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

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