Cartesian Products of Two Paracompact Spaces – Continued

Consider the real line \mathbb{R} with a topology finer than the usual topology obtained by isolating each point in \mathbb{P} where \mathbb{P} is the set of all irrational numbers. The real line with this finer topology is called the Michael line and we use \mathbb{M} to denote this topological space. It is a classic result that \mathbb{M} \times \mathbb{P} is not normal (see “Michael Line Basics”). Even though the Michael line \mathbb{M} is paracompact (it is in fact hereditarily paracompact), \mathbb{M} is not perfectly normal. Result 3 below will imply that the Michael line cannot be perfectly normal. Otherwise \mathbb{M} \times \mathbb{P} would be paracompact (hence normal). Result 3 is the statement that if X is paracompact and perfectly normal and Y is a metric space then X \times Y is paracompact and perfectly normal. We also use this result to show that if X is hereditarily Lindelof and Y is a separable metric space, then X \times Y 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 X is paracompact and Y is compact, then X \times Y is paracompact.

Result 2

    If X is paracompact and Y is \sigma-compact, then X \times Y is paracompact.

Result 3

    If X is paracompact and perfectly normal and Y is metrizable, then X \times Y is paracompact and perfectly normal.

Result 4

    If X is hereditarily Lindelof and Y is a separable metric space, then X \times Y 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.

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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 X be a regular space. Then X is paracompact if and only if every open cover \mathcal{U} of X has a \sigma-locally finite open refinement.

Proposition 2
Every F_\sigma-subset of a paracompact space is paracompact.

Proposition 3
Any paracompact space with a dense Lindelof subspace is Lindelof.

Proof of Proposition 3
Let L be a paracompact space. Let M \subset L be a dense Lindelof subspace. Let \mathcal{U} be an open cover of L. Since we are working with a regular space, let \mathcal{V} be an open cover of L such that \left\{\overline{V}: V \in \mathcal{V} \right\} refines \mathcal{U}. Let \mathcal{W} be a locally finite open refinement of \mathcal{V}. Choose \left\{W_1,W_2,W_3,\cdots \right\} \subset \mathcal{W} such that it is a cover of M. Since M \subset \bigcup \limits_{i=1}^\infty W_i, \overline{\bigcup \limits_{i=1}^\infty W_i}=L.

Since the sets W_i come from a locally finite collection, they are closure preserving. Hence we have:

    \overline{\bigcup \limits_{i=1}^\infty W_i}=\bigcup \limits_{i=1}^\infty \overline{W_i}=L

For each i, choose some U_i \in \mathcal{U} such that \overline{W_i} \subset U_i. Then \left\{U_1,U_2,U_3,\cdots \right\} is a countable subcollection of \mathcal{U} covering the space L. \blacksquare

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 G_\delta-set in the space (equivalently every open subset is an F_\sigma-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 L be a space. Then L is hereditarily Lindelof if and only if every open subspace of L is Lindelof.

Proposition 5
Let L be a Lindelof space. Then L is hereditarily Lindelof if and only if L is perfectly normal.

Proof of Proposition 5
\Rightarrow Suppose L is hereditarily Lindelof. It is well known that regular Lindelof space is normal. Thus L is normal. It remains to show that every open subset of L is F_\sigma. Let U \subset L be an non-empty open set. For each x \in U, let V_x be open such that x \in V_x and \overline{V_x} \subset U (the space is assumed to be regular). By assumption, the open set U is Lindelof. The open sets V_x form an open cover of U. Thus U is the union of countably many \overline{V}_x.

\Leftarrow Suppose L is perfectly normal. To show that L is hereditarily Lindelof, it suffices to show that every open subset of L is Lindelof (by Proposition 4). Let U \subset L be non-empty open. By assumption, U=\bigcup \limits_{i=1}^\infty F_i where each F_i is a closed set in L. Since the Lindelof property is hereditary with respect to closed subsets, U is Lindelof. \blacksquare

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 \sigma-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 X be a space. Then X is metrizable if and only if X has a \sigma-locally finite base.

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Result 3

Result 3 is the statement that:

    If X is paracompact and perfectly normal and Y is a metric space then X \times Y 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 X is an F_\sigma-set in X,
  • Y is a metric space,

then every open subset of X \times Y is an F_\sigma-set in X \times Y.

Proof of Lemma 7
Let U be a open subset of X \times Y. If U=\varnothing, then U is certainly the union of countably many closed sets. So assume U \ne \varnothing. Let \mathcal{B}=\bigcup \limits_{i=1}^\infty \mathcal{B}_i be a base for Y such that each \mathcal{B}_i is locally finite in Y (by Theorem 6, such a base exists since Y is metrizable).

Consider all non-empty B \in \mathcal{B} such that we can choose nonempty open set W_B \subset X with W_B \times \overline{B} \subset U. Since U is non-empty open, such pairs (B, W_B) exist. Let \mathcal{B}^* be the collection of all non-empty B \in \mathcal{B} for which there is a matching non-empty W_B. For each i, let \mathcal{B}_i^*=\mathcal{B}^* \cap \mathcal{B}_i. Of course, each \mathcal{B}_i^* is still locally finite.

Since every open subset of X is an F_\sigma-set in X, for each W_B, we can write W_B as

    W_B=\bigcup \limits_{j=1}^\infty W_{B,j}

where each W_{B,i} is closed in X.

For each i=1,2,3,\cdots and each j=1,2,3,\cdots, consider the following collection:

    \mathcal{V}_{i,j}=\left\{W_{B,j} \times \overline{B}: B \in \mathcal{B}_i^* \right\}

Each element of \mathcal{V}_{i,j} is a closed set in X \times Y. Since \mathcal{B}_i^* is a locally finite collection in Y, \mathcal{V}_{i,j} is a locally finite collection in X \times Y. Define V_{i,j}=\bigcup \mathcal{V}_{i,j}. The set V_{i,j} is a union of closed sets. In general, the union of closed sets needs not be closed. However, V_{i,j} is still a closed set in X \times Y since \mathcal{V}_{i,j} is a locally finite collection of closed sets. This is because a locally finite collection of sets is closure preserving. Note the following:

    \overline{V_{i,j}}=\overline{\bigcup \mathcal{V}_{i,j}}=\overline{\bigcup \left\{W_{B,j} \times \overline{B}: B \in \mathcal{B}_i^* \right\}}=\bigcup \left\{\overline{W_{B,j} \times \overline{B}}: B \in \mathcal{B}_i^* \right\}

      =\bigcup \left\{W_{B,j} \times \overline{B}: B \in \mathcal{B}_i^* \right\}=V_{i,j}

Finally, we have U=\bigcup \limits_{i=1}^\infty \bigcup \limits_{j=1}^\infty V_{i,j}, which is the union of countably many closed sets. \blacksquare

Lemma 8
If X is a paracompact space satisfying the following two conditions:

  • every open subset of X is an F_\sigma-set in X,
  • Y is a metric space,

then X \times Y is paracompact.

Proof of Lemma 8
As in the proof of the above lemma, let \mathcal{B}=\bigcup \limits_{i=1}^\infty \mathcal{B}_i be a base for Y such that each \mathcal{B}_i is locally finite in Y. Let \mathcal{U} be an open cover of X \times Y. Assume that elements of \mathcal{U} are of the form A \times B where A is open in X and B \in \mathcal{B}.

