Let me give some examples motivating the question.

**The use of forcing instead of induction:** For this consider Cantor's theorem:

**Theorem 1.** Any two countable dense linear orders $I, J$ without end points are order isomorphic.

**Proof.** Let $\mathbb{P}$ be the partial order consisting of partial finite order preserving maps from $I$ to $J$. For $i\in I, j\in J$ the sets $D_i=\{ p: i\in dom(p)\}$ and $R_j=\{p: j\in range(p) \}$ are dense, and since we have only countably many dense sets, we can get a filter $G$ which meets all of these dense sets. Then $\bigcup G: I \to J$ is the required isomorphism.

**The use of forcing instead of diagonal arguments:** This time let's consider another result of Cantor:

**Theorem 2.** For any regular cardinal $\kappa, 2^\kappa > \kappa.$

**Proof.** Let $F$ be a family of functions from $\kappa\to 2$ of size $\kappa.$ Let $Add(\kappa, 1)$ be the Cohen forcing for adding a new subset of $\kappa,$ and for $f\in F,$ let $D_f=\{p: p\neq f \}.$ Each $D_f$ is easily seen to be dense in $Add(\kappa, 1)$, and since the forcing is $\kappa-$closed, there is an $F-$generic filter $G$ over $Add(\kappa, 1).$ Then $\bigcup G: \kappa \to 2$ is different from all $f\in G.$

**Theorem 3.** If $\lambda=cf(\kappa) < \kappa,$ then $\kappa^\lambda > \kappa.$

**Proof.** Let $F$ be family of size $\kappa$ of functions from $\lambda\to \kappa.$ Let $\mathbb{P}=\{p:p$ is a partial function of size $<\lambda$ from $\lambda\to \kappa \}.$ It is $\lambda-$closed. Let $(\kappa_\alpha: \alpha<\lambda)$ be increasing cofinal in $\kappa,$ let $F_0 \subseteq F_1 \subseteq ... (\alpha<\lambda)$ with $|F_\alpha|=\kappa_\alpha$ and $F=\bigcup_\alpha F_\alpha.$ Now consider dense sets $D_\alpha=\{p: \forall f\in F_\alpha, p\neq f \}.$ If a filter $G$ meets all $D_\alpha$'s, then $\bigcup G$ is different from all elements of $F$.

Below are a few more examples of the results that can be proved in a similar way:

**(A)** If $\mathcal{A}$ and $\mathcal{B}$ are countable families of subsets of
$\mathbb{N}$ such that $A \cap B$ is finite for each $A\in \mathcal{A}$ and $B\in \mathcal{B}$, then there is
a $C$ such that $A \subseteq^* C$ for every $A\in \mathcal{A}$ and $B\cap C$ is finite for every $B\in \mathcal{B}.$

**(B)** There exists a continuous, nowhere dieffrentiable
function on $[0,1].$

**(C)** Let's call a set $A$ of natural numbers (not including 0) is small, if $\Sigma_{n\in A}1/n < \infty.$ Given a countable family $\mathcal{F}$ of small sets, there exists a small set $J$ such that $I\subseteq^* J,$ for all $I\in \mathcal{F}.$

**(ِِD)** All consequences of $MA(\aleph_0),$ in particular the Baire category theorem,....

Question.Are there any more non-trivial results which are proved using forcing in the following way: we produce a forcing notion $\mathbb{P}$ and a family $\mathcal{D}$ of dense subsets of it. Then we argue that there must be a $\mathcal{D}-$generic filter $G$ over $\mathbb{P},$ and use $G$ to conclude our required result.

forcingur just an application of existence of generic filters? By "forcing" I understand the whole rigamarole of building models (usually of ZF). These examples seem like a very good introduction to generic filters, so students can see them first without the model-building machinery. $\endgroup$ – Andrej Bauer Jan 14 '15 at 14:083more comments