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  • Set theory - Chapter 14: Forcing

Set Theory - Chapter 14: Forcing

14.1

• ⇒ If $p,q\in G$ then by (14.1) (iii) there exists $r\in G$ smaller than $p,q$. So $p,q$ are compatible.
• ⇐ Let $D$ the set of elements of $P$ that are either incompatible with $p$ or with $q$ or smaller than $p,q$. $D\in M$. Let $t\in P$. We shall prove that there exists $x\in D$ such that $x\le t$. If $t$ is incompatible with $p$ then take $x=t$. Otherwise, let $r$ a refinement of $p,t$. If $r$ is incompatible with $q$ then take $x=r$. Otherwise, let $s$ a refinement of $r,q$ and take $x=s$. So $D$ is dense. If $G$ is generic, let $y\in G\cap D$. Then $y\in G$ is compatible with both $p,q$ by the weaker property. Hence $y\in D$ is actually smaller than $p,q$.

14.2

• ⇒ trival (Any dense set is dense below any $p\in G$).
• ⇐ Let $D\in M$ be dense below $p$. $D\text{'}=D\cup \left\{x\in P|x\text{ incompatible with }p\right\}$is in $M$ and dense in $P$. Let $G$ be generic, then $D\text{'}\cap G$ is nonempty and contains an element compatible with $p\in G$. So such an element is actually in $D$.

14.3

• ⇒ trivial (Any open dense set is dense)
• ⇐ Let $D\in M$ be a dense set and define $D\text{'}\in M$ open dense by $D\text{'}=\left\{p\in P|\exists d\in Dp\le d\right\}$. Let $g\in D\text{'}\cap G$ and let $d\in D$ greater. Then by 14.1 (ii) $d\in G$.

14.4

• ⇒ trivial (Any dense set is predense)
• ⇐ Let$D\in M$ be a predense set and define $D\text{'}\in M$ dense by $D\text{'}=\left\{p\in P|\exists d\in Dp\le d\right\}$. Let $g\in D\text{'}\cap G$ and let $d\in D$ greater. Then by 14.1 (ii) $d\in G$.

14.5

• ⇒ trivial (Any maximal antichain is predense)
• ⇐ Let$D\in M$ be a predense set and define by induction $D\text{'}\in M$ a maximal antichain in $D$ by picking pairwise incompatible elements. Because any $p\notin D$ is compatible with some $d\in D$ and any $d\in D$ is compatible with some $d\text{'}\in D\text{'}$, then $D\text{'}$ is actually a maximal antichain in $P$. Then $G\cap D\supseteq G\cap D\text{'}\ne \varnothing$.

14.6

Let $F$ be a filter and $D=\left\{p\in P|p\notin F\right\}$. Let $p\in P$ and $q,r$ incompatible elements smaller than $p$. Then at least one of them is not in $F$ so is in $D$. Hence $D$ is dense.

Now, let $G$ be generic over $M$. If $G\in M$ then we can define the set above for $F=G$ and this set is in $M$. But $G\cap D$ is empty. A contradiction.

This page is W3C-compliant - Author: Frédéric WANG