Hereditarily supercompact spaces (Q386207)

A topological space \(X\) is said to be \textit{supercompact} if it has a subbase such that each cover of \(X\) by elements of this subbase has a two-element subcover (so, by the well-known Alexander Lemma, supercompact spaces are compact). Compact metrizable spaces are supercompact [\textit{M. Strok} and \textit{A. Szymanski}, Fundam. Math. 89, 81--91 (1975; Zbl 0316.54030)].NEWLINENEWLINEIn the paper under review, the authors investigate \textit{hereditarily supercompact spaces} -- i.e., spaces with the property that every closed subspace is supercompact. The main results of the paper are:NEWLINENEWLINE\noindent (1) A dyadic compact space is hereditarily supercompact if, and only if, it is metrizable.NEWLINENEWLINE\noindent (2) Under \({MA} + \neg CH\), each separable hereditarily supercompact space is hereditarily separable and hereditarily Lindelöf. This implies that under \({MA} + \neg CH\) a scattered compact space is metrizable if, and only if, it is separable and hereditarily supercompact.NEWLINENEWLINE\noindent (3) Hereditary supercompactness is not productive: the product \([0,1] \times \alpha D\) of the closed interval and the one-point compactification \(\alpha D\) of a discrete space \(D\) of cardinality \(|D| \geqslant \text{non}(\mathcal{M})\) is not hereditarily supercompact. In this result, \(\text{non}(\mathcal{M})\) denotes the \textit{uniformity} of the ideal of meager sets, meaning that \(\text{non}(\mathcal{M})\) is the cardinality of the smallest non-meager subset of \(\mathbb{R}\).