Why $\forall{f\in \omega^{\omega}\cap V}$ $ \forall^{\infty}n f(n)\neq f_{G}(n)$

Question

Eventually different forcing, $\mathbb{E}=\{\langle s,H \rangle:s \in \omega^{\omega}\wedge H\subseteq [\omega^{\omega}]^{<\omega}\}$.

ordered by $(s',H')\leq (s,H)$ iff $s \subseteq s'$ and $H \subseteq H'$ and $\forall{i \in \text{dom}(s'\setminus s)}\forall{g \in H'}(s(i)\neq g(i))$. It generically adds a new real $f_{G}=\bigcup\{s:\exists{H}(s,H)\in G\}$.

If $f_{G}=\bigcup\{s:\exists{H}(s,H)\in G\}$. Why $\forall{f\in \omega^{\omega}\cap V}$ $ \forall^{\infty}n f(n)\neq f_{G}(n)$.

Also. I do not understand why $\mathbb{E}$ adds a function $f_{G}\in \omega^{\omega}$.

Any contribution, thank you very much.

Answer 1

HINT: For every $f\in\omega^\omega\cap V$, consider the set as defined in $V$, $D_f=\{(p,H)\in\mathbb E\mid f\in H\}$. Show it's dense, and use genericity to conclude that $f_G$ is new and has the wanted property.