In some previous posts we discuss examples surrounding the Michael line showing that the product of a paracompact space and a complete metric space needs not be normal (see “Michael Line Basics”) and that the product of a Lindelof space and a separable metric space need not be normal (see “Bernstein Sets and the Michael Line”). These examples are classic counterexamples demonstrating that both paracompactness and Lindelofness are not preserved by taking twofactor cartesian products even when one of the factors is nice (complete metric space in the first example and separable metric space in the second example). We now show some positive results. Of course, these results require additional conditions on one or both of the factors. We prove the following results.
Result 1

If is paracompact and is compact, then is paracompact.
Result 2

If is paracompact and is compact, then is paracompact.
Result 3

If is paracompact and perfectly normal and is metrizable, then is paracompact and perfectly normal.
Result 4

If is hereditarily Lindelof and is a separable metric space, then is hereditarily Lindelof.
With Results 1 and 2, compact spaces and compact spaces can be called productively paracompact since the product of each of these spaces with any paracompact space is paracompact. We prove Result 1 and Result 2 below.
Result 3 and Result 4 are proved in another post Cartesian Products of Two Paracompact Spaces – Continued.
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Paracompact Spaces
First, recall some definitions. All spaces are at least regular (to us regular implies Hausdorff). Let be a space. A collection of subsets of is said to be a cover of if (in words every point of the space belongs to one set in the collection). Furthermore, is an open cover of is it is a cover of consisting of open subsets of .
Let and be covers of the space . The cover is said to be a refinement of ( is said to refine ) if for every , there is some such that . The cover is said to be an open refinement of if refines and is an open cover.
A collection of subsets of is said to be a locally finite collection if for each point , there is a nonempty open subset of such that and has nonempty intersection with at most finitely many sets in . An open cover of is said to have a locally finite open refinement if there exists an open cover of such that refines and is a locally finite collection. We have the following definition.
Definition

The space is said to be paracompact if every open cover of has a locally finite open refinement.
A collection of subsets of the space is said to be a locally finite collection if such that each is a locally finite collection of subsets of . Consider the property that every open cover of has a locally finite open refinement. This on the surface is a stronger property than paracompactness. However, Theorem 1 below shows that it is actually equivalent to paracompactness. The proof of Theorem 1 can be found in [1] (Theorem 5.1.11 in page 302) or in [2] (Theorem 20.7 in page 146).
Theorem 1
Let be a regular space. Then is paracompact if and only if every open cover of has a locally finite open refinement.
Theorem 2 below is another characterization of paracompactness that is useful. For a proof of Theorem 2, see “Finite and Countable Products of the Michael Line”.
Theorem 2
Let be a regular space. Then is paracompact if and only if the following holds:

For each open cover of , there exists a locally finite open cover such that for each .
Theorem 3 below shows that paracompactness is hereditary with respect to subsets.
Theorem 3
Every subset of a paracompact space is paracompact.
Proof of Theorem 3
Let be paracompact. Let such that where each is a closed subset of . Let be an open cover of . For each , let be open in such that .
For each , let be the set of all such that . Let be a locally finite refinement of . Let be the following:
It is clear that each is a locally finite collection of open set in covering . All the together form a refinement of . Thus is a locally finite open refinement of . By Theorem 1, the set is paracompact.
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Result 1
Result 1 is the statement that:

If is paracompact and is compact, then is paracompact.
To prove Result 1, we use the Tube lemma (for a proof, see “The Tube Lemma”).
The Tube Lemma
Let be any space and be compact. For each and for each open set such that , there is an open set such that .
Proof of Result 1
Let be an open cover of . For each , choose a finite such that is a cover of . By the Tube Lemma, for each , there is an open set such that . Since is paracompact, by Theorem 2, let be a locally finite open refinement of such that for each .
Let . We claim that is a locally finite open refinement of . First, this is an open cover of . To see this, let . Then for some . Furthermore, and . Thus, for some . Secondly, it is clear that is a refinement of the original cover .
It remains to show that is locally finite. To see this, let . Then there is an open in such that and can meets only finitely many . Then can meet only finitely many sets in .
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Result 2
Result 2 is the statement that:

If is paracompact and is compact, then is paracompact.
Proof of Result 2
Note that the compact space is Lindelof. Since regular Lindelof are normal, is normal and is thus completely regular. So we can embed into a compact space . For example, we can let , which is the StoneCech compactification of (see “Embedding Completely Regular Spaces into a Cube”). For our purpose here, any compact space containing will do. By Result 1, is paracompact. Note that can be regarded as a subspace of .
Let where each is compact in . Note that and each is a closed subset of . Thus the product is an subset of . According to Theorem 3, subsets of any paracompact space is paracompact space. Thus is paracompact.
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Reference
 Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
 Willard, S., General Topology, AddisonWesley Publishing Company, 1970.
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