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 such that the product of 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 ).
Let be the Michael line. Let be the set of irrational numbers with the usual topology inherited from the real line. It is a classical result that the product is not normal (see “Michael Line Basics”). The Lindelof example we wish to discuss is an uncountable Lindelof subspace of such that contains the set of rational numbers. The same proof that is not normal will show that is not normal.
See the following posts for a basic discussion of the Michael line:
“Michael Line Basics”
“Finite and Countable Products of the Michael Line”
“Bernstein Sets and the Michael Line”
The Lindelof space 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 be a space. Let . The set is said to be nowhere dense in if for every non-empty open subset of , there is a non-empty open subset of such that and misses (equivalently, the closure of has no interior). The set is of first category in if it is the union of countably many nowhere dense sets.
To define Luzin sets, we focus on the Euclidean space . Let . The set is said to be a Luzin set if for every set that is of first category in the real line, 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 . We have the following theorem.
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 . Let be the set of all closed nowhere dense sets in the real line. Choose a real number to start. For each with , choose a real number not in the following set:
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 can always be selected at each . Then is a Luzin set.
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 be an uncountable Luzin set. Then if is a closed nowhere dense set in the real line, then contains all but countably many points of . Furthermore, if are closed nowhere dense subsets of the real line, then contains all but countably many points of the Luzin set .
Note that the set 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 in the preceding paragraph is a dense -set. Thus the complement of the union of countably many closed nowhere dense sets is a dense -set. Thus the observation in the preceding paragraph gives the following proposition:
Given an uncountable Luzin set and given a dense subset of the real line, contains all but countably many points of .
In fact, Proposition 2 not only hold in the real line, it also holds in any uncountable dense subset of the real line.
Let be an uncountable Luzin set. Let be uncountable and dense in the real line such that is uncountable. Given a dense subset of , contains all but countably many points of .
Proof of Proposition 3
We want to show that can only contain countably many points of . Let where each is open and dense in . Then for each , let be open in the real line such that . Each is open and dense in the real line. Thus contains all but countably many points of the Luzin set . Note the following set inclusion:
Suppose that contains uncountably many points of . Then these points, except for countably many points, must belong to . The above set inclusion shows that these points must belong to too, a contradiction. Thus can only contain countably many points of , equivalently the -set contains all but countably many points of .
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 is not normal.
Let be an uncountable Luzin set such that . Then cannot be an -set in the Euclidean space , equivalently cannot be a -set in the space .
Proof of Proposition 4
By Proposition 3, any dense -subset of must be co-countable.
The following proposition is another useful observation about Luzin sets. Let . Let be a countable dense subset of the real line. The set is said to be concentrated about if for every open subset of the real line such that , contains all but countably many points of . The following proposition can be readily checked based on the definition of Luzin sets.
For any , is a Luzin set if and only if is concentrated about every countable dense subset of the real line.
Lindelof Subspace of The Michael Line
Let be an uncountable Luzin set. We can assume that is dense in the real line. If not, just add a countble subset of that is dense in the real line. Let . It is clear that adding countably many points to a Luzin set still results in a Luzin set. Thus is also a Luzin set. Now consider as a subspace of the Michael line . Then points of are discrete and points in have Euclidean open neighborhoods. By Proposition 5, the set is concentrated about every countable dense subset of the real line. In particular, it is concentrated about . Thus as a subspace of the Michael line, is a Lindelof space, since every open set containing contains all but countably many points of .
The Non-Normal Product
We highlight the following two facts about the Luzin set as discussed in the preceding section.
- is not an -set in (as Euclidean space).
- 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 we start with is dense. Recall that when thinking of as a subspace of the Michael line, are isolated and retains the usual real line open sets. Because of the above two bullet points, is not normal. The proof that is not normal is the corollary of the proof that is not normal. Note that in the proof for showing is not normal, the two crucial points about the proof are that the isolated points of the Michael line cannot be an -set and are dense in the real line (found in “Michael Line Basics”).
The example that we construct here was hinted in footnote 4 in . In a later publication, E. Michael constructed an uncountable Lindelof subspace of the Michael line (see Lemma 3.1 in ). That construction should produce a similar set as the Luzin sets since the approach in  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  was to pick points in dense -sets in a transfinite induction process.
A Michael space is a Lindelof space whose product with 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  (page 160). A Michael space can also be constructed using Martin’s axiom (see ).
A space is said to be a productively Lindelof space if its product with every Lindelof space is Lindelof. Is 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 is not productively Lindelof in ZFC alone?
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