For each B \in \mathcal{B}, consider the following two items:

    \mathcal{W}_B=\left\{A: A \times B \in \mathcal{U} \right\}

    W_B=\bigcup \mathcal{W}_B

To simplify matter, we only consider B \in \mathcal{B} such that \mathcal{W}_B \ne \varnothing. Each W_B is open in X and hence by assumption an F_\sigma-set in X. Thus by Proposition 2, each W_B is paracompact. Note that \mathcal{W}_B is an open cover of W_B. Let \mathcal{H}_B be a locally finite open refinement of \mathcal{W}_B. Consider the following two items:

    For each j=1,2,3,\cdots, let \mathcal{V}_j=\left\{A \times B: A \in \mathcal{H}_B \text{ and } B \in \mathcal{B}_j \right\}

    \mathcal{V}=\bigcup \limits_{j=1}^\infty \mathcal{V}_j

We observe that \mathcal{V} is an open cover of X \times Y and that \mathcal{V} refines \mathcal{U}. Furthermore each \mathcal{V}_j is a locally finite collection. The open cover \mathcal{U} we start with has a \sigma-locally finite open refinement. Thus X \times Y is paracompact. \blacksquare

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Result 4

Result 4 is the statement that:

    If X is hereditarily Lindelof and Y is a separable metric space, then X \times Y is hereditarily Lindelof.

Proof of Result 4
Suppose X is hereditarily Lindelof and that Y is a separable metric space. It is well known that regular Lindelof spaces are paracompact. Thus X is paracompact. By Proposition 5, X is perfectly normal. By Result 3, X \times Y is paracompact and perfectly normal.

Let D be a countable dense subset of Y. We can think of D as a \sigma-compact space. The product of any Lindelof space with a \sigma-compact space is Lindelof (see Corollary 3 in the post “The Tube Lemma”). Thus X \times D is Lindelof. Furthermore X \times D is a dense Lindelof subspace of X \times Y. By Proposition 3, X \times Y is Lindelof. By Proposition 5, X \times Y is hereditarily Lindelof. \blacksquare

Remark
In the previous post “Bernstein Sets and the Michael Line”, a non-normal 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 non-normal product space, the Lindelof factor cannot be hereditarily Lindelof.

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Reference

  1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
  2. Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

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\copyright \ \ 2012

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Cartesian Products of Two Paracompact Spaces

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 two-factor 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 X is paracompact and Y is compact, then X \times Y is paracompact.

Result 2

    If X is paracompact and Y is \sigma-compact, then X \times Y is paracompact.

Result 3

    If X is paracompact and perfectly normal and Y is metrizable, then X \times Y is paracompact and perfectly normal.

Result 4

    If X is hereditarily Lindelof and Y is a separable metric space, then X \times Y is hereditarily Lindelof.

With Results 1 and 2, compact spaces and \sigma-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 X be a space. A collection \mathcal{A} of subsets of X is said to be a cover of X if X=\bigcup \mathcal{A} (in words every point of the space belongs to one set in the collection). Furthermore, \mathcal{A} is an open cover of X is it is a cover of X consisting of open subsets of X.

Let \mathcal{A} and \mathcal{B} be covers of the space X. The cover \mathcal{B} is said to be a refinement of \mathcal{A} (\mathcal{B} is said to refine \mathcal{A}) if for every B \in \mathcal{B}, there is some A \in \mathcal{A} such that B \subset A. The cover \mathcal{B} is said to be an open refinement of \mathcal{A} if \mathcal{B} refines \mathcal{A} and \mathcal{B} is an open cover.

A collection \mathcal{A} of subsets of X is said to be a locally finite collection if for each point x \in X, there is a non-empty open subset V of X such that x \in V and V has non-empty intersection with at most finitely many sets in \mathcal{A}. An open cover \mathcal{A} of X is said to have a locally finite open refinement if there exists an open cover \mathcal{C} of X such that \mathcal{C} refines \mathcal{A} and \mathcal{C} is a locally finite collection. We have the following definition.

Definition

    The space X is said to be paracompact if every open cover of X has a locally finite open refinement.

A collection \mathcal{U} of subsets of the space X is said to be a \sigma-locally finite collection if \mathcal{U}=\bigcup \limits_{i=1}^\infty \mathcal{U}_i such that each \mathcal{U}_i is a locally finite collection of subsets of X. Consider the property that every open cover of X has a \sigma-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 X be a regular space. Then X is paracompact if and only if every open cover \mathcal{U} of X has a \sigma-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 X be a regular space. Then X is paracompact if and only if the following holds:

    For each open cover \left\{U_t: t \in T \right\} of X, there exists a locally finite open cover \left\{V_t: t \in T \right\} such that \overline{V_t} \subset U_t for each t \in T.

Theorem 3 below shows that paracompactness is hereditary with respect to F_\sigma-subsets.

Theorem 3
Every F_\sigma-subset of a paracompact space is paracompact.

Proof of Theorem 3
Let X be paracompact. Let Y \subset X such that Y=\bigcup \limits_{i=1}^\infty Y_i where each Y_i is a closed subset of X. Let \mathcal{U} be an open cover of Y. For each U \in \mathcal{U}, let U^* be open in X such that U^* \cap Y=U.

For each i, let \mathcal{U}_i^* be the set of all U^* such that U \cap Y_i \ne \varnothing. Let \mathcal{V}_i^* be a locally finite refinement of \mathcal{U}_i^* \cup \left\{X-Y_i \right\}. Let \mathcal{V}_i be the following:

    \mathcal{V}_i=\left\{V \cap Y: V \in \mathcal{V}_i^* \text{ and } V \cap Y_i \ne \varnothing \right\}

It is clear that each \mathcal{V}_i is a locally finite collection of open set in Y covering Y_i. All the \mathcal{V}_i together form a refinement of \mathcal{U}. Thus \mathcal{V}=\bigcup \limits_{i=1}^\infty \mathcal{V}_i is a \sigma-locally finite open refinement of \mathcal{U}. By Theorem 1, the F_\sigma-set Y is paracompact. \blacksquare
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Result 1

Result 1 is the statement that:

    If X is paracompact and Y is compact, then X \times Y is paracompact.

To prove Result 1, we use the Tube lemma (for a proof, see “The Tube Lemma”).

The Tube Lemma
Let X be any space and Y be compact. For each x \in X and for each open set U \subset X \times Y such that \left\{x \right\} \times Y \subset U, there is an open set O \subset X such that \left\{x \right\} \times Y \subset O \times Y \subset U.

Proof of Result 1
Let \mathcal{U} be an open cover of X \times Y. For each x \in X, choose a finite \mathcal{U}_x \subset \mathcal{U} such that \mathcal{U}_x is a cover of \left\{x \right\} \times Y. By the Tube Lemma, for each x \in X, there is an open set O_x \subset X such that \left\{x \right\} \times Y \subset O_x \times Y \subset \cup \mathcal{U}_x. Since X is paracompact, by Theorem 2, let \left\{W_x: x \in X \right\} be a locally finite open refinement of \left\{O_x: x \in X \right\} such that W_x \subset O_x for each x \in X.

