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	filtration of σ-algebras Filtrator Filtrator FindingAnotherParticularSolutionOfLinearODE finding another particular solution of linear ODE FindingEigenvalues finding eigenvalues FindingTheOrderOfAGroup finding the order of a group FineAndWilfsTheoremOnWords Fine and Wilf’s theorem on words Finite finite FiniteAndCountableDiscreteSpaces finite and countable discrete spaces FiniteChangesInConvergentSeries finite changes in convergent series FiniteCharacter finite character FiniteComplementTopology finite complement topology FiniteDifference finite difference FinitedimensionalLinearProblem finite-dimensional linear problem