Two footnotes in a paper of E. Michael

Posted on April 16, 2011 by Dan Ma

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

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.

The Lexicographic Order and The Double Arrow Space

Posted on October 7, 2009 by Dan Ma

This post is a basic discussion on the unit square with the lexicographic order and one of its subspace called the double arrow space. The goal is to establish several basic facts of these two spaces. Both of these spaces are compact non-metrizable spaces. The double arrow space consists of the top and bottom edges of the unit square. The double arrow has a strong connection to the Sorgenfrey Line and its square is an example of a space that is normal that is not completely normal. The unit square with the lexicographic order is an example of a completely normal space that is not perfectly normal.

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Defining the spaces

Let $I$ be the unit interval $[0,1]$ . Let $\mathcal{X}$ be $I\times I$ . Consider the lexicographic order on $\mathcal{X}$ . This is the linear order $<$ defined by letting $(a,b)<(c,d)$ whenever $a<c$ or $(a=c$ and $b<d)$ . The goal is to consider the square with the topology induced by this linear order. For $p=(a,b)$ and $q=(c,d)$ , $<p,q>$ is the notation in this post for open intervals in the unit square.

To facilitate further discussion, let’s look at the following subsets of $\mathcal{X}$ .

$\mathcal{A}_0=I \times \lbrace{0}\rbrace$

$\mathcal{A}_1=I \times \lbrace{1}\rbrace$

$\mathcal{B}_0=\lbrace{0}\rbrace \times I$

$\mathcal{B}_1=\lbrace{1}\rbrace \times I$

$\mathcal{S}_0=(0,1) \times \lbrace{0}\rbrace$

$\mathcal{S}_1=(0,1) \times \lbrace{1}\rbrace$

$\mathcal{A}=\mathcal{A}_0 \cup \mathcal{A}_1$

The subspace $\mathcal{A}$ is called the double arrow space. In this post, my aim is to establish some basic facts about the double arrow space $\mathcal{A}$ and then the unit square $\mathcal{X}$ .

Double arrow space

$\mathcal{A}$

$\mathcal{A}^2$

Unit square with lexicographic order

$\mathcal{X}$

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Open intervals in these spaces

Before proving the results stated above, let’s make sense of the open intervals in $\mathcal{X}$ and then in $\mathcal{A}$ . The open intervals for points in $\mathcal{X}-\mathcal{A}$ are essentially the usual open intervals. For example, for the point $(0.5,0.5)$ , the following is an open interval in $\mathcal{X}$ .

$\lbrace{0.5}\rbrace \times (0.25,0.75)$

Because of this, $\mathcal{X}$ has an uncountable collection of pairwise disjoint open intervals (thus does not have ccc and is not separable). Also, a vertical line in $\mathcal{X}$ of the form $\lbrace{t}\rbrace \times I$ is a homeomorphic copy of the unit interval $[0,1]$ with the usual topology.

The more interesting open neighborhoods are for the points in the top and bottom lines (i.e. the double arrow space). Let’s look an example of an open interval in $\mathcal{X}$ for the point $(0.5,1)$ . Consider the open interval $<p,q>$ where $p=(0.5,0.9)$ and $q=(0.6,0.2)$ . It follows that the open interval $<p,q>=L \cup M \cup R$ where

$L=\lbrace{0.5}\rbrace \times (0.9,1]$

$M=(0.5,0.6) \times I$

$R=\lbrace{0.6}\rbrace \times [0,0.2)$

Note that $M$ is the vertical strip in the middle, $L$ is the left edge, and $R$ is the right edge. If you look at $<p,q>\cap{\mathcal{A}}$ , then the open interval for the point $(0.5,1)$ becomes

$\biggl((0.5,0.6] \times \lbrace{0}\rbrace \biggr) \cup \biggl( [0.5,0.6) \times \lbrace{1}\rbrace \biggr)$

The above is also an open interval in $\mathcal{A}$ containing the point $(0.6,0)$ . Similarly, an example of an open interval containing the point $(0.5,0)$ is:

$\biggl( (0.4,0.5] \times \lbrace{0}\rbrace \biggr) \cup \biggl([0.4,0.5) \times \lbrace{1}\rbrace\biggr)$

Based on the above examples of open intervals, open intervals in $\mathcal{A}$ have the following form, hence the term “double arrow”.

$\biggl((a,b] \times \lbrace{0}\rbrace\biggr) \cup \biggl([a,b) \times \lbrace{1}\rbrace\biggr)$

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Proofs

The double arrow space contains two copies of the Sorgenfrey line. So knowing the basic properties of the Sorgenfrey line will help follow the proofs of A through D below. Information about the Sorgenfrey line can be found in the following blog post.

