Intuitive explanation of generic extension

Question

I am attempting to understand cohen's proof by forcing that the continuum hypotheses is independent of ZFC . However,I am having trouble understanding intuitively the concept of a generic extension of a model. Can anyone give me an intuitive explanation (with examples if possible)?. Thanks in advance

Answer 1

Assuming you're talking about set theory, the main theorem is that $V[G]$ is the minimal model of set theory containing $V$ with $G$ as an element.

More precisely, if $V\models \mathsf{ZFC}$ is a transitive model, $\mathbb{P}\in V$ is a poset, and $G$ (usually not in $V$) is $\mathbb{P}$-generic over $V$, then

1. $V[G]\models\mathsf{ZFC}$,
2. $V[G]$ is transitive with $V\subseteq V[G]$, $G\in V[G]$, and
3. If $M\models\mathsf{ZFC}$ is transitive with $V\subseteq M$ and $G\in M$, $V[G]\subseteq M$.

The elements of the generic extension come from so-called $\mathbb{P}$-names. To give some context, consider the constructible universe $L$ defined in stages where we take $\{\in\}$-definable subsets at each stage. There's a notion of $L[x]$ where we take $\{\in,x\}$-definable subsets at each stage. In essence, we allow ourselves ask questions about membership in $x$. If $x\subseteq \mathbb{P}$ for some $\mathbb{P}\in L$, then $x\in L[x]$ and we basically consider everything constructible from $x$ in the same way we construct the elements of $L$.

This all works out nicely for $L$, which has a nice definition. But how should we generalize this to such an arbitrary $V$? Well the idea is that we consider all sorts of potential constructions that $V$ can carry out. Then when we have access to $V[G]$, we just thin out these potential constructions by our (outside of $V$) access to $G$. These potential constructions are just sets tagged with members of our poset $\mathbb{P}$. Once we know what's in $G$, we can include the elements tagged with things in $G$ and throw out the members tagged with things not in $G$. For example, if $G=\{p\}$, $\{\langle 0,p\rangle ,\langle 1,q\rangle \}$ would be thinned out to just $\{0\}$. If $G=\{p,q\}$, the same set would be thinned out to $\{0,1\}$.

So this is basically the setup: we iteratively form these potential constructions and then with access to $G$, we thing things out and end up with $V[G]$:

• $V_0^{\mathbb{P}}=\emptyset$
• $V_{\alpha + 1}^{\mathbb{P}}$ is the set of potential constructions using elements of $V_{\alpha}^{\mathbb{P}}$, i.e. $V_{\alpha + 1}^{\mathbb{P}}=\mathcal{P}(V_{\alpha}^{\mathbb{P}}\times \mathbb{P})$ = $V_{\alpha}^{\mathbb{P}}=\bigcup_{\xi<\alpha}V_{\xi}^{\mathbb{P}}$ for limit $\alpha$.

As described above, we thin these constructions our (or interpret) these names by setting $$\tau_G=\{\sigma_G:\langle \sigma,p\rangle\in \tau\text{ for some }p\in G\}\text{.}$$ Then we take $$V[G]=\{\tau_G:\tau\in V^{\mathbb{P}}=\bigcup_{\alpha\in\mathrm{Ord}}V_\alpha^{\mathbb{P}}\}$$ Why exactly does this work? For example, how do we know $G\in V[G]$? It's easy to see that $\tau = \{\langle p,p\rangle:p\in\mathbb{P}\}$ gives $\tau_G=G$. How do we know $V\subseteq V[G]$? One can show that there are canonical names for $x\in V$ basically just by iteratively setting $\check \emptyset = \emptyset$, and $$\check x = \{\langle \check y, p\rangle:y\in x\text{ and }p\in\mathbb{P}\}\text{.}$$ It's not too difficult to show that this works.

Again, the main idea is that $V[G]$ is what $V$ would look like if it had access to $G$ because we are basing $V[G]$ on what constructions $V$ allows.