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:

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
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.
2 For any point y \in Y, let W_y be the union of this 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.
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, hence X \times \{ y \} \subseteq X \times 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.
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.
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) Since P_js refine V_ys (Step (4)), there exists y \in Y such that P_j \subseteq V_y. In turn, by the definition of V_y (Step (2)), we have X \times V_y \subseteq W_y, which in turn is a union of finitely many U_is (Step (1)). Thus, X \times P_j is contained in a union of finitely many U_is, and hence, is the union of its intersection with those U_is. Since U_i are all open, the intersections (X \times P_j) \cap U_i are all open.
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 P_j, and for any j, there are finitely many k \in K with k = (i,j). Steps (4), (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 the (X \times P_j) \cap U_i, so the latter also form an open cover of X \times Y.
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), (6) Since \{ P_j \}_{j \in J} form a locally finite open cover of Y (Step (4)), there exists an open subset S of Y such that S contains y and S intersects only finitely many of the P_js. Set R  = X \times S, so R is open in X \times Y. R therefore intersects only finitely many of the X \times P_js. For any Q_k, with k = (i,j), we have Q_k \subseteq P_j by construction (Step (6)), so if Q_k intersects R so does X \times P_j. Thus, the Q_ks that intersect R must correspond to the finitely many js for which R intersects X \times P_j. Since there are finitely many ks for each j, this gives that there are finitely many Q_ks intersecting R.
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.