A Note On The Sorgenfrey Line

A Short Note About The Sorgenfrey Line

Proof of A. I now proof that the double arrow space $\mathcal{A}$ is compact. I do so by showing that it is Lindelof and countably compact. A space is countably compact if any countable open cover has a finite subcover (equivalently, any infinite subspace has a limit point). Here, limit point is in the topological sense (i.e. the point $p$ is a limit point of the set $K$ if any open set containing $p$ contains a point of $K$ different from $p$ ).

Note that $\mathcal{S}_0$ and $\mathcal{S}_1$ as defined above are homeomorphic copies of the Sorgenfrey Line. Thus $\mathcal{A}$ is made of of two copies of the Sorgenfrey Line plus 4 points. This shows that $\mathcal{A}$ is hereditarily Lindelof.

Pick an infinite subset $W$ of $\mathcal{A}$ . Then either the top or the bottom contains infinitely many points of $W$ . Assume the top does. There is a point $(q,1)$ that is a limit point of $W \cap \mathcal{A}_1$ in the usual topology. If the point $(q,1)$ is a right-sided limit point of $W \cap \mathcal{A}_1$ in the usual topology, then the point $(q,1)$ is a limit point of $W$ in the order topology. If the point $(q,1)$ is a left-sided limit point of $W \cap \mathcal{A}_1$ in the usual topology, then the point $(q,0)$ is a limit point of $W$ in the order topology.

Proof of B, C and D. As observed in the proof of A, the double arrow space is the union of 2 copies of the Sorgenfrey Line plus 4 points. Thus it is hereditarily Lindelof, hereditarily separable and perfectly normal.

Note that $\mathcal{A}^2$ contains a copy of the Sorgenfrey Plane. Thus it is not hereditarily normal. This also implies that both the double arrow space and the unit square with the lexicographic order are not metrizable.

Proof of E. Let $\mathcal{U}$ be an open cover for $\mathcal{X}$ that consists of open intervals. Since $\mathcal{A}$ is compact, there exists a finite collection of open intervals $\lbrace{U_0,U_1,...,U_n}\rbrace$ from $\mathcal{U}$ that cover $\mathcal{A}$ . Based on the above observation about open intervals of points in the double arrow space, the open intervals $U_i$ cover all the points in $\mathcal{X}-\mathcal{A}$ except possibly the left and right edges of the open intervals $U_i$ . With the topology inherited from the order topology, each of the edges is homeomorphic to the unit interval with the usual topology. Thus there are finitely many open intervals from $\mathcal{U}$ that cover these left and right edges. Thus $\mathcal{U}$ has a finite subcover and $\mathcal{X}$ is compact.

Proof of F, G, H. These are clear based on the above observation made about the open intervals in the unit square.

Proof of I. I show here that $\mathcal{A}$ cannot be a $G_\delta$ set in the unit square $\mathcal{X}$ . For each $n<\omega$ , let $U_n$ be an open set in $\mathcal{X}$ such that $\mathcal{A} \subset U_n$ . Since $\mathcal{A}$ is compact, we can assume that $U_n$ is the union of finitely many open intervals. Based on the observation given above, each $U_n$ covers all of $\mathcal{X}-\mathcal{A}$ except for finitely many vertical lines (the left and right edges of these open intervals). Pick one vertical line that is not one of the vertical edges from the open intervals $U_n$ . Clearly this vertical line is covered by $U_n$ for each $n<\omega$ . Thus $\mathcal{A}$ is not a $G_\delta$ set in $\mathcal{X}$ .

Comment. The double arrow space is made up of 2 copies of the Sorgenfrey Line and the 4 corner points. It follows that it is hereditarily Lindelof and perfectly normal. However, since the square of the double arrow space contains a copy of the Sorgenfrey Plane, the square of the double arrow space is not hereditarily normal, thus showing that normality is not a hereditary notion. The square of the double arrow space $\mathcal{A}^2$ is a handy example of a normal space that is not completely normal. This implies both the double arrow space and the unit square with the lexicographic order not metrizable. The space $\mathcal{A}^2$ also demonstrates that hereditarily normality is not preserved by Cartesian product.

The unit square with the lexicographic order topology is completely normal ([Steen & Seebach]). Thus it is an example of a space that completely normal (T4) but not perfectly normal (T5).

Reference

Counterexamples In Topology

Dan Ma's Topology Blog

Expository Articles In General Topology

Tag Archives: Lexicographic order

Two footnotes in a paper of E. Michael

The Lexicographic Order and The Double Arrow Space