloneqq \emptyset \,.

Then assume that for nn \in \mathbb{N} we have constructed a set {U iX} i{1,,n}\{U_i \subset X\}_{i \in \{1, \cdots, n\}} with the required properties.

In particular this implies that

Q nCl(U n)K n1X Q_n \coloneqq Cl(U_n) \cup K_{n-1} \;\subset X

is a compact subspace. We now construct an open neighbourhood U n+1U_{n+1} of this union as follows:

Let {U xX} xQ n\{U_x \subset X\}_{x \in Q_n} be a set of open neighbourhood around each of the points in Q nQ_n. By local compactness of XX, for each xx there is a smaller open neighbourhood V xV_x with

{x}V xCl(V x)compactU x. \{x\} \subset V_x \subset \underset{\text{compact}}{\Cl(V_x)} \subset U_x \,.

So {V xX} xQ n\{V_x \subset X\}_{x \in Q_n} is still an open cover of Q nQ_n. By compactness of Q nQ_n, there exists a finite set J nQ nJ_n \subset Q_n such that {V xX} xJ n\{V_x \subset X\}_{x \in J_n} is a finite open cover. The union

U n+1xJV x U_{n + 1} \coloneqq \underset{x \in J}{\cup} V_x

is an open neighbourhood of Q nQ_n, hence in particular of Cl(U n)Cl(U_n). Moreover, since finite unions of compact spaces are compact (this prop.) and since the closure of a finite union is the union of the closures (this prop.), the closure of U n+1U_{n+1} is compact:

Cl(U n+1) =Cl(xJ nV x) =xJ nCl(V x)compact. \begin{aligned} Cl(U_{n+1}) &= Cl\left( \underset{x\in J_n}{\cup} V_x \right) \\ & = \underset{x \in J_n}{\cup} \underset{\text{compact}}{Cl( V_x )} \end{aligned} \,.

This produces by induction a set {U iX} i\{U_i \subset X\}_{i \in \mathbb{N}} with Cl(U i)Cl(U_i) compact and Cl(U i)U i+1Cl(U_i) \subset U_{i+1} for all ii \in \mathbb{N}. It remains to see that this is a cover. This follows since by construction each U i+1U_{i+1} is an open neighbourhood not just of Cl(U i)Cl(U_{i}) but in fact of Q iQ_i, hence in particular of K iK_i, and since the K iK_i form a cover:

iU iiK i=X. \underset{i \in \mathbb{N}}{\cup} U_i \supset \underset{i \in \mathbb{N}}{\cup} K_i = X \,.
Proposition

(locally compact and sigma-compact spaces are paracompact)

Let XX be a topological space which is

  1. locally compact;

  2. sigma-compact.

Then XX is also paracompact.

Proof

Let {U iX} iI\{U_i \subset X\}_{i \in I} be an open cover of XX. We need to show that this has a refinement by a locally finite cover.

By lemma 1 there exists a countable open cover {V nX} n\{V_n \subset X\}_{n \in \mathbb{N}} of XX such that for all nn \in \mathbb{N}

  1. Cl(V n)Cl(V_n) is compact;

  2. Cl(V n)V n+1Cl(V_n) \subset V_{n+1}.

Notice that the complement Cl(V n+1)V nCl(V_{n+1}) \setminus V_n is compact, since Cl(V n+1)Cl(V_{n+1}) is compact and V nV_n is open (by this lemma).

By this compactness, the cover {U iX} iI\{U_i \subset X\}_{i \in I} regarded as a cover of the subspace Cl(V n+1)V nCl(V_{n+1})\setminus V_n has a finite subcover {U iX} iJ n\{U_i \subset X\}_{i \in J_n} indexed by a finite set J nIJ_n \subset I, for each nn \in \mathbb{N}.

We consider the sets of intersections

𝒰 n{U i(V n+2Cl(V n1))}. \mathcal{U}_n \coloneqq \{ U_i \cap ( V_{n+2} \setminus Cl(V_{n-1}) ) \} \,.

Since V n+2Cl(V n1)V_{n+2} \setminus Cl(V_{n-1}) is open, and since Cl(V n+1)V n+2Cl(V_{n+1}) \subset V_{n+2} by construction, this is still an open cover of Cl(V n+1)V nCl(V_{n+1})\setminus V_n. We claim now that

𝒰n𝒰 n \mathcal{U} \coloneqq \underset{n\in \mathbb{N}}{\cup} \mathcal{U}_n

is a locally finite refinement of the original cover, as required:

  1. 𝒰\mathcal{U} is a refinement, since by construction each element in 𝒰 n\mathcal{U}_n is contained in one of the U iU_i;

  2. 𝒰\mathcal{U} is still a covering because by construction it covers Cl(V n+1)V nCl(V_{n+1}) \setminus V_n for all nn \in \mathbb{N}, and since by the nested nature of the cover {V nX} n\{V_n \subset X\}_{n \in \mathbb{N}} also {Cl(V n+1)V n} n\{Cl(V_{n+1}) \setminus V_n\}_{n \in \mathbb{N}} is a cover of XX.

  3. 𝒰\mathcal{U} is locally finite because each point xXx \in X has an open neighbourhood of the form V n+2Cl(V n1)V_{n+2} \setminus Cl(V_{n-1}) (since these also form an open cover, by the nestedness) and since by construction this has trivial intersection with 𝒰 n+3\mathcal{U}_{\geq n+3} and since all 𝒰 n\mathcal{U}_n are finite, so that also k<n+3𝒰 k\underset{k \lt n+3}{\cup} \mathcal{U}_k is finite.

References

Revised on May 23, 2017 06:01:17 by Urs Schreiber (131.220.184.222)