Let be the Michael line and let be the set of all irrational numbers with the Euclidean topology. In the post called “Michael Line Basics”, we show that the product 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 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 of the real line such that every uncountable closed subset of the real line has non-empty intersection with both and the complement of .

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

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

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

In the post “Michael Line Basics”, we give a proof that is not normal. This proof hinges on the same two facts about the set in the hypothesis in the above proposition. Thus the proof for the above proposition is just like the one for . Whenever we topologize the modified Michael line by using a non--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 be any Bernstein set. The set is clearly not an -set in the real line and is clearly dense in the real line. Then is not normal. Note that in , the set is discrete and its complement has the usual topology. To see that is Lindelof, note that any open cover of has a countable subcollection that covers . 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 (otherwise there would be an uncountable Euclidean closed set that misses ).

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 -set in the real line. Furthermore any uncountable subset of the real line must contain a Cantor set and thus cannot be a Bernstein set.

A similar example to 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 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**

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

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