Covering compacta by discrete and other separated sets (Q1013825)

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scientific article; zbMATH DE number 5546618
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Covering compacta by discrete and other separated sets
scientific article; zbMATH DE number 5546618

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    Covering compacta by discrete and other separated sets (English)
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    23 April 2009
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    A space is said to be crowded if it has no isolated points. In [Topology Appl. 154, No.~2, 283--286 (2007; Zbl 1107.54004)], \textit{I. Juhász} and \textit{J. van Mill} asked: If \(X\) is a compact Hausdorff crowded space, is it true that \(X\) cannot be covered by fewer than \(\mathfrak c\)-many discrete subspaces? The author proves: (2.2) If \(\kappa\) is an infinite cardinal, and \(X\) is the union of \(\kappa\)-many discrete subspaces, then so is any perfect image of \(X\). As a corollary to 2.2, he obtains an affirmative answer to the preceding question. He also notes that for \(\kappa=\omega\), 2.2 is due to [\textit{D. Burke} and \textit{R. Hansell}, Ann. N. Y. Acad. Sci. 788, 54--56 (1996; Zbl 0914.54029)]. In the final section the author studies coverings by right-separated and left-separated subspaces. Recall that a space \(Y\) is right (resp., left)-separated if there is a well-order on \(Y\) such that each point \(y\in Y\) has a neighborhood missing its successors (resp., predecessors). Two theorems he obtains are the following. If the continuum \(\mathfrak c\) is regular, then no crowded first-countable compact Hausdorff space can be covered by fewer than \(\mathfrak c\)-many right-separated subspaces. If the continuum \(\mathfrak c\) is regular, then no crowded first-countable compact Hausdorff space can be covered by fewer than \(\mathfrak c\)-many left-separated subspaces. He remarks that it is unsolved whether or not either first countability or the restriction on \(\mathfrak c\) is a necessary assumption.
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    compact spaces
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    discrete subspaces
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    left-separated
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    right-separated
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