\(P_{\lambda}\)-sets and skeletal mappings (Q2839280)
From MaRDI portal
| This is the item page for this Wikibase entity, intended for internal use and editing purposes. Please use this page instead for the normal view: \(P_{\lambda}\)-sets and skeletal mappings |
scientific article; zbMATH DE number 6184253
| Language | Label | Description | Also known as |
|---|---|---|---|
| English | \(P_{\lambda}\)-sets and skeletal mappings |
scientific article; zbMATH DE number 6184253 |
Statements
4 July 2013
0 references
sequence
0 references
filter
0 references
skeletal mapping
0 references
0 references
0.70996845
0 references
0.7036836
0 references
0.70081294
0 references
0 references
\(P_{\lambda}\)-sets and skeletal mappings (English)
0 references
The paper studies the topology on the set \(Seq\) of all finite sequences of natural numbers determined by \(P_\lambda\) filters. Given a collection of free filters \(\{\mathcal F_s\}_{s\in Seq}\), there is an induced topology on \(Seq\) obtained by using filters \(\mathcal F_s\) to define sections of open sets on the successors of \(s\). A review on the subject is given in [\textit{A. Błaszczyk}, Topol. Proc. 41, 65--84 (2013; Zbl 1286.54031)].NEWLINENEWLINEThe paper is divided in two parts. In the first part the authors prove that (under certain conditions) \(Seq\) is a \(P_\lambda\)-set in its Čech-Stone compactification \(\beta Seq\). The result is a generalization of several previously established results of the same kind.NEWLINENEWLINEIn the second part they study the connection to skeletal maps and nowhere constant maps. A continuous mapping is skeletal if the preimage of every open dense subset is dense. A mapping is nowhere constant if the preimage of every point is nowhere dense. The authors prove (under certain conditions) that if \(f: \beta Seq \to X\) is a skeletal map onto a Hausdorff space \(X\) then the set of isolated points in \(X\) is dense. Furthermore, they provide a partial answer to Problem 3.12 of [\textit{M. R. Burke}, Topology Appl. 103, No. 1, 95--110 (2000; Zbl 0958.54009)] by showing that \(Seq\) has a nowhere constant mapping into \(\mathbb{R}\) but does not have a skeletal mapping into \(\mathbb{R}\).
0 references