## Franklin's Notes

### Forcing notion

Let $\mathrm{M}$ be a countable transitive model of ZFC. (The downward Löwenheim-Skolem Theorem implies the existence of such a model.) Then any set $P\in\mathrm{M}$, partially ordered by $\supset$, can be used as a forcing notion of $\mathrm{M}$.

If $p,q\in P$ with $p\subset q$, then $q$ is called an extension of $p$. Two elements $p,q\in P$ with a common extension are called compatible. A subset $D\subset P$ is called dense if every $p\in P$ has an extension in $D$, and it is called dense above $p$ if every extension of $p$ has an extension in $D$.

An ideal of $P$ is a subset $G\subset P$ which satisfies the following two conditions:

1. If $q\in G$ and $q\supset p\in P$ then $p\in G$. (This condition is called downward stability, and it means that $G$ contains all "reductions" of its elements.)
2. If $p_1,p_2\in G$ then there exists $q\in G$ with $p_1,p_2\subset q$. (This is called directedness, and it means that any two elements have a common extension.)

An ideal $G$ is called generic (relative to $\mathrm{M}$) if it has nontrivial intersection with every dense subset $D\subset P$ that lies in $\mathrm{M}$. For any forcing notion $P$ and $p_0\in P$, there exists a generic ideal $G\subset P$ which contains $p_0$. This can be constructed by enumerating the dense subsets of $P$ that lie in $\mathrm{M}$ (there are countably many, since $\mathrm{M}$ is countable) and building the ideal $G$ element by element in such a way that it intersects each of these dense subsets in at least one point.

Say that $P$ is trivial if it contains an element any two of whose extensions are compatible. If such an element exists, then it is easy to see that the set of all elements compatible with it comprises a generic ideal. Otherwise, if $P$ is nontrivial, then any generic ideal $G\subset P$ is not in $\mathrm{M}$. This is because if $P$ is nontrivial, then the complement of any ideal of $P$ must be dense. (For any $p\in P$ and ideal $I\subset P$, $p$ must have two incompatible extensions $p_1,p_2\in P$, and since they are compatible they cannot both be in $I$, meaning that $P\backslash I$ contains an extension of $p$ for all $p\in P$, and $P\backslash I$ is dense.) So if $P$ were nontrivial and $G$ were in $\mathrm{M}$, then $P\backslash G\in \mathrm{M}$ would be dense, contradicting the fact that $G$ intersects all dense subsets of $P$ in $\mathrm{M}$. This reasoning, reminiscent of the "diagonalization" technique used in the proof of Cantor's Theorem , gives us the following theorem:

Proposition 1. If $\mathrm{M}$ is a countable transitive model of ZFC and $P$ is a nontrivial forcing notion of $\mathrm{M}$ with generic ideal $G$, then $G\notin\mathrm{M}$.

Since $G$ is necessarily not an element of $\mathrm{M}$, it is a candidate for "extending" the model $\mathrm{M}$. There is a technique for extending $\mathrm{M}$ to form a countable transitive extension, denoted $\mathrm{M}[G]$, containing $G$, which uses P-names to determine which other sets must also be added to $\mathrm{M}$ simultaneously with $G$ in order to preserve satisfaction of the axioms of ZFC.

Below are the proofs of some general-purpose lemmas regarding forcing notions that are useful in the proof of the Fundamental Theorem of Forcing , taken from Nik Weaver's phenomenal book Forcing for Mathematicians.

Proposition 2. Let $G$ be a generic ideal of $P$ and let $D\in \mathrm{M}$ be a subset of $P$. If each element of $G$ is compatible with some element of $D$, then $G$ has nontrivial intersection with $D$.

Proof. Let $D'$ be the set of $p\in P$ which either extend some element of $D$ or are incompatible with all elements of $D$. Then $D'\in \mathrm{M}$ (since $P\in \mathrm{M}$ and $D'$ can be defined from $P$ using ZFC's Axiom of Subsets) and $D'$ is dense. Thus, since $G$ is generic, it must intersect $D'$, meaning that some element $g\in G$ extends an element of $D$, because (by assumption) no element of $G$ is incompatible with all elements of $D$. But if $g\in G$ extends $g'\in D$, then $g'\in G$ by downward stability. $\blacksquare$

Proposition 3. If $G$ is a generic ideal of $P$, then
1. If $D\in\mathrm{M}$ is dense above some $p\in G$, then $G$ intersects $D$.
2. If $G\subset A\in\mathrm{M}$ then there exists $p\in G$ such that every extension of $p$ lies in $A$.

Proof. To prove $(1)$, we show that each element of $G$ is compatible with some element of $D$ and then defer to the previous proposition. Suppose $q\in G$. By directedness, there exists $r$ extending both $p$ and $q$. Since $D$ is dense above $p$ and $r$ extends $p$, there exists $r'\in D$ extending $r$ and therefore also extending $q$. Hence, every element $q\in G$ is compatible with some $r'\in D$, and we have by the previous proposition that $G\cap D\ne\varnothing$.

Now for $(2)$. Suppose for contradiction that every element $p\in G$ has some extension $p'\notin A$. Consider the set $D$ of all extensions of elements of $p$ that are not in $A$. Then every element of $G$ is compatible with some element of $D$, namely one of its own extensions. Hence $G$ intersects $D$ by the previous proposition, but this is impossible since $G\subset A$ and $D$ is disjoint from $A$ by definition. $\blacksquare$