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 $G\cap D\neq\varnothing$ 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. ∎

Title	a shorter proof: Martin’s axiom and the continuum hypothesis
Canonical name	AShorterProofMartinsAxiomAndTheContinuumHypothesis
Date of creation	2013-03-22 13:53:58
Last modified on	2013-03-22 13:53:58
Owner	x_bas (2940)
Last modified by	x_bas (2940)
Numerical id	11
Author	x_bas (2940)
Entry type	Proof
Classification	msc 03E50
Defines	$\mathcal{D}$ -generic
Defines	generic
Defines	dense