# Two footnotes in a paper of E. Michael

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]:

• Footnote 2: The reader should recall that paracompact spaces are normal, and that regular Lindelof spaces are paracompact. A Lindelof space $X$ with all open subsets $F_\sigma$ is hereditarily Lindelof, and conversely if $X$ is regular.
• Footnote 4: This example is new for $n=1$. In contrast to this example, S. Willard has shown that, if $X \times Y$ is paracompact with $X$ Lindelof and $Y$ separable, then $X \times Y$ must be Lindelof.

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 $X$, the space $X$ is hereditarily Lindelof property if and only if $X$ is perfectly normal.

A space is perfectly normal if it is normal and that every open subspace is an $F_\sigma$ set. Thus, an alternative way to check whether a Lindelof space is hereditarily Lindelof is to check whether every open subset is $F_\sigma$ (or every closed subset is $G_\delta$). In particular, any compact space with a closed subset (or even a singleton set) that is not $G_\delta$ cannot be hereditarily Lindelof. Some examples: the unit square with the lexicographic order, the ordinal $\omega_1+1$, and $[0,1]^{\mathcal{K}}$ where $\mathcal{K}$ is any uncountable cardinal.

To prove Theorem 1, we need the following proposition.

Proposition 1
Any space $X$ is hereditarily Lindelof if and only if every open subspace of $X$ is Lindelof.

Proof. The direction $\Rightarrow$ is clear. To see $\Leftarrow$, let $Y \subset X$ and let $\mathcal{U}$ be an open cover of $Y$. Let $\mathcal{U}^*$ be a collection of open subsets of $X$ such that for each $U^* \in \mathcal{U}^*$, $U^* \cap Y=U$ for some $U \in \mathcal{U}$. Then $\bigcup \mathcal{U}^*$ is Lindelof. We can find countably many sets in $\mathcal{U}^*$ whose union equals $\bigcup \mathcal{U}^*$. It follows that we can find a countable subcollection of $\mathcal{U}$ that covers $Y$.

Proof of Theorem 1. $\Rightarrow$ Suppose $X$ is hereditarily Lindelof. The normality of $X$ comes from the fact that $X$ is regular and Lindelof. Let $U \subset X$ be an open subset. For each $x \in U$, let $V_x$ be open such that $x \in V_x \subset \overline{V_x} \subset U$ (this comes from the fact that $X$ is a regular space). Since $U$ is Lindelof, we can find countably many $V_x$ such that the union of these countably many $\overline{V_x}$ equals $U$. This shows that every open subset of $X$ is an $F_\sigma$ set.

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

Discussion of Footnote 4
Example 1.4 in [2] provides, under the Continuum Hypothesis, for each positive integer $n$, a regular space $Y$ such that $Y^n$ is Lindelof and $Y^{n+1}$ is paracompact, but $Y^{n+1}$ is not Lindelof. For $n=1$, Example 1.4 is essentially a negative answer to the question: if $X \times Y$ is paracompact and each of the factors is Lindelof, must $X \times Y$ be Lindelof? Footnote 4 in [2] says that if one of the factors is separable, then $X \times Y$ must be Lindelof. We have the following theorem.

Theorem 2
If $X \times Y$ is paracompact such that $X$ is Lindelof anf $Y$ is separable, then $X \times Y$ is Lindelof.

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

Theorem 3
If $X$ is paracompact and has a dense Lindelof subspace, then $X$ must be Lindelof.

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

Let $\mathcal{U}$ be a locally finite open cover of $X$. For each $y \in Y$, choose open $O_y$ such that $y \in O_y$ and $O_y$ only meets finitely many sets in $\mathcal{U}$. For each such $O_y$, let $U_y$ be the union of the finitely many $U \in \mathcal{U}$ that intersect $O_y$.

Since $Y$ is Lindelof, we can find countably many $O_y$ whose union contains $Y$. Consider the countably many corresponding $U_y$. We claim that the countably many $U \in \mathcal{U}$ that are associated with these countably many $U_y$ form a cover of $X$. Let $x \in X$. Choose some $U \in \mathcal{U}$ such that $x \in U$. Since $Y$ is dense in $X$, choose some $z \in U \cap Y$. Then $z$ belongs to one of the countably many $O_y$ that cover $Y$. Thus, $U$ is associated with one of the corresponding $U_y$. This completes the proof of Theorem 3.

