disjunction property of Wallman

A partially ordered set $\\mathfrak{A}$ with a least element $0$ has the disjunction property of Wallman if for every pair $(a,b)$ of elements of the poset, either $b\\leq a$ or there exists an element $c\\leq b$ such that $c\eq 0$ and $c$ has no nontrivial common predecessor with $a$ . That is, in the latter case, the only $x$ with $x\\leq a$ and $x\\leq c$ is $x=0$ .

For the case if the poset $\\mathfrak{A}$ is a $\\cap$ -semilattice disjunction property of Wallman is equivalent to every of the following three formulas:

1.
$\\forall a,b\\in\\mathfrak{A}:(\\{c\\in\\mathfrak{A}|c\\cap a\eq 0\\}=\\{c\\in\\mathfrak%{A}|c\\cap b\eq 0\\}\\Rightarrow a=b)$ ;
2.
$\\forall a,b\\in\\mathfrak{A}:(\\{c\\in\\mathfrak{A}|c\\cap a\eq 0\\}\\subseteq\\{c\\in%\\mathfrak{A}|c\\cap b\eq 0\\}\\Rightarrow a\\subseteq b)$ ;
3.
$\\forall a,b\\in\\mathfrak{A}:(a\\subset b\\Rightarrow\\{c\\in\\mathfrak{A}|c\\cap a%\eq 0\\}\\subset\\{c\\in\\mathfrak{A}|c\\cap b\eq 0\\})$ .

The proof of this equivalence can be found in http://www.mathematics21.org/binaries/filters.pdfthis online article.

单词	DisjunctionPropertyOfWallman
释义	disjunction property of Wallman A partially ordered set $\\mathfrak{A}$ with a least element $0$ has the disjunction property of Wallman if for every pair $(a,b)$ of elements of the poset, either $b\\leq a$ or there exists an element $c\\leq b$ such that $c\eq 0$ and $c$ has no nontrivial common predecessor with $a$ . That is, in the latter case, the only $x$ with $x\\leq a$ and $x\\leq c$ is $x=0$ . For the case if the poset $\\mathfrak{A}$ is a $\\cap$ -semilattice disjunction property of Wallman is equivalent to every of the following three formulas: 1. $\\forall a,b\\in\\mathfrak{A}:(\\{c\\in\\mathfrak{A}\|c\\cap a\eq 0\\}=\\{c\\in\\mathfrak%{A}\|c\\cap b\eq 0\\}\\Rightarrow a=b)$ ; 2. $\\forall a,b\\in\\mathfrak{A}:(\\{c\\in\\mathfrak{A}\|c\\cap a\eq 0\\}\\subseteq\\{c\\in%\\mathfrak{A}\|c\\cap b\eq 0\\}\\Rightarrow a\\subseteq b)$ ; 3. $\\forall a,b\\in\\mathfrak{A}:(a\\subset b\\Rightarrow\\{c\\in\\mathfrak{A}\|c\\cap a%\eq 0\\}\\subset\\{c\\in\\mathfrak{A}\|c\\cap b\eq 0\\})$ . The proof of this equivalence can be found in http://www.mathematics21.org/binaries/filters.pdfthis online article.
随便看	extravagant number Extremal extremal ExtremallyDisconnected extremally disconnected ExtremePoint extreme point ExtremeSubsetOfConvexSet extreme subset of convex set ExtremeValueTheorem extreme value theorem Extremum extremum ExtremumPointsOfFunctionOfSeveralVariables extremum points of function of several variables FaaDiBrunosFormula FaceOfAConvexSet face of a convex set FaceOfAConvexSetAlternativeDefinitionOf face of a convex set, alternative definition of FactorEmbeddable factor embeddable Factorial factorial FactorialBase