Absolutely strongly star-Menger spaces (Q1939205)

Following \textit{L. Kočinac} [Publ. Math. 55, No. 3--4, 421--431 (1999; Zbl 0932.54022)] a topological space \(X\) is called a strongly star-Menger space if, for each sequence \(({\mathcal U})_{n<\omega}\) of open covers of \(X\), there exists a sequence \((F_n)_{n<\omega}\) in \([X]^{<\omega}\) such that \(\{St(F_n,{\mathcal U}_n)\mid n<\omega\}\) is an open cover of \(X\). If, for each sequence \(({\mathcal U})_{n<\omega}\) of open covers of \(X\) and for each dense set \(D\) in \(X\), there exists a sequence \((F_n)_{n<\omega}\) in \([D]^{<\omega}\) such that \(\{St(F_n,{\mathcal U}_n)\mid n<\omega\}\) is an open cover of \(X\), then \(X\) is said to be an absolutely strongly star-Menger space (see \textit{A. Caserta}, \textit{G. Di Maio} and \textit{L. Kočinac} [Topology Appl. 158, No. 12, 1360--1368 (2011; Zbl 1229.54029)]). The main theorem of this paper states that, for a strongly star-Menger \(T_1\)-space \(X\), the Alexandroff duplicate \(A(X)\) of \(X\) is an absolutely strongly star-Menger space if and only if \(e(X)<\omega_1\). Every absolutely countably compact space in the sense of \textit{M. V. Matveev} [Topology Appl. 58, No. 1, 81--92 (1994; Zbl 0801.54021)] is an absolutely strongly star-Menger space. It is shown that the well-known Tychonoff plank is an absolutely strongly star-Menger space that is not absolutely countably compact. Moreover, an example is given of a countably compact Tychonoff space that is not absolutely strongly star-Menger. Every absolutely strongly star-Menger space is absolutely star-Lindelöf in the sense of \textit{M. Bonanzinga} [Quest. Answers Gen. Topology 16, No. 2, 79--104 (1998; Zbl 0931.54019)]. It is shown that there exists a Tychonoff absolutely star-Lindelöf space that is not absolutely strongly star-Menger. Closed subspaces of absolutely strongly star-Menger spaces need not be absolutely strongly star-Menger. In fact, an example is given of a Tychonoff countably compact, absolutely strongly star-Menger space with a regular-closed subspace that is not absolutely strongly star-Menger. The product of an absolutely strongly star-Menger space and a compact space need not be absolutely strongly star-Menger. However, if \(X\times Y\) is absolutely strongly star-Menger, then \(X\) and \(Y\) are absolutely strongly star-Menger.