Let \mathcal{W}=\left\{(W_x \times Y) \cap U: x \in X, U \in \mathcal{U}_x \right\}. We claim that \mathcal{W} is a locally finite open refinement of \mathcal{U}. First, this is an open cover of X \times Y. To see this, let (a,b) \in X \times Y. Then a \in W_x for some x \in X. Furthermore, a \in O_x and (a,b) \in \cup \mathcal{U}_x. Thus, (a,b) \in (W_x \times Y) \cap U for some U \in \mathcal{U}_x. Secondly, it is clear that \mathcal{W} is a refinement of the original cover \mathcal{U}.

It remains to show that \mathcal{W} is locally finite. To see this, let (a,b) \in X \times Y. Then there is an open V in X such that x \in V and V can meets only finitely many W_x. Then V \times Y can meet only finitely many sets in \mathcal{W}. \blacksquare

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Result 2

Result 2 is the statement that:

    If X is paracompact and Y is \sigma-compact, then X \times Y is paracompact.

Proof of Result 2
Note that the \sigma-compact space Y is Lindelof. Since regular Lindelof are normal, Y is normal and is thus completely regular. So we can embed Y into a compact space K. For example, we can let K=\beta Y, which is the Stone-Cech compactification of Y (see “Embedding Completely Regular Spaces into a Cube”). For our purpose here, any compact space containing Y will do. By Result 1, X \times K is paracompact. Note that X \times Y can be regarded as a subspace of X \times K.

Let Y=\bigcup \limits_{i=1}^\infty Y_i where each Y_i is compact in Y. Note that X \times Y=\bigcup \limits_{i=1}^\infty X \times Y_i and each X \times Y_i is a closed subset of X \times K. Thus the product X \times Y is an F_\sigma-subset of X \times K. According to Theorem 3, F_\sigma-subsets of any paracompact space is paracompact space. Thus X \times Y is paracompact. \blacksquare

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Reference

  1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
  2. Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

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\copyright \ \ 2012

The Michael Line and the Continuum Hypothesis

There exist a Lindelof space and a separable metric space such that their Cartesian product is not normal (discussed in the post “Bernstein Sets and the Michael Line”). The separable metric space is a Bernstein set, a subspace of the real line that is far from being a complete metric space. However, this example is constructed without using any additional set theory axiom beyond the Zermelo-Fraenkel axioms plus the axiom of choice (abbreviated ZFC). A natural question is whether there exists a Lindelof space and a complete metric space such that their product is not normal. In particular, does there exist a Lindelof space L such that the product of L with the space of all irrational numbers is not normal? As of the writing of this post, it is still unknown that such a Lindelof space can exist in just ZFC alone without applying additional set theory axiom. However, such a Lindelof space can be constructed from various additional axioms (e.g. continuum hypothesis or Martin’s axiom). In this post, we present an example of such construction using the continuum hypothesis (the statement that the cardinality of the real line is the same as the first uncountable cardinal \aleph_1).

Let \mathbb{M} be the Michael line. Let \mathbb{P} be the set of irrational numbers with the usual topology inherited from the real line. It is a classical result that the product \mathbb{M} \times \mathbb{P} is not normal (see “Michael Line Basics”). The Lindelof example we wish to discuss is an uncountable Lindelof subspace L of \mathbb{M} such that L contains the set \mathbb{Q} of rational numbers. The same proof that \mathbb{M} \times \mathbb{P} is not normal will show that L \times \mathbb{P} is not normal.

See the following posts for a basic discussion of the Michael line:

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Luzin Sets

The Lindelof space X we want to find is a subset of the real line that is called a Luzin set. Before defining Luzin sets, recall some definitions. Let Y be a space. Let A \subset Y. The set A is said to be nowhere dense in Y if for every non-empty open subset U of Y, there is a non-empty open subset V of Y such that V \subset U and V misses A (equivalently, the closure of A has no interior). The set A is of first category in Y if it is the union of countably many nowhere dense sets.

To define Luzin sets, we focus on the Euclidean space \mathbb{R}. Let A \subset \mathbb{R}. The set A is said to be a Luzin set if for every set W \subset \mathbb{R} that is of first category in the real line, A \cap W is at most countable. The Russian mathematician Luzin in 1914 constructed such an uncountable Luzin set using continuum hypothesis (CH). A good reference for Luzin sets is [4]. We have the following theorem.

Theorem 1
Assume CH. There exists an uncountable Luzin set.

Proof of Theorem 1
There are continuum many closed nowhere dense subsets of the real line. Since we assume the continuum hypothesis, we can enumerate these sets in a sequence of length \omega_1. Let \left\{F_\alpha: \alpha < \omega_1 \right\} be the set of all closed nowhere dense sets in the real line. Choose a real number x_0 \notin F_0 to start. For each \alpha with 0 < \alpha <\omega_1, choose a real number x_\alpha not in the following set:

    \left\{x_\beta: \beta<\alpha \right\} \cup \bigcup \limits_{\beta<\alpha} F_\beta

The above set is a countable union of closed nowhere dense sets of the real line. As a complete metric space, the real line cannot be of first category. In fact, according to the Baire category theorem, the complement of a set of first category (such as the one described above) is dense in the real line. So such an x_\alpha can always be selected at each \alpha<\omega_1. Then X=\left\{x_\alpha: \alpha<\omega_1 \right\} is a Luzin set. \blacksquare

Now that we have a way of constructing an uncountable Luzin sets, the following observations provide some useful facts for our problem at hand.

Nowhere dense sets and sets of first category are "thin" sets. Any "thin" set can intersect with a Luzin set with only countably many points. Thus any "co-thin" set contains all but countably many points of a Luzin set. For example, let A be an uncountable Luzin set. Then if F is a closed nowhere dense set in the real line, then \mathbb{R}-F contains all but countably many points of A. Furthermore, if F_1,F_2,F_3,\cdots, are closed nowhere dense subsets of the real line, then \mathbb{R}- \bigcup \limits_{i=1}^\infty F_i contains all but countably many points of the Luzin set A.

Note that the set \mathbb{R}-F in the preceding paragraph is a dense open set. Thus the complement of a closed nowhere dense set is a dense open set. Note that the set \mathbb{R}- \bigcup \limits_{i=1}^\infty F_i in the preceding paragraph is a dense G_\delta-set. Thus the complement of the union of countably many closed nowhere dense sets is a dense G_\delta-set. Thus the observation in the preceding paragraph gives the following proposition:

Proposition 2
Given an uncountable Luzin set A and given a dense G_\delta subset H of the real line, H contains all but countably many points of A.

In fact, Proposition 2 not only hold in the real line, it also holds in any uncountable dense subset of the real line.

Proposition 3
Let A be an uncountable Luzin set. Let Y \subset \mathbb{R} be uncountable and dense in the real line such that A \cap Y is uncountable. Given a dense G_\delta subset H of Y, H contains all but countably many points of A \cap Y.

