Indiscrete rational extension for $\mathbb R$ (examp 66 in "Counterexamples in topology")

Question

As a continuation to this question:

Let $X$ be the Indiscrete rational extension for $\mathbb R$ (examp 66 page 88 in "Counterexamples in topology").

Let $\langle \mathcal{U}_n: n \in \mathbb{N} \rangle$ be a sequence of open covers of $X$. Can we always find a sequence $\langle F_n: n \in \mathbb{N} \rangle$ with each $F_n \in \mathcal{U}_n$ such that $\cup F_n$ is an open cover of $X$?

If this is not true for $\mathbb R$ with the regular topology. Will this imply that this is also not true for $X$?

Thank you!!

Answer 1

What you are asking about is called by many names in the literature. Subsets of $\mathbb{R}$ (usual topology) with this property were introduced by F. Rothberger as having property $C^{\prime\prime}$, and other places have since referred to it as have the Rothberger property. In Marion Scheepers's scheme of selection principles for open covers, it may be denoted $\mathsf{S}_1 ( \mathcal{O} , \mathcal{O} )$.

It is an easy observation that if $A \subseteq \mathbb{R}$ (with the usual topology) has the Rothberger property, then $A$ is strongly measure zero (meaning that for any sequence $\langle \varepsilon_n \rangle_{n \in \mathbb{N}}$ of positive reals there is a sequence $\langle I_n \rangle_{n \in \mathbb{N}}$ of open intervals such that $A \subseteq \bigcup_n I_n$ and $\mathrm{length} ( I_n ) \leq \varepsilon_n$ for all $n$).

proof. Let $\langle \varepsilon_n \rangle_n$ be given, and for each $n$ let $\mathcal{U}_n$ be the family of all open intervals in $\mathbb{R}$ of length $\varepsilon_n$. Clearly each $\mathcal{U}_n$ covers $A$, and so by the Rothberger property we may choose $I_n \in \mathcal{U}_n$ such that $A \subseteq \bigcup_n I_n$.

It follows immediately that $\mathbb{R}$ (with the usual topology) does not have the Rothberger property. (A result of R. Laver states that it is consistent with $\mathsf{ZFC}$ that every strong measure zero subset of $\mathbb{R}$ is countable, so it's a fool's errand to find non-$\sigma$-compact subsets of $\mathbb{R}$ with the Rothberger property.)

As the Indiscrete Rational Extension topology on $\mathbb{R}$ is finer than the usual topology on $\mathbb{R}$, the same result will show that this space also fails to have the desired proeprty.