# A ZFC example of a Menger space which is not Scheepers

$$\Omega$$: The collection of all $$\omega$$-covers of a space $$X$$. An open cover $$\mathcal U$$ of $$X$$ is said to be $$\omega$$-cover if $$X\notin\mathcal U$$ and for each finite $$F\subseteq X$$ there exists a $$U\in\mathcal U$$ such that $$F\subseteq U$$.

1. A space $$X$$ is said to have the Menger property if for each sequence $$(\mathcal U_n)$$ of open covers of $$X$$ there exists a sequence $$(\mathcal V_n)$$ such that for each $$n$$ $$\mathcal V_n$$ is a finite subset of $$\mathcal U_n$$ and $$\{\cup\mathcal V_n : n\in\mathbb N\}$$ covers $$X$$.

2. A space $$X$$ is said to have the Scheepers property if for each sequence $$(\mathcal U_n)$$ of open covers of $$X$$ there exists a sequence $$(\mathcal V_n)$$ such that for each $$n$$ $$\mathcal V_n$$ is a finite subset of $$\mathcal U_n$$ and $$\{\cup\mathcal V_n : n\in\mathbb N\}\in\Omega$$ or $$\cup\mathcal V_n=X$$ for some $$n$$.

A space $$X$$ is said to be Menger (respectively, Scheepers) if $$X$$ has the Menger (respectively, Scheepers) property. It is well known that every Scheepers space is Menger but the coverse is not true (see here). We are looking for a ZFC example of a Menger space which is not Scheepers (here, it means without any additional set-theoretic assumption, is there a Menger space which is not Scheepers?).

Spaces with all finite powers Menger are $$S_\text{fin}(\Omega,\Omega)$$, and in particular Scheepers.