Some authors of papers introduce or motivate their results by giving basic facts either in the body of the papers or in footnotes. The basic information can be a treasure trove of information for those who study or review topology. In this post I discuss two footnotes in a paper of E. Michael (see [2]) and how they relate to the results/examples in the paper. Here’s the footnote 2 and footnote 4 in [2]:

: The reader should recall that paracompact spaces are normal, and that regular Lindelof spaces are paracompact. A Lindelof space with all open subsets is hereditarily Lindelof, and conversely if is regular.**Footnote 2**: This example is new for . In contrast to this example, S. Willard has shown that, if is paracompact with Lindelof and separable, then must be Lindelof.**Footnote 4**

The title of the paper is *Paracompactness and the Lindelof Property in Finite and Countable Cartesian Products*. The example of the Sorgenfrey Line shows that the product of two Lindelof spaces needs not be normal. The goal of [2] is to present several examples demonstrating that higher powers of paracompact or Lindelof spaces can behave unpredictably too.

Footnote 2 gives background information about paracompact, Lindelof spaces, and hereditarily Lindelof spaces. The last sentence in the footnote is that any Lindelof space is hereditarily Lindelof if and only if it is perfectly normal. Footnote 4 provides some contrasting information to Example 1.4 in [2]. In the following discussion, all spaces are assumed to be Hausdorff and regular.

**Discussion of Footnote 2**

The last sentence in the footnote is essentially the following theorem:

**Theorem 1**

For any Lindelof space , the space is hereditarily Lindelof property if and only if is perfectly normal.

A space is perfectly normal if it is normal and that every open subspace is an set. Thus, an alternative way to check whether a Lindelof space is hereditarily Lindelof is to check whether every open subset is (or every closed subset is ). In particular, any compact space with a closed subset (or even a singleton set) that is not cannot be hereditarily Lindelof. Some examples: the unit square with the lexicographic order, the ordinal , and where is any uncountable cardinal.

To prove Theorem 1, we need the following proposition.

**Proposition 1**

Any space is hereditarily Lindelof if and only if every open subspace of is Lindelof.

* Proof*. The direction is clear. To see , let and let be an open cover of . Let be a collection of open subsets of such that for each , for some . Then is Lindelof. We can find countably many sets in whose union equals . It follows that we can find a countable subcollection of that covers .

* Proof of Theorem 1*. Suppose is hereditarily Lindelof. The normality of comes from the fact that is regular and Lindelof. Let be an open subset. For each , let be open such that (this comes from the fact that is a regular space). Since is Lindelof, we can find countably many such that the union of these countably many equals . This shows that every open subset of is an set.

Suppose the Lindelof space is perfectly normal. To show that is hereditarily Lindelof, it suffices to show that every open subset of is Lindelof. This follows from that fact that every open subset of is an set and that the Lindelof property is hereditary with respect to subsets.

**Discussion of Footnote 4**

Example 1.4 in [2] provides, under the Continuum Hypothesis, for each positive integer , a regular space such that is Lindelof and is paracompact, but is not Lindelof. For , Example 1.4 is essentially a negative answer to the question: if is paracompact and each of the factors is Lindelof, must be Lindelof? Footnote 4 in [2] says that if one of the factors is separable, then must be Lindelof. We have the following theorem.

**Theorem 2**

If is paracompact such that is Lindelof anf is separable, then is Lindelof.

To prove Theorem 2, we need the following two results.

**Theorem 3**

If is paracompact and has a dense Lindelof subspace, then must be Lindelof.

* Proof*. Suppose that is paracompact. Let be a dense Lindelof subspace. To show that is Lindelof, it suffices to show that every locally finite open cover of has a countable subcover.

Let be a locally finite open cover of . For each , choose open such that and only meets finitely many sets in . For each such , let be the union of the finitely many that intersect .

Since is Lindelof, we can find countably many whose union contains . Consider the countably many corresponding . We claim that the countably many that are associated with these countably many form a cover of . Let . Choose some such that . Since is dense in , choose some . Then belongs to one of the countably many that cover . Thus, is associated with one of the corresponding . This completes the proof of Theorem 3.

**Theorem 4**

If is Lindelof and is a compact space, then is Lindelof.

* Proof*. It is known that is Lindelof if one factor is Lindelof and the other factor is compact (see this previous post). As a corollary, if one of the factor is the union of countably many compact spaces, is Lindelof.

* Proof of Theorem 2*. Suppose that is paracompact and that is Lindelof and is separable. Let be a countable dense subset of . Then is Lindelof by Theorem 4. Furthermore, is a dense Lindelof subspace of , By Theorem 3, must be Lindelof.

*Reference*

- Engelking, R.,
*General Topology, Revised and Completed edition*, 1989, Heldermann Verlag, Berlin. - Michael, E.,
*Paracompactness and the Lindelof property in Finite and Countable Cartesian Products*, Compositio Math. 23 (1971) 199-214. - Willard, S.,
*General Topology*, 1970, Addison-Wesley Publishing Company.