Proof of Proposition 3
We want to show that Y-H can only contain countably many points of A. Let H=\bigcap \limits_{i=1}^\infty O_i where each O_i is open and dense in Y. Then for each i, let U_i be open in the real line such that U_i \cap Y=O_i. Each U_i is open and dense in the real line. Thus H^*=\bigcap \limits_{i=1}^\infty U_i contains all but countably many points of the Luzin set A. Note the following set inclusion:

    H=\bigcap \limits_{i=1}^\infty U_i \cap Y=\bigcap \limits_{i=1}^\infty O_i \subset \bigcap \limits_{i=1}^\infty U_i=H^*

Suppose that Y-H contains uncountably many points of A. Then these points, except for countably many points, must belong to H^*=\bigcap \limits_{i=1}^\infty U_i. The above set inclusion shows that these points must belong to H too, a contradiction. Thus Y-H can only contain countably many points of A, equivalently the G_\delta-set H contains all but countably many points of A \cap Y. \blacksquare

The following proposition follows from Proposition 3 and is a useful fact that will help us see that the product of an uncountable Luzin set and \mathbb{P} is not normal.

Proposition 4
Let Y be an uncountable Luzin set such that \mathbb{Q} \subset Y. Then Y-\mathbb{Q} cannot be an F_\sigma-set in the Euclidean space Y, equivalently \mathbb{Q} cannot be a G_\delta-set in the space Y.

Proof of Proposition 4
By Proposition 3, any dense G_\delta-subset of Y must be co-countable. \blacksquare

The following proposition is another useful observation about Luzin sets. Let A \subset \mathbb{R}. Let D \subset \mathbb{R} be a countable dense subset of the real line. The set A is said to be concentrated about D if for every open subset O of the real line such that D \subset O, O contains all but countably many points of A. The following proposition can be readily checked based on the definition of Luzin sets.

Proposition 5
For any A \subset \mathbb{R}, A is a Luzin set if and only if A is concentrated about every countable dense subset of the real line.

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Lindelof Subspace of The Michael Line

Let A be an uncountable Luzin set. We can assume that A is dense in the real line. If not, just add a countble subset of \mathbb{P} that is dense in the real line. Let L=A \cup \mathbb{Q}. It is clear that adding countably many points to a Luzin set still results in a Luzin set. Thus L is also a Luzin set. Now consider L as a subspace of the Michael line \mathbb{M}. Then points of L-\mathbb{Q} are discrete and points in \mathbb{Q} have Euclidean open neighborhoods. By Proposition 5, the set L is concentrated about every countable dense subset of the real line. In particular, it is concentrated about \mathbb{Q}. Thus as a subspace of the Michael line, L is a Lindelof space, since every open set containing \mathbb{Q} contains all but countably many points of L.

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The Non-Normal Product L \times \mathbb{P}

We highlight the following two facts about the Luzin set L=A \cup \mathbb{Q} as discussed in the preceding section.

  • L-\mathbb{Q} is not an F_\sigma-set in L (as Euclidean space).
  • A=L-\mathbb{Q} is dense in the real line.

The first bullet point follows from Proposition 4. The second bullet point is clear since we assume the Luzin set A we start with is dense. Recall that when thinking of L as a subspace of the Michael line, L-\mathbb{Q} are isolated and \mathbb{Q} retains the usual real line open sets. Because of the above two bullet points, L \times \mathbb{P} is not normal. The proof that L \times \mathbb{P} is not normal is the corollary of the proof that \mathbb{M} \times \mathbb{P} is not normal. Note that in the proof for showing \mathbb{M} \times \mathbb{P} is not normal, the two crucial points about the proof are that the isolated points of the Michael line cannot be an F_\sigma-set and are dense in the real line (found in “Michael Line Basics”).

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Michael Space

The example L \times \mathbb{P} that we construct here was hinted in footnote 4 in [6]. In a later publication, E. Michael constructed an uncountable Lindelof subspace of the Michael line (see Lemma 3.1 in [5]). That construction should produce a similar set as the Luzin sets since the approach in [5] is a mirror image of the Luzin set construction. The approach in the Luzin set construction in Theorem 1 is to pick points not in the union of countably many closed nowhere dense sets, while the approach in [5] was to pick points in dense G_\delta-sets in a transfinite induction process.

A Michael space is a Lindelof space whose product with \mathbb{P} is not normal. The example shown here shows that under CH, there exists a Michael space. However, the question of whether there exists a Michael space in ZFC is still unsolved. This is called the Michael problem. A recent mention of this unsolved problem is [3] (page 160). A Michael space can also be constructed using Martin’s axiom (see [1]).

A space is said to be a productively Lindelof space if its product with every Lindelof space is Lindelof. Is \mathbb{P} a productively Lindelof space? As we see here, under CH the answer is no. Another way of looking at the Michael problem: is it possible to show that \mathbb{P} is not productively Lindelof in ZFC alone?

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Reference

  1. Alster, K., The product of a Lindelof space with the space of irrationals under Martin’s Axiom, Proc. Amer. Math. Soc., 110 (1990) 543-547.
  2. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
  3. Hart, K. P., Nagata J. I., Vaughan, J. E., editors, Encyclopedia of General Topology, First Edition, Elsevier Science Publishers B. V, Amsterdam, 2003.
  4. Miller, A. W., Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 201-233, 1984.
  5. Michael, E., Paracompactness and the Lindelof property in Finite and Countable Cartesian Products, Compositio Math. 23 (1971) 199-214.
  6. Michael, E., The product of a normal space and a metric space need not be normal, Bull. Amer. Math. Soc., 69 (1963) 375-376.
  7. Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

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\copyright \ \ 2012

Bernstein Sets and the Michael Line

Let \mathbb{M} be the Michael line and let \mathbb{P} be the set of all irrational numbers with the Euclidean topology. In the post called “Michael Line Basics”, we show that the product \mathbb{M} \times \mathbb{P} is not normal. This is a classic counterexample showing that the product of two paracompact spaces need not be normal even when one of the factors is a complete metric space. The Michael line \mathbb{M} is not Lindelof. A natural question is: can the first factor be made a Lindelof space? In this post, as an application of Bernstein sets, we present a non-normal product space where one factor is Lindelof and the other factor is a separable metric space. It is interesting to note that while one factor is upgraded (from paracompact to Lindelof), the other factor is downgraded (from a complete metric space to just a separable metric space).

Bernstein sets have been discussed previously in this blog. They are special subsets of the real line and with the Euclidean subspace topology, they are spaces in which the Banach-Mazur game is undecidable (see the post “Bernstein Sets Are Baire Spaces”). A Bernstein set is a subset B of the real line such that every uncountable closed subset of the real line has non-empty intersection with both B and the complement of B.

Bernstein sets are constructed by transfinite induction. The procedure starts by ordering all uncountable closed subsets of the real line in a sequence of length that is as long as the cardinality of continuum. To see how Bernstein sets are constructed, see the post “Bernstein Sets Are Baire Spaces”.

After we discuss a generalization of the definition of the Michael line, we discuss the non-normal product space based on Bernstein sets.
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Generalizing the Michael Line

Let \mathbb{R} be the real number line. Let \mathbb{P} be the set of all irrational numbers and let \mathbb{Q}=\mathbb{R}-\mathbb{P}. Recall that the Michael line is the real line \mathbb{R} topologized by letting points in \mathbb{P} discrete and letting points in \mathbb{Q} retain their usual open neighborhoods. We can carry out the same process on any partition of the real number line.

