The Suslinian number and other cardinal invariants of continua

T. Banakh¹ ; V. V. Fedorchuk² ; J. Nikiel³ ; M. Tuncali⁴

¹ Uniwersytet Humanistyczno-Przyrodniczy Jana Kochanowskiego Kielce, Poland and Department of Mathematics Ivan Franko Lviv National University Lviv, Ukraine
² Faculty of Mechanics and Mathematics Lomonosov Moscow State University Vorob'evy Gory, 1 Moscow, Russia
³ Instytut Matematyki i Informatyki Uniwersytet Opolski Oleska 48 45-052 Opole, Poland
⁴ Nipissing University North Bay, Ontario, Canada

Fundamenta Mathematicae, Tome 209 (2010) no. 1, pp. 43-57

Résumé

By the Suslinian number $\mathop{\rm Sln}(X)$ of a continuum $X$ we understand the smallest cardinal number $\kappa$ such that $X$ contains no disjoint family $\mathbb C$ of non-degenerate subcontinua of size $|\mathbb C|>\kappa$. For a compact space $X$, $\mathop{\rm Sln}(X)$ is the smallest Suslinian number of a continuum which contains a homeomorphic copy of $X$. Our principal result asserts that each compact space $X$ has weight $\le\mathop{\rm Sln}(X)^+$ and is the limit of an inverse well-ordered spectrum of length $\le \mathop{\rm Sln}(X)^+$, consisting of compacta with weight $\le\mathop{\rm Sln}(X)$ and monotone bonding maps. Moreover, $w(X)\le\mathop{\rm Sln}(X)$ if no $\mathop{\rm Sln}(X)^+$-Suslin tree exists. This implies that under the Suslin Hypothesis all Suslinian continua are metrizable, which answers a question of Daniel et al. [Canad. Math. Bull. 48 (2005)]. On the other hand, the negation of the Suslin Hypothesis is equivalent to the existence of a hereditarily separable non-metrizable Suslinian continuum. If $X$ is a continuum with $\mathop{\rm Sln}(X)2^{\aleph_0}$, then $X$ is 1-dimensional, has rim-weight $\le\mathop{\rm Sln}(X)$ and weight $w(X)\ge\mathop{\rm Sln}(X)$. Our main tool is the inequality $w(X)\le\mathop{\rm Sln}(X)\cdot w(f(X))$ holding for any light map $f:X\to Y$.

Détail

DOI : 10.4064/fm209-1-4

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