Theorem 4
If $X$ is Lindelof and $Y$ is a $\sigma \text{-}$compact space, then $X \times Y$ is Lindelof.

Proof. It is known that $X \times Y$ 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, $X \times Y$ is Lindelof.

Proof of Theorem 2. Suppose that $X \times Y$ is paracompact and that $X$ is Lindelof and $Y$ is separable. Let $D$ be a countable dense subset of $Y$. Then $X \times D$ is Lindelof by Theorem 4. Furthermore, $X \times D$ is a dense Lindelof subspace of $X \times Y$, By Theorem 3, $X \times Y$ must be Lindelof.

Reference

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

# An observation about hereditarily separable function spaces

For any completely regular space $X$, by $C_p(X)$ we mean the space of all real-valued continuous functions on $X$ endowed with the pointwise convergent topology. It is known that $C_p(X)$ is hereditarily separable if and only if $X^n$ is hereditarily Lindelof for all positive integer $n$ if and only if $X^\omega$ is hereditarily Lindelof where $\omega$ is the first infinite ordinal (a result that follows from a theorem of Zenor in [4]). This result points to a duality between hereditary separability of the function space $C_p(X)$ and the hereditary Lindelof property of the domain space $X$ and is restated below.

Theorem
Let $X$ be a completely regular space. Then the following conditions are equivalent:

1. $C_p(X)$ is hereditarily separable.
2. $X^n$ is hereditarily Lindelof for all positive integer $n$.
3. $X^\omega$ is hereditarily Lindelof.

As an introduction to this theorem, we present the proof to one direction of this theorem for $n=1$.

Observation
Let $X$ be any completely regular space. We have the following obervation:

If $C_p(X)$ is hereditarily separable, then $X$ is hereditarily Lindelof.

Suppose $X$ is not hereditarily Lindelof. We aim to show that $C_p(X)$ is not hereditarily separable by producing a non-separable subspace $\mathcal{F}$ of $C_p(X)$.

Let $Y \subset X$ be a subspace that is not Lindelof. Let $\mathcal{U}$ be a collection of open subsets of $X$ such that $\mathcal{U}$ covers $Y$ and no countable subcollection of $\mathcal{U}$ covers $Y$.

For each $y \in Y$, choose $U_y \in \mathcal{U}$ such that $y \in U_y$. By the completely regularity of $X$, choose a continuous $f_y: X \rightarrow \mathbb{R}$ such that $f_y$ maps $X-U_y$ to $0$ and $f_y(y)=1$. Let $\mathcal{F}=\left\{f_y:y \in X \right\}$. It can be shown that $\mathcal{F}$ is a non-separable subspace of $C_p(X)$. That is, no countable subset of $\mathcal{F}$ can be dense in $\mathcal{F}$. $\blacksquare$

For any completely regular space $X$, it is also known (see [2]) that $C_p(X)$ is separable if and only if $X$ has a weaker separable metrizable topology (i.e. $X$ has a weaker topology such that $X$ with this weaker topology is a separable metrizable space). The result in [2] combined with the observation presented here provides a way to obtain sepearable $C_p(X)$ that is not hereditarily separable. Look for any $X$ that is not hereditarily Lindelof but has a weaker separable metrizable topology. One such example is the Michael Line.

The observation we make here is a rather weak result. The double arrow space $Z$ is hereditarily Lindelof. Yet $C_p(Z)$ is not even separable since $Z$ is compact space that is not metrizable. Note that $Z^2$ is not hereditarily Lindelof since it contains a copy of the Sorgenfrey plane (see the previous post on double arrow space).

Reference

1. Engelking, R., General Topology, Revised and Completed edition, 1989, Heldermann Verlag, Berlin.
2. Noble, N., The density character of function spaces, Proc. Amer. Math. Soc. 42:1 (1974) 228-233.
3. Willard, S., General Topology, 1970, Addison-Wesley Publishing Company.
4. Zenor, P., Hereditarily m-separability and the hereditarily m-lindelof property in product spaces and function spaces, Fund. Math. 106 (1980), 175-180