Let D and E be disjoint sets such that \mathbb{R}=D \cup E where the set E is dense in the real line. The intention is to make D the discrete part and E the Euclidean part. In other words, we topologize \mathbb{R} be letting points in D discrete and letting points in E retain their Euclidean open sets. Let X_D denote the resulting topological space. For the lack of a better term, we call the space X_D the modified Michael line. An open set in the space X_D is of the form U \cup V where U is a Euclidean open subset of the real line and V \subset D. We have the following result:

    Proposition
    Suppose that D is not an F_\sigma-set in the Euclidean real line and that D is dense in the Euclidean real line. Then the product space X_D \times D is not normal (the second factor D is considered a subspace of the Euclidean real line).

In the post “Michael Line Basics”, we give a proof that \mathbb{M} \times \mathbb{P} is not normal. This proof hinges on the same two facts about the set D in the hypothesis in the above proposition. Thus the proof for the above proposition is just like the one for \mathbb{M} \times \mathbb{P}. Whenever we topologize the modified Michael line by using a non-F_\sigma-set as the discrete part, we can always be certain that we have a non-normal product as indicated here.

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Non-Normal Product Space

Let B be any Bernstein set. The set B is clearly not an F_\sigma-set in the real line and is clearly dense in the real line. Then X_B \times B is not normal. Note that in X_B, the set B is discrete and its complement \mathbb{R}-B has the usual topology. To see that X_B is Lindelof, note that any open cover of X_B has a countable subcollection that covers \mathbb{R}-B. This countable subcollection consists of Euclidean open sets. Furthermore, the complement of the union of these countably many Euclidean open sets must contain all but countably many points of the Bernstein set B (otherwise there would be an uncountable Euclidean closed set that misses B).

As commented at the beginning, in obtaining this non-normal product space, one factor is enhanced at the expense of the other factor (one is made Lindelof while the other is no longer a complete metric space). Even though any Bernstein set (with the Euclidean topology) is a separable metric space, it cannot be completely metrizable. Any completely metrizable subset of the real line must be a G_\delta-set in the real line. Furthermore any uncountable G_\delta subset of the real line must contain a Cantor set and thus cannot be a Bernstein set.

A similar example to X_B \times B is presented in E. Michael’s paper (see [3]). It is hinted in footnote 4 of that paper that with the additional assumption of continuum hypothesis (CH), one can have a non-normal product space where one factor is a Lindelof space and the second factor is the space of irrationals. So with an additional set-theoretic assumption, we can keep one factor from losing complete metrizability. For this construction, see point (d) in Example 3.2 of [2].

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A Brief Remark

Note that the Lindelof space X_B presented here is not hereditarily Lindelof, since it has uncountably many isolated points. Can a hereditarily Lindelof example be constructed such that its product with a particular separable metric space is not normal? The answer is no. The product of a hereditarily Lindelof space and any separable metric space is hereditarily Lindelof (see Result 4 in the post Cartesian Products of Two Paracompact Spaces – Continued).

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Reference

  1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
  2. Michael, E., Paracompactness and the Lindelof property in Finite and Countable Cartesian Products, Compositio Math. 23 (1971) 199-214.
  3. Michael, E., The product of a normal space and a metric space need not be normal, Bull. Amer. Math. Soc., 69 (1963) 375-376.
  4. Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

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\copyright \ \ 2012

Finite and Countable Products of the Michael Line

Consider the real number line \mathbb{R} with a topology stronger than the Euclidean topology such that the irrational numbers are isolated and the rational numbers retain their Euclidean open neighborhoods. When the real number line is endowed with this topology, the resulting topological space is called the Michael line and is denoted by \mathbb{M}. It is well known that \mathbb{M} \times \mathbb{P} is not normal where \mathbb{P} is the space of irrational numbers with the Euclidean topology. This and other basic results about the Michael line are discussed in the post Michael Line Basics. In this post, we show that \mathbb{M}^n is paracompact for any positive integer n and that \mathbb{M}^\omega (the product of countably and infinitely many copies of \mathbb{M}) is not normal. Thus the Michael line is an example demonstrating that even when paracompactness is preserved by taking finite products, it can be destroyed by taking infinite product.

The results discussed in this post are from a paper by E. Michael (Example 1.1 in [2]). This paper had been discussed previously in this blog (see Two footnotes in a paper of E. Michael).

As discussed before, let \mathbb{R} be the real number line. Let \mathbb{P} be the set of all irrational numbers. Let \mathbb{Q}=\mathbb{R}-\mathbb{P}, the set of all rational numbers. Let \tau be the usual topology of the real line \mathbb{R}. The following is a base that defines the topology for the Michael line \mathbb{M}.

    \mathcal{B}=\tau \cup \left\{\left\{ x \right\}: x \in \mathbb{P}\right\}

Other basic results about the Michael line are discussed in Michael Line Basics.

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Paracompactness

A space X is paracompact if every open cover of X has a locally finite open refinement. In proving \mathbb{M}^n is paracompact, we need two basic results about paracompactness. The proof of Theorem 1 can be found in [1] (Theorem 5.1.11 in page 302) or in [3] (Theorem 20.7 in page 146). We prove Theorem 2.

Theorem 1
Let X be a regular space. Then X is paracompact if and only if every open cover of X has an open \sigma-locally finite refinement, i.e., the following holds:

    every open cover \mathcal{U} of X has an open refinement \bigcup \limits_{j=1}^\infty \mathcal{V}_j such that each \mathcal{V}_j is a locally finite collection of open subsets of X.

Theorem 2
Let X be a regular space. Then X is paracompact if and only if the following hold:

    For each open cover \left\{U_t: t \in T \right\} of X, there exists a locally finite open cover \left\{V_t: t \in T \right\} such that \overline{V_t} \subset U_t for each t \in T.

Proof of Theorem 2
The direction \Longleftarrow is clear.

\Longrightarrow Let X be paracompact. Let \mathcal{U}=\left\{U_t: t \in T \right\} be an open cover of X. By regularity, there is an open cover \mathcal{W} of X such that \left\{\overline{W}: W \in \mathcal{W} \right\} refines \mathcal{U}. Since X is paracompact, \mathcal{W} has an open locally finite refinement \mathcal{H}=\left\{H_a: a \in A \right\}.

We now tie \mathcal{H} to the original open cover \mathcal{U}. For each a \in A, choose f(a) \in T such that \overline{H_a} \subset U_{f(a)}. Now, we go the opposite direction, i.e., for each t \in T, consider all a \in A such that \overline{H_a} \subset U_{f(a)}=U_t. For each t \in T, let V_t be defined by:

    V_t=\bigcup \left\{H_a: a \in A \text { and } \overline{H_a} \subset U_{f(a)}=U_t \right\}=\bigcup \limits_{f(a)=t} H_a

Each V_t is open since it is a union of open sets. Since \mathcal{H} is locally finite, any subcollection of \mathcal{H} is closure preserving. We have:

    \overline{V_t}=\bigcup \left\{\overline{H_a}: a \in A \text { and } \overline{H_a} \subset U_{f(a)}=U_t \right\}=\bigcup \limits_{f(a)=t} \overline{H_a}

Thus we have \overline{V_t} \subset U_t for all t \in T. Since \mathcal{H} is locally finite, \left\{V_t: t \in T \right\} is locally finite. Furthermore, \left\{V_t: t \in T \right\} is clearly a cover of X. Thus Theorem 2 is established. \blacksquare

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Finite Products

What makes the Michael line finitely productive for paracompactness is that all but countably many points in \mathbb{M} are isolated. The paracompactness of the finite products of the Michael line follows from Theorem 4 (see Corollary 5 below). Lemma 3 is used in proving Theorem 4.

