Compact times paracompact implies paracompact

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This article states and proves a result of the following form: the product of two topological spaces, the first satisfying the property Compact space (?) and the second satisfying the property Paracompact space (?), is a topological space satisfying the property Paracompact space (?).
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Statement

Let X be a compact space and Y a paracompact space. Then X \times Y, the Cartesian product endowed with the product topology, is paracompact.

Related facts

Other results using the same proof technique:

Facts used

  1. Tube lemma: Suppose X is a compact space and Y is a topological space. Then, given any open subset U of X \times Y containing X \times \{ y \} for some y \in Y, there exists an open subset V of Y such that y \in V and X \times V \subseteq U.

Proof

Given: A compact space X, a paracompact space Y. \{ U_i \}_{i \in I} form an open cover of X \times Y.

To prove: There exists a locally finite open refinement of the U_is, i.e., an open cover \{ Q_k \}_{k \in K} of X \times Y such that:

  • It is locally finite: For any point (x,y) \in X \times Y, there exists an open set R containing (x,y) that intersects only finitely many of the Q_ks.
  • It refines \{ U_i \}_{i \in I}: Every Q_k is contained in one of the U_is.

Proof: Note that in the proof below, Step (4) has three aspects (locally finite, cover, and refinement) and different later steps use different aspects of Step (4), with the specific aspect used indicated parenthetically in the Previous steps used column.

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation Commentary
1 For any point y \in Y, there is a finite collection of U_i that cover X \times \{ y \}. X is compact Since X is compact, the subspace X \times \{ y \} of X \times Y is also compact, so the cover by the open subsets U_i has a finite subcover. Begin focusing on a slice.
2 For any point y \in Y, let W_y be the union of the finite collection of open subsets U_i as obtained in Step (1). There exists an open subset V_y of Y such that y \in V_y and X \times V_y \subseteq W_y. Fact (1) X is compact Step (1) Follows from Fact (1), setting the U of Fact (1) to be W_y. Use compactness to get a tube around the slice.
3 The open subsets V_y, y \in Y obtained in Step (2) form an open cover of Y. Step (2) By Step (2), y \in V_y. Since \bigcup_{y \in Y} \{ y \} = Y, and y \in V_y \subseteq Y, we get \bigcup_{y \in Y} V_y = Y. Project down to the paracompact space.
4 There exists a locally finite open refinement \{ P_j \}_{j \in J} of the V_y in Y. Y is paracompact Step (3) Step-given combination direct. Use paracompactness to get the locally finite open refinement, in the projected-down setting.
5 For each P_j, X \times P_j is a union of finitely many intersections (X \times P_j) \cap U_i, all of which are open subsets. Steps (1), (2), (4) (refinement aspect) [SHOW MORE] Go back to the big space, reversing the projection step.
6 The open subsets of the form (X \times P_j) \cap U_i of Step (5) form an open cover of X \times Y that refines the U_is (note that not every combination of P_j and U_i is included -- only the finitely many U_is needed as in Step (5)). We will index this open cover by indexing set K \subseteq I \times J, and call it \{ Q_k \}_{k \in K}, where Q_k = (X \times P_j) \cap U_i. In particular, if k = (i,j), then Q_k \subseteq X \times P_j and Q_k \subseteq U_i, and for any j, there are finitely many k \in K with the second coordinate of k equal to j. Steps (4) (cover aspect), (5) \{ P_j \}_{j \in J} cover Y, so \{ X \times P_j \}_{j \in J} cover X \times Y. By Step (5), X \times P_j is the union of finitely many (X \times P_j) \cap U_i, so the latter also form an open cover of X \times Y. Wrap things up.
7 The open cover \{ Q_k \}_{k \in K} of Step (6) is a locally finite open cover. In other words, for any (x, y) \in X \times Y, there is an open subset R \ni (x,y) such that R intersects only finite many Q_ks. Steps (4) (locally finite aspect), (6) [SHOW MORE] Wrap things up.
8 The open cover \{ Q_k \}_{k \in K} is as desired Steps (6), (7) Combine the two steps to get what we wanted to prove. --
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