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Revision difference : a shorter proof: Martin's axiom and the continuum hypothesis |
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Version 6 |
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| This is another, shorter, proof for the fact that $MA_{\aleph_0}$ always holds. |
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 $(P,\leq)$, than 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 $\mathcal D$ 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 $(P,\leq)$, than 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 $p \in G$. |
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 $p \in G$. |
| \begin{proof} |
\begin{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. |
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. |
| \end{proof} |
\end{proof} |
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