A Universe of Sorts

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

§ Forcing to Add a Function