# Lindelof x Metric Needs Not Be Normal

In a previous post, an example for Paracompact x Metric needs not be normal was given. The example is product of the Michael Line and the space of irrationals. In another post, I showed that as long as the set of isolated points in the “Michael Line” construction is not an $F_\sigma-$set, we can obtain a non-normal product. I present here another example using the same technique, but this time the set of isolated points is a Bernstein set. This produces a Lindelof space and the cross product of this Lindelof space with the Bernstein set with the usual topology is not normal.

As in the previous post, let $H \subset \mathbb{R}$. Define a new topology on the real line $\mathbb{R}$ by using open sets of the form $U \cup V$ where $U$ is an open set in the usual topology and $V \subset H$. In this notation, the Michael Line is $\mathbb{R}(\mathbb{P})$ where $\mathbb{P}$ is the space of all irrational numbers. I proved in the previous post that the space $\mathbb{R}(H) \times H$ is normal if and only if the set $H$ is an $F_\sigma-$set in the real line. Thus $\mathbb{R}(\mathbb{P}) \times \mathbb{P}$ is not normal.

I consider the example $\mathbb{R}(S)$ where $S \subset \mathbb{R}$ and both $S$ and its complement $\mathbb{R}-S$ contain no uncountable compact subset of the line. Such a set is called a Bernstein set. The space $\mathbb{R}(S)$ is Lindelof. Let $\mathcal{W}$ be an open cover of $\mathbb{R}-S$ consisting of open sets in the usual topology. Then $\mathcal{W}$ has to cover all of the real line except for countably many points. Otherwise $\mathbb{R}-\cup \mathcal{W}$ is an uncountable closed set, which contains an uncountable compact set that is contained in $S$. Thus any open cover of $\mathbb{R}-S$ made up of usual open sets has to cover the real line except for countably many points. Now let $\mathcal{U}$ be an open cover of $\mathbb{R}(S)$. The usual open sets in $\mathcal{U}$ that covers $\mathbb{R}-S$ has a countable subcover. This countable subcover covers all of the real line except for countably many points. Thus $\mathcal{U}$ has a countable subcover.

Since a Berstein set cannot be an $F_\sigma-$set, $\mathbb{R}(S) \times S$ is not normal. Thus the cross product of a Lindelof space and a separable metric space needs not be normal. Is there a non-nonmal example of hereditarily Lindelof x separable metric space? If there is, it will not be using this “Michael Line” type construction. It is easy to see that $\mathbb{R}(H)$ is hereditarily Lindelof if and only if $H$ is countable.