## § Forcing to add a function

- Let $M$ be a countable transitive model of ZFC.
- We will add a new function $c: \aleph_0^M \to \{0, 1\}^M$ into $M$ by creating $M[G]$.
- Let $P$ be the set of all finite partial functions from $\aleph_0$ to $\{0, 1\}$ in $M$.
- Let $G$ be a generic maximal ideal of $P$. That is, $G$ intersects every dense set of $M$.
- Also, since it is a maximal ideal, taking the full union $\cup G \equiv c$ will give us a well defined total function.
- It will be well defined since no two elements of $G$ disagree, and it will be total because if it were not, we could extend $G$, contradicting the maximality of $G$.
- Great, so if we can construct $M[G]$, we will also have $c = \cup G \in M[G]$.
- But how do we know that $c$ is new? Ie, how do we know that $c \not in M$?
- Well, consider for any function $h \in M$, the subset of $P$ that disagrees with $h$. That is, the subset $D_h \equiv \{ p \in P : \exists i, p(i) \neq h(i) \}$.
- See that $D_h$ is dense in $M$: Suppose $p \in P$, and $p$ is well-defined on some subset $S$. Either $p$ disagrees with $h$ on $S$, that is, there is some $s \in S$ such that $p(s) \neq h(s)$, in which case $p \in D_h$ and we are done.
- On the other hand, maybe $h|S = p$ (that is, $h$ restricted to $S$ fully agrees with $p$). Then we pick some point $s' \not in S$and extend $p$ into $p'$ to disagree with $h$ at $s'$. So define $p'(s') \equiv h(s') + 1$ or something. Now we have $p \leq p'$ and $p' \in D_h$.
- Since $D_h$ is generic, we have that $G \cap D_h \neq \emptyset$, thus $f$ disagrees with $h$ at some point!
- Thinking intuitively, it would be a CRAZY coincidence for it to agree with a function $h$ fully in $M$. If we build it "randomly", or "generically", one
*would * expect it to disagree with stuff in $M$ at some point in the construction!. - Cool, we've now seen how to enlarge the universe to add a
*single * function of interest. - Reference