Question on $[\mathbb{N}]^{\omega}$ I'm reading Fremlin's article about $\mathfrak{p} = \mathfrak{t}$, and in 4B proposition, he shows that $\Vdash_{\mathbb{P}}\,\mathcal{P}(\mathbb{N}) = \mathcal{P}(\mathbb{N})\check{}$, where $\mathbb{P} = ([\mathbb{N}]^{\omega},\supseteq^*,\mathbb{N})$, the forcing of infinite subsets of $\mathbb{N}$, ordered by almost containment, and $\mathcal{P}(\mathbb{N})\check{}$ is the canonical name of $\mathcal{P}(\mathbb{N})$. He says that the only thing we need to see is that $\mathbb{P}$ is countable closed, but I can't see this fact.
Thanks a lot!
 A: Countable closure implies that no new subsets of $\omega$ are added. 
The usual first application of this is proving the consistency of CH via forcing (as opposed to inner models): the set of countable partial injections $\omega_1\rightarrow\mathcal{P}(\omega)$ is countably closed under reverse extension, and so in the forcing extension $V[G]$ we get $\mathcal{P}(\omega)^{V[G]}=\mathcal{P}(\omega)^V$, but we also clearly have in $V[G]$ a bijection between $\mathcal{P}(\omega)^V$ and $\omega_1^V$. 
Informally: Suppose $\mathbb{P}$ is countably closed, and $\nu$ is a name for a set of natural numbers. Fixing an arbitrary condition $r$, we can - in $V$ - find a sequence of conditions $r\ge p_0\ge p_1\ge ...$ such that $p_i$ decides $\nu(i)$. By countable closure, we get some $q$ stronger than each $p_i$; then $q$ decides all of $\nu$, and by definability of forcing we get $\nu[G]\in V$ for any generic $G\ni q$. Since this was forcible below any $r$, it holds for all generics.
A hint towards a complete proof: Suppose $\mathbb{P}$ is countably closed, $G$ is $\mathbb{P}$-generic over $V$, and $\nu\in V^\mathbb{P}$ with $\nu[G]\subseteq\omega$. Some $c\in G$ forces $\nu\subseteq\omega$; WLOG, take $c=\mathbb{1}$. Now given $d\le c$, let $$Set(d)=\{i\in\omega: (d\Vdash\nu(i)=0)\vee (d\Vdash\nu(i)=1)\};$$ can you show that $$\{d\le c: Set(d)=\omega\}$$ is dense below $c$? 
