a shorter proof: Martin’s axiom and the continuum hypothesis

This is another, shorter, proof for the fact that $MA_{\\aleph_{0}}$ always holds.

Let $(P,\\leq)$ be a partially ordered set and $\\mathcal{D}$ be a collection of subsets of $P$ . We remember that a filter $G$ on $(P,\\leq)$ is $\\mathcal{D}$ -generic if $G\\cap D\eq\\varnothing$ for all $D\\in\\mathcal{D}$ which are dense in $(P,\\leq)$ . (In this context “dense” means: If $D$ is dense in $(P,\\leq)$ , then for every $p\\in P$ there’s a $d\\in D$ such that $d\\leq p$ .)

Let $(P,\\leq)$ be a partially ordered set and $\\mathcal{D}$ a countable collection of dense subsets of $P$ . Then there exists a $\\mathcal{D}$ -generic filter $G$ on $P$ . Moreover, it could be shown that for every $p\\in P$ there’s such a $\\mathcal{D}$ -generic filter $G$ with $p\\in G$ .

Proof.

Let $D_{1},\\dots,D_{n},\\dots$ be the dense subsets in $\\mathcal{D}$ . Furthermore let $p_{0}=p$ . Now we can choose for every $1\\leq n<\\omega$ an element $p_{n}\\in P$ such that $p_{n}\\leq p_{n-1}$ and $p_{n}\\in D_{n}$ . If we now consider the set $G:=\\{q\\in P\\mid\\exists\\;n<\\omega\\text{ s.t. }p_{n}\\leq q\\}$ , then it is easy to check that $G$ is a $\\mathcal{D}$ -generic filter on $P$ and $p\\in G$ obviously. This completes the proof.∎

单词	AShorterProofMartinsAxiomAndTheContinuumHypothesis
释义	a shorter proof: Martin’s axiom and the continuum hypothesis This is another, shorter, proof for the fact that $MA_{\\aleph_{0}}$ always holds. Let $(P,\\leq)$ be a partially ordered set and $\\mathcal{D}$ be a collection of subsets of $P$ . We remember that a filter $G$ on $(P,\\leq)$ is $\\mathcal{D}$ -generic if $G\\cap D\eq\\varnothing$ for all $D\\in\\mathcal{D}$ which are dense in $(P,\\leq)$ . (In this context “dense” means: If $D$ is dense in $(P,\\leq)$ , then for every $p\\in P$ there’s a $d\\in D$ such that $d\\leq p$ .) Let $(P,\\leq)$ be a partially ordered set and $\\mathcal{D}$ a countable collection of dense subsets of $P$ . Then there exists a $\\mathcal{D}$ -generic filter $G$ on $P$ . Moreover, it could be shown that for every $p\\in P$ there’s such a $\\mathcal{D}$ -generic filter $G$ with $p\\in G$ . Proof. Let $D_{1},\\dots,D_{n},\\dots$ be the dense subsets in $\\mathcal{D}$ . Furthermore let $p_{0}=p$ . Now we can choose for every $1\\leq n<\\omega$ an element $p_{n}\\in P$ such that $p_{n}\\leq p_{n-1}$ and $p_{n}\\in D_{n}$ . If we now consider the set $G:=\\{q\\in P\\mid\\exists\\;n<\\omega\\text{ s.t. }p_{n}\\leq q\\}$ , then it is easy to check that $G$ is a $\\mathcal{D}$ -generic filter on $P$ and $p\\in G$ obviously. This completes the proof.∎
随便看	NormalSection normal section NormalSubgroup normal subgroup NormalSubgroupLatticeIsModular normal subgroup lattice is modular NormalSubgroupsFormSublatticeOfASubgroupLattice normal subgroups form sublattice of a subgroup lattice NormalSubgroupsOfTheSymmetricGroups normal subgroups of the symmetric groups NormalVariety normal variety NormAndSpectralRadiusInCalgebras norm and spectral radius in C^*-algebras NormAndTraceOfAlgebraicNumber norm and trace of algebraic number NormedAlgebra normed algebra NormedPlane normed plane NormedVectorSpace normed vector space NormEuclideanNumberField norm-Euclidean number field NotANumber