CMS/SMC
Canadian Mathematical Society
www.cms.math.ca
Canadian Mathematical Society
  location:  Publicationsjournals
Publications        
Search results

Search: MSC category 03E17 ( Cardinal characteristics of the continuum )

  Expand all        Collapse all Results 1 - 3 of 3

1. CMB 2013 (vol 57 pp. 119)

Mildenberger, Heike; Raghavan, Dilip; Steprans, Juris
Splitting Families and Complete Separability
We answer a question from Raghavan and Steprāns by showing that $\mathfrak{s} = {\mathfrak{s}}_{\omega, \omega}$. Then we use this to construct a completely separable maximal almost disjoint family under $\mathfrak{s} \leq \mathfrak{a}$, partially answering a question of Shelah.

Keywords:maximal almost disjoint family, cardinal invariants
Categories:03E05, 03E17, 03E65

2. CMB 2012 (vol 57 pp. 61)

Geschke, Stefan
2-dimensional Convexity Numbers and $P_4$-free Graphs
For $S\subseteq\mathbb R^n$ a set $C\subseteq S$ is an $m$-clique if the convex hull of no $m$-element subset of $C$ is contained in $S$. We show that there is essentially just one way to construct a closed set $S\subseteq\mathbb R^2$ without an uncountable $3$-clique that is not the union of countably many convex sets. In particular, all such sets have the same convexity number; that is, they require the same number of convex subsets to cover them. The main result follows from an analysis of the convex structure of closed sets in $\mathbb R^2$ without uncountable 3-cliques in terms of clopen, $P_4$-free graphs on Polish spaces.

Keywords:convex cover, convexity number, continuous coloring, perfect graph, cograph
Categories:52A10, 03E17, 03E75

3. CMB 2009 (vol 52 pp. 303)

Shelah, Saharon
A Comment on ``$\mathfrak{p} < \mathfrak{t}$''
Dealing with the cardinal invariants ${\mathfrak p}$ and ${\mathfrak t}$ of the continuum, we prove that ${\mathfrak m}={\mathfrak p} = \aleph_2\ \Rightarrow\ {\mathfrak t} =\aleph_2$. In other words, if ${\bf MA}_{\aleph_1}$ (or a weak version of this) holds, then (of course $\aleph_2\le {\mathfrak p}\le {\mathfrak t}$ and) ${\mathfrak p}=\aleph_2\ \Rightarrow\ {\mathfrak p}={\mathfrak t}$. The proof is based on a criterion for ${\mathfrak p}<{\mathfrak t}$.

Categories:03E17, 03E05, 03E50

© Canadian Mathematical Society, 2014 : https://cms.math.ca/