Cohen's independence of CH

Remarks

One starts from a countable transitive model $M$ of ZFC and uses forcing with the set $P$ of finite partial functions from $\omega_2^M \times \omega$ to $\set{0, 1}$, ordered by reverse inclusion. A generic filter $G$ gives a function from $\omega_2^M \times \omega$ to $\set{0, 1}$, and so $\aleph_2^M$ different new reals. Every antichain of $P$ is countable, and this chain condition implies that $M$ and $M[G]$ have the same cardinals. So $M[G] \models 2^{\aleph_0} \ge \aleph_2$. The existence of a countable transitive model of ZFC does not follow from the consistency of ZFC. The standard way around this uses countable transitive models of large finite fragments of ZFC, or Boolean-valued models.

Going deeper

For a student

The proof is the first use of forcing, and its pattern is used in almost all independence proofs. The first key step is to choose a partial order whose generic filter adds the desired objects. The second is to check that the forcing does not collapse cardinals, here by the countable chain condition. Changing $\omega_2$ to other cardinals gives many possible values of $2^{\aleph_0}$; Easton's theorem describes the possible values of $2^\kappa$ for all regular $\kappa$. Cohen also used forcing to show that the axiom of choice is independent of ZF.