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  AShorterProofMartinsAxiomAndTheContinuumHypothesis 2013-03-22 13:53:58 2013-03-22 13:53:58 x_bas (2940) x_bas (2940) 11 x_bas (2940) Proof msc 03E50 $\mathcal{D}$-generic generic dense