Lemma 3
Let X be a space such that all but countably many points of X are isolated. Let A be the set of all isolated points of X. Then for each n=2,3,4,\cdots, X-A^n can be expressed as the following:

    X-A^n=\bigcup \limits_{k=1}^\infty Y_k satisfying the following:

    • For each k, Y_k is homeomorphic to X^{n-1}.
    • For each k, there exists a continuous map F_k:X^n \rightarrow Y_k such that F_k \upharpoonright Y_k is the indentity map.

Proof of Lemma 3
Note that X-A is countable. Fix n \in \left\{2,3,4,\cdots \right\}. For each x \in X^n, express x=(x_1,x_2,\cdots,x_n). For each i<n and for each a \in X-A, let Y_{i,a}=\left\{x \in X^n: x_i=a \right\} (the i^{th} coordinate is fixed and the other n-1 coordinates are free to vary). There are only countably many such Y_{i,a}. Clearly X-A^n is the union of all Y_{i,a}. Furthermore, each Y_{i,a} is homeomorphic to X^{n-1}.

Define F_{i,a}:X^n \rightarrow Y_{i,a} by mapping each (x_1,x_2,\cdots,x_i,\cdots,x_n) to (x_1,x_2,\cdots,a,\cdots,x_n). In other words, the i^{th} coordinate of each point is mapped to the fixed point a. This is a continuous map since it is a projection map. It is clear that when this map is restricted to Y_{i,a}, it is the identity map.

When we order all Y_{i,a} in a sequence Y_1,Y_2,Y_3,\cdots, the lemma is established. \blacksquare

Theorem 4
Let X be a regular space such that all but countably many points of X are isolated. Then X^n is paracompact for each n=1,2,3,\cdots.

Proof of Theorem 4
We prove X^n is paracompact by induction on n. Let A be the set of all isolated points of X. Let B=X-A.

First we show X is paracompact. Let \mathcal{U} be an open cover of X. Enumerate B by \left\{b_1,b_2,b_3,\cdots \right\}. For each i, choose U_i \in \mathcal{U} with b_i \in U_i. Let \mathcal{U}_i=\left\{U_i \right\}. Let \mathcal{A} be the set of all \left\{ x \right\} where x \notin \bigcup \limits_{i=1}^\infty U_i. Then \mathcal{A} \cup \mathcal{U}_1 \cup \mathcal{U}_2 \cup \mathcal{U}_3 \cup \cdots is an open \sigma-locally finite refinement of \mathcal{U}. By Theorem 1, X is paracompact.

Suppose that X^{n-1} is paracompact where n \ge 2. Let \mathcal{U}=\left\{U_t: t \in T \right\} be an open cover of X^n. By Lemma 3, there exist Y_1,Y_2,Y_3,\cdots, all subspaces of X^n, such that:

    X-A^n=\bigcup \limits_{k=1}^\infty Y_k satisfying the following:

    • For each k, Y_k is homeomorphic to X^{n-1}.
    • For each k, there exists a continuous map F_k:X^n \rightarrow Y_k such that F_k \upharpoonright Y_k is the indentity map.

Fix k where k=1,2,3,\cdots. Note that \left\{U_t \cap Y_k: t \in T \right\} is an open cover of Y_k. Since each Y_k is paracompact, using Theorem 2, we can find a locally finite open refinement \left\{V_t: t \in T \right\} (open in Y_k) of \left\{U_t \cap Y_k: t \in T \right\} such that V_t \subset U_t \cap Y_k for each t \in T. For each t, let W_{k,t}=F_k^{-1}(V_t) \cap U_t.

Then \left\{W_{k,t}: t \in T \right\} is a locally finite collection of open subsets of X^n covering Y_k. Since the map F_k is identity on Y_k, V_t \subset F_k^{-1}(V_t). Thus \left\{W_{k,t}: t \in T \right\} is a cover of Y_k. To see that it is locally finite, let z \in X^n. We have F_k(z) \in Y_k. There exists V (open in Y_k) such that F_k(z) \in V and V only meets finitely many V_t, say, V_{t_1},V_{t_2},\cdots,V_{t_m}. Consider the following the open sets:

    W_{k,t_1}=F_k^{-1}(V_{t_1}) \cap U_{t_1}
    W_{k,t_2}=F_k^{-1}(V_{t_2}) \cap U_{t_2}

      \cdot
      \cdot
      \cdot

    W_{k,t_m}=F_k^{-1}(V_{t_m}) \cap U_{t_m}

F_k^{-1}(V) is an open set containing z. It follows that the open sets W_{k,t} that F_k^{-1}(V) can meet are limited to ones listed above. For any s \in T where s \notin \left\{t_1,t_2,\cdots,t_m \right\}, V_s \cap V=\varnothing. Thus F_k^{-1}(V) \cap F_k^{-1}(V_s)=\varnothing and F_k^{-1}(V) \cap W_s=\varnothing. Thus \left\{W_{k,t}: t \in T \right\} is locally finite in X^n.

For each k=1,2,3,\cdots, let \mathcal{W}_k be \left\{W_{k,t}: t \in T \right\}, which is an open locally finite collection covering Y_k (as shown above). All together \bigcup \limits_{k=1}^\infty \mathcal{W}_k is an open \sigma-locally finite collection covering X^n-A^n. Let \mathcal{A} be the set of all \left\{ x \right\} where x \in A^n. Then \mathcal{A} \cup \bigcup \limits_{k=1}^\infty \mathcal{W}_k is an open \sigma-locally finite refinement of \mathcal{U}. By Theorem 1, X^n is paracompact. \blacksquare

Corollary 5
For each n=1,2,3,\cdots, \mathbb{M}^n is paracompact.

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Infinite Products

Let \mathbb{P} be the space of the irrational numbers with the Euclidean topology. Let \omega be the set of all nonnegative integers. We now show that \mathbb{M}^\omega, the product of countably and infinitely many copies of the Michael line, is not normal. Before doing that, we point out that when \omega is considered a discrete space, \omega^\omega, the product of countably and infinitely many copies of \omega, is homeomorphic to \mathbb{P} (Thinking about the Space of Irrationals Topologically).

Let D be a countably infinite subset of the Michael line \mathbb{M} such that D is closed and discrete. As discussed above, D^\omega is a homeomorphic copy of \mathbb{P}. Furthermore D^\omega is a closed subset of \mathbb{M}^\omega. Thus \mathbb{M} \times \mathbb{M}^\omega contains \mathbb{M} \times D^\omega \cong \mathbb{M} \times \mathbb{P} as a closed subspace. Since \mathbb{M} \times \mathbb{P} is not normal, \mathbb{M} \times \mathbb{M}^\omega is not normal. On the other hand, \mathbb{M}^\omega is homeomorphic to \mathbb{M} \times \mathbb{M}^\omega. Thus \mathbb{M}^\omega is not normal. \blacksquare

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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.

