Consider the real line with a topology finer than the usual topology obtained by isolating each point in where is the set of all irrational numbers. The real line with this finer topology is called the Michael line and we use to denote this topological space. It is a classic result that is not normal (see “Michael Line Basics”). Even though the Michael line is paracompact (it is in fact hereditarily paracompact), is not perfectly normal. Result 3 below will imply that the Michael line cannot be perfectly normal. Otherwise would be paracompact (hence normal). Result 3 is the statement that if is paracompact and perfectly normal and Y is a metric space then is paracompact and perfectly normal. We also use this result to show that if is hereditarily Lindelof and is a separable metric space, then is hereditarily Lindelof (see Result 4 below).
This post is a continuation of the post “Cartesian Products of Two Paracompact Spaces”. In that post, four results are listed. They are:
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.
Result 1 and Result 2 are proved in the previous post “Cartesian Products of Two Paracompact Spaces”. Result 3 and Result 4 are proved in this post. All spaces are assumed to be regular.
___________________________________________________________________________________
Paracompact Spaces, Lindelof Spaces and Other Information
A paracompact space is one in which every open cover has a locally finite open refinement. The previous post “Cartesian Products of Two Paracompact Spaces” has a basic discussion on paracompact spaces. For the sake of completeness, we repeat here some of the results discussed in that post. A proof of Proposition 1 can be found in [1] (Theorem 5.1.11 in page 302) or in [2] (Theorem 20.7 in page 146).. For a proof of Proposition 2, see Theorem 3 in the previous post “Cartesian Products of Two Paracompact Spaces”. We provide a proof for Proposition 3.
Proposition 1
Let be a regular space. Then is paracompact if and only if every open cover of has a locally finite open refinement.
Proposition 2
Every subset of a paracompact space is paracompact.
Proposition 3
Any paracompact space with a dense Lindelof subspace is Lindelof.
Proof of Proposition 3
Let be a paracompact space. Let be a dense Lindelof subspace. Let be an open cover of . Since we are working with a regular space, let be an open cover of such that refines . Let be a locally finite open refinement of . Choose such that it is a cover of . Since , .
Since the sets come from a locally finite collection, they are closure preserving. Hence we have:
For each , choose some such that . Then is a countable subcollection of covering the space .
A space is said to be a perfectly normal if it is a normal space with the additional property that every closed subset is a set in the space (equivalently every open subset is an set). We need two basic results about hereditarily Lindelof spaces. A space is Lindelof if every open cover of that space has a countable subcover. A space is hereditarily Lindelof if every subspace of that space is Lindelof. Proposition 4 below, stated without proof, shows that to prove a space is hereditarily Lindelof, we only need to show that every open subspace is Lindelof.
Proposition 4
Let be a space. Then is hereditarily Lindelof if and only if every open subspace of is Lindelof.
Proposition 5
Let be a Lindelof space. Then is hereditarily Lindelof if and only if is perfectly normal.
Proof of Proposition 5
Suppose is hereditarily Lindelof. It is well known that regular Lindelof space is normal. Thus is normal. It remains to show that every open subset of is . Let be an nonempty open set. For each , let be open such that and (the space is assumed to be regular). By assumption, the open set is Lindelof. The open sets form an open cover of . Thus is the union of countably many .
Suppose is perfectly normal. To show that is hereditarily Lindelof, it suffices to show that every open subset of is Lindelof (by Proposition 4). Let be nonempty open. By assumption, where each is a closed set in . Since the Lindelof property is hereditary with respect to closed subsets, is Lindelof.
Another important piece of information that we need is the following metrization theorem. It shows that being a metrizable space is equivalent to have a base that is locally finite. In proving Result 3, we will assume that the metric factor has such a base. This is a classic metrization theorem (see [1] or [2] or any other standard topology text).
Theorem 6
Let be a space. Then is metrizable if and only if has a locally finite base.
___________________________________________________________________________________
Result 3
Result 3 is the statement that:

If is paracompact and perfectly normal and Y is a metric space then is paracompact and perfectly normal.
Result 3 follows from the following two lemmas.
Lemma 7
If the following two conditions hold:
 every open subset of is an set in ,
 is a metric space,
then every open subset of is an set in .
Proof of Lemma 7
Let be a open subset of . If , then is certainly the union of countably many closed sets. So assume . Let be a base for such that each is locally finite in (by Theorem 6, such a base exists since is metrizable).
Consider all nonempty such that we can choose nonempty open set with . Since is nonempty open, such pairs exist. Let be the collection of all nonempty for which there is a matching nonempty . For each , let . Of course, each is still locally finite.
Since every open subset of is an set in , for each , we can write as
where each is closed in .
For each and each , consider the following collection:
Each element of is a closed set in . Since is a locally finite collection in , is a locally finite collection in . Define . The set is a union of closed sets. In general, the union of closed sets needs not be closed. However, is still a closed set in since is a locally finite collection of closed sets. This is because a locally finite collection of sets is closure preserving. Note the following:
Finally, we have , which is the union of countably many closed sets.
Lemma 8
If is a paracompact space satisfying the following two conditions:
 every open subset of is an set in ,
 is a metric space,
then is paracompact.
Proof of Lemma 8
As in the proof of the above lemma, let be a base for such that each is locally finite in . Let be an open cover of . Assume that elements of are of the form where is open in and .
For each , consider the following two items:
To simplify matter, we only consider such that . Each is open in and hence by assumption an set in . Thus by Proposition 2, each is paracompact. Note that is an open cover of . Let be a locally finite open refinement of . Consider the following two items:

For each , let
We observe that is an open cover of and that refines . Furthermore each is a locally finite collection. The open cover we start with has a locally finite open refinement. Thus is paracompact.
___________________________________________________________________________________
Result 4
Result 4 is the statement that:

If is hereditarily Lindelof and is a separable metric space, then is hereditarily Lindelof.
Proof of Result 4
Suppose is hereditarily Lindelof and that is a separable metric space. It is well known that regular Lindelof spaces are paracompact. Thus is paracompact. By Proposition 5, is perfectly normal. By Result 3, is paracompact and perfectly normal.
Let be a countable dense subset of . We can think of as a compact space. The product of any Lindelof space with a compact space is Lindelof (see Corollary 3 in the post “The Tube Lemma”). Thus is Lindelof. Furthermore is a dense Lindelof subspace of . By Proposition 3, is Lindelof. By Proposition 5, is hereditarily Lindelof.
Remark
In the previous post “Bernstein Sets and the Michael Line”, a nonnormal product space where one factor is Lindelof and the other factor is a separable metric space is presented. That Lindelof space is not hereditarily Lindelof (it has uncountably many isolated points). Note that by Result 4, for any such nonnormal product space, the Lindelof factor cannot be hereditarily Lindelof.
___________________________________________________________________________________
Reference
 Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
 Willard, S., General Topology, AddisonWesley Publishing Company, 1970.
___________________________________________________________________________________