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\copyright \ \ 2012

Michael Line Basics

Like the Sorgenfrey line, the Michael line is a classic counterexample that is covered in standard topology textbooks and in first year topology courses. This easily accessible example helps transition students from the familiar setting of the Euclidean topology on the real line to more abstract topological spaces. One of the most famous results regarding the Michael line is that the product of the Michael line with the space of the irrational numbers is not normal. Thus it is an important example in demonstrating the pathology in products of paracompact spaces. The product of two paracompact spaces does not even have be to be normal, even when one of the factors is a complete metric space. In this post, we discuss this classical result and various other basic results of the Michael line.

Let \mathbb{R} be the real number line. Let \mathbb{P} be the set of all irrational numbers. Let \mathbb{Q}=\mathbb{R}-\mathbb{P}, the set of all rational numbers. Let \tau be the usual topology of the real line \mathbb{R}. The following is a base that defines a topology on \mathbb{R}.

    \mathcal{B}=\tau \cup \left\{\left\{ x \right\}: x \in \mathbb{P}\right\}

The real line with the topology generated by \mathcal{B} is called the Michael line and is denoted by \mathbb{M}. In essense, in \mathbb{M}, points in \mathbb{P} are made isolated and points in \mathbb{Q} retain the usual Euclidean open sets.

The Euclidean topology \tau is coarser (weaker) than the Michael line topology (i.e. \tau being a subset of the Michael line topology). Thus the Michael line is Hausdorff. Since the Michael line topology contains a metrizable topology, \mathbb{M} is submetrizable (submetrized by the Euclidean topology). It is clear that \mathbb{M} is first countable. Having uncountably many isolated points, the Michael line does not have the countable chain condition (thus is not separable). The following points are discussed in more details.

  1. The space \mathbb{M} is paracompact.
  2. The space \mathbb{M} is not Lindelof.
  3. The extent of the space \mathbb{M} is c where c is the cardinality of the real line.
  4. The space \mathbb{M} is not locally compact.
  5. The space \mathbb{M} is not perfectly normal, thus not metrizable.
  6. The space \mathbb{M} is not a Moore space, but has a G_\delta-diagonal.
  7. The product \mathbb{M} \times \mathbb{P} is not normal where \mathbb{P} has the usual topology.
  8. The product \mathbb{M} \times \mathbb{P} is metacompact.
  9. The space \mathbb{M} has a point-countable base.
  10. For each n=1,2,3,\cdots, the product \mathbb{M}^n is paracompact.
  11. The product \mathbb{M}^\omega is not normal.
  12. There exist a Lindelof space L and a separable metric space W such that L \times W is not normal.

Results 10, 11 and 12 are shown in some subsequent posts.

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Baire Category Theorem

Before discussing the Michael line in greater details, we point out one connection between the Michael line topology and the Euclidean topology on the real line. The Michael line topology on \mathbb{Q} coincides with the Euclidean topology on \mathbb{Q}. A set is said to be a G_\delta-set if it is the intersection of countably many open sets. By the Baire category theorem, the set \mathbb{Q} is not a G_\delta-set in the Euclidean real line (see the section called “Discussion of the Above Question” in the post A Question About The Rational Numbers). Thus the set \mathbb{Q} is not a G_\delta-set in the Michael line. This fact is used in Result 5.

The fact that \mathbb{Q} is not a G_\delta-set in the Euclidean real line implies that \mathbb{P} is not an F_\sigma-set in the Euclidean real line. This fact is used in Result 7.

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Result 1

Let \mathcal{U} be an open cover of \mathbb{M}. We proceed to derive a locally finite open refinement \mathcal{V} of \mathcal{U}. Recall that \tau is the usual topology on \mathbb{R}. Assume that \mathcal{U} consists of open sets in the base \mathcal{B}. Let \mathcal{U}_\tau=\mathcal{U} \cap \tau. Let Y=\cup \mathcal{U}_\tau. Note that Y is a Euclidean open subspace of the real line (hence it is paracompact). Then there is \mathcal{V}_\tau \subset \tau such that \mathcal{V}_\tau is a locally finite open refinement \mathcal{V}_\tau of \mathcal{U}_\tau and such that \mathcal{V}_\tau covers Y (locally finite in the Euclidean sense). Then add to \mathcal{V}_\tau all singleton sets \left\{ x \right\} where x \in \mathbb{M}-Y and let \mathcal{V} denote the resulting open collection.

The resulting \mathcal{V} is a locally finite open collection in the Michael line \mathbb{M}. Furthermore, \mathcal{V} is also a refinement of the original open cover \mathcal{U}. \blacksquare

A similar argument shows that \mathbb{M} is hereditarily paracompact.

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Result 2

To see that \mathbb{M} is not Lindelof, observe that there exist Euclidean uncountable closed sets consisting entirely of irrational numbers (i.e. points in \mathbb{P}). For example, it is possible to construct a Cantor set entirely within \mathbb{P}.

Let C be an uncountable Euclidean closed set consisting entirely of irrational numbers. Then this set C is an uncountable closed and discrete set in \mathbb{M}. In any Lindelof space, there exists no uncountable closed and discrete subset. Thus the Michael line \mathbb{M} cannot be Lindelof. \blacksquare

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Result 3

The argument in Result 2 indicates a more general result. First, a brief discussion of the cardinal function extent. The extent of a space X is the smallest infinite cardinal number \mathcal{K} such that every closed and discrete set in X has cardinality \le \mathcal{K}. The extent of the space X is denoted by e(X). When the cardinal number e(X) is e(X)=\aleph_0 (the first infinite cardinal number), the space X is said to have countable extent, meaning that in this space any closed and discrete set must be countably infinite or finite. When e(X)>\aleph_0, there are uncountable closed and discrete subsets in the space.

It is straightforward to see that if a space X is Lindelof, the extent is e(X)=\aleph_0. However, the converse is not true.

The argument in Result 2 exhibits a closed and discrete subset of \mathbb{M} of cardinality c. Thus we have e(\mathbb{M})=c. \blacksquare

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Result 4

The Michael line \mathbb{M} is not locally compact at all rational numbers. Observe that the Michael line closure of any Euclidean open interval is not compact in \mathbb{M}. \blacksquare

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Result 5

A set is said to be a G_\delta-set if it is the intersection of countably many open sets. A space is perfectly normal if it is a normal space with the additional property that every closed set is a G_\delta-set. In the Michael line \mathbb{M}, the set \mathbb{Q} of rational numbers is a closed set. Yet, \mathbb{Q} is not a G_\delta-set in the Michael line (see the discussion above on the Baire category theorem). Thus \mathbb{M} is not perfectly normal and hence not a metrizable space. \blacksquare

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Result 6

The diagonal of a space X is the subset of its square X \times X that is defined by \Delta=\left\{(x,x): x \in X \right\}. If the space is Hausdorff, the diagonal is always a closed set in the square. If \Delta is a G_\delta-set in X \times X, the space X is said to have a G_\delta-diagonal. It is well known that any metric space has G_\delta-diagonal. Since \mathbb{M} is submetrizable (submetrized by the usual topology of the real line), it has a G_\delta-diagonal too.

Any Moore space has a G_\delta-diagonal. However, the Michael line is an example of a space with G_\delta-diagonal but is not a Moore space. Paracompact Moore spaces are metrizable. Thus \mathbb{M} is not a Moore space. For a more detailed discussion about Moore spaces, see Sorgenfrey Line is not a Moore Space. \blacksquare

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Result 7

We now show that \mathbb{M} \times \mathbb{P} is not normal where \mathbb{P} has the usual topology. In this proof, the following two facts are crucial:

  • The set \mathbb{P} is not an F_\sigma-set in the real line.
  • The set \mathbb{P} is dense in the real line.

Let H and K be defined by the following:

    H=\left\{(x,x): x \in \mathbb{P} \right\}
    K=\mathbb{Q} \times \mathbb{P}.

The sets H and K are disjoint closed sets in \mathbb{M} \times \mathbb{P}. We show that they cannot be separated by disjoint open sets. To this end, let H \subset U and K \subset V where U and V are open sets in \mathbb{M} \times \mathbb{P}.

To make the notation easier, for the remainder of the proof of Result 7, by an open interval (a,b), we mean the set of all real numbers t with a<t<b. By (a,b)^*, we mean (a,b) \cap \mathbb{P}. For each x \in \mathbb{P}, choose an open interval U_x=(a,b)^* such that \left\{x \right\} \times U_x \subset U. We also assume that x is the midpoint of the open interval U_x. For each positive integer k, let P_k be defined by:

    P_k=\left\{x \in \mathbb{P}: \text{ length of } U_x > \frac{1}{k} \right\}

Note that \mathbb{P}=\bigcup \limits_{k=1}^\infty P_k. For each k, let T_k=\overline{P_k} (Euclidean closure in the real line). It is clear that \bigcup \limits_{k=1}^\infty P_k \subset \bigcup \limits_{k=1}^\infty T_k. On the other hand, \bigcup \limits_{k=1}^\infty T_k \not\subset \bigcup \limits_{k=1}^\infty P_k=\mathbb{P} (otherwise \mathbb{P} would be an F_\sigma-set in the real line). So there exists T_n=\overline{P_n} such that \overline{P_n} \not\subset \mathbb{P}. So choose a rational number r such that r \in \overline{P_n}.

Choose a positive integer j such that \frac{2}{j}<\frac{1}{n}. Since \mathbb{P} is dense in the real line, choose y \in \mathbb{P} such that r-\frac{1}{j}<y<r+\frac{1}{j}. Now we have (r,y) \in K \subset V. Choose another integer m such that \frac{1}{m}<\frac{1}{j} and (r-\frac{1}{m},r+\frac{1}{m}) \times (y-\frac{1}{m},y+\frac{1}{m})^* \subset V.

Since r \in \overline{P_n}, choose x \in \mathbb{P} such that r-\frac{1}{m}<x<r+\frac{1}{m}. Now it is clear that (x,y) \in V. The following inequalities show that (x,y) \in U.

    \lvert x-y \lvert \le \lvert x-r \lvert + \lvert r-y \lvert < \frac{1}{m}+\frac{1}{j} \le \frac{2}{j} < \frac{1}{n}

The open interval U_x is chosen to have length > \frac{1}{n}. Since \lvert x-y \lvert < \frac{1}{n}, y \in U_x. Thus (x,y) \in \left\{ x \right\} \times U_x \subset U. We have shown that U \cap V \ne \varnothing. Thus \mathbb{M} \times \mathbb{P} is not normal. \blacksquare

Remark
As indicated above, the proof of Result 7 hinges on two facts about \mathbb{P}, namely that it is not an F_\sigma-set in the real line and it is dense in the real line. We can modify the construction of the Michael line by using other partition of the real line (where one set is isolated and its complement retains the usual topology). As long as the set D that is isolated is not an F_\sigma-set in the real line and is dense in the real line, the same proof will show that the product of the modified Michael line and the space D (with the usual topology) is not normal. This will be how Result 12 is derived.

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Result 8

The product \mathbb{M} \times \mathbb{P} is not paracompact since it is not normal. However, \mathbb{M} \times \mathbb{P} is metacompact.

A collection of subsets of a space X is said to be point-finite if every point of X belongs to only finitely many sets in the collection. A space X is said to be metacompact if each open cover of X has an open refinement that is a point-finite collection.

Note that \mathbb{M} \times \mathbb{P}=(\mathbb{P} \times \mathbb{P}) \cup (\mathbb{Q} \times \mathbb{P}). The first \mathbb{P} in \mathbb{P} \times \mathbb{P} is discrete (a subspace of the Michael line) and the second \mathbb{P} has the Euclidean topology.

Let \mathcal{U} be an open cover of \mathbb{M} \times \mathbb{P}. For each a=(x,y) \in \mathbb{Q} \times \mathbb{P}, choose U_a \in \mathcal{U} such that a \in U_a. We can assume that U_a=A \times B where A is a usual open interval in \mathbb{R} and B is a usual open interval in \mathbb{P}. Let \mathcal{G}=\lbrace{U_a:a \in \mathbb{Q} \times \mathbb{P}}\rbrace.

Fix x \in \mathbb{P}. For each b=(x,y) \in \lbrace{x}\rbrace \times \mathbb{P}, choose some U_b \in \mathcal{U} such that b \in U_b. We can assume that U_b=\lbrace{x}\rbrace \times B where B is a usual open interval in \mathbb{P}. Let \mathcal{H}_x=\lbrace{U_b:b \in \lbrace{x}\rbrace \times \mathbb{P}}\rbrace.

As a subspace of the Euclidean plane, \bigcup \mathcal{G} is metacompact. So there is a point-finite open refinement \mathcal{W} of \mathcal{G}. For each x \in \mathbb{P}, \mathcal{H}_x has a point-finite open refinement \mathcal{I}_x. Let \mathcal{V} be the union of \mathcal{W} and all the \mathcal{I}_x where x \in \mathbb{P}. Then \mathcal{V} is a point-finite open refinement of \mathcal{U}.

Note that the point-finite open refinement \mathcal{V} may not be locally finite. The vertical open intervals in \lbrace{x}\rbrace \times \mathbb{P}, x \in \mathbb{P} can “converge” to a point in \mathbb{Q} \times \mathbb{P}. Thus, metacompactness is the best we can hope for. \blacksquare

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Result 9

A collection of sets is said to be point-countable if every point in the space belongs to at most countably many sets in the collection. A base \mathcal{G} for a space X is said to be a point-countable base if \mathcal{G}, in addition to being a base for the space X, is also a point-countable collection of sets. The Michael line is an example of a space that has a point-countable base and that is not metrizable. The following is a point-countable base for \mathbb{M}:

    \mathcal{G}=\mathcal{H} \cup \left\{\left\{ x \right\}: x \in \mathbb{P}\right\}

where \mathcal{H} is the set of all Euclidean open intervals with rational endpoints. One reason for the interest in point-countable base is that any countable compact space (hence any compact space) with a point-countable base is metrizable (see Metrization Theorems for Compact Spaces).

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Reference

  1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
  2. Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

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\copyright \ \ 2012

A proof about the Michael Line

The original post about the Michael line is now replaced by the following newer posts. You can go to these newer posts using the following links.

“Michael Line Basics”

“Finite and Countable Products of the Michael Line”

“Bernstein Sets and the Michael Line”

“The Michael Line and the Continuum Hypothesis”