Questions in Jech's presentation of the Independence of CH (p. 219) I just had a few questions regarding Jech's proof of the Independence of CH. This is on page 219 of the most recent edition. There's just a couple lines in here I'm not quite seeing. Maybe they're obvious, but I figured I'd ask anyways.

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*For a generic set of conditions $G$, let $f=\bigcup G$. We claim that $f$ is a function. Jech says that $f$ is a function because $G$ is a filter. Am I missing something obvious...? Why is this?

*Secondly, he claims $\text{dom}(f)=\omega_2\times\omega,$ and gives a reason why using some argument about dense sets and $G$ meeting all of them. This... is going a bit over my head. I'm not 100% comfortable with dense/generic/filter arguments, so could someone give a little bit of a more thorough explanationn of this part?

 A: *

*If $f$ is not a function, then there is some $a\in\omega_2\times\omega$ such that both $(a,0)\in f$ and $(a,1)\in f$. Since $f=\bigcup G$, we can find two conditions $p,p'\in G$ such that $(a,0)\in p$ and $(a,1)\in p'$.
In a filter, any two elements have a common extension, so there needs to be some $p''\supset p$ and $p''\supset p'$. But this is impossible, since such a $p''$ would also include both $(a,0)$ and $(a,1)$, and therefore not be a function (and all elements of $P$ are functions).


*We can argue that any arbitrary $a\in \omega_2\times \omega$ is in the domain of $f$. We do this by defining a dense set $$D_a=\{ p\in P\mid a\in\mathrm{dom}(p)\}.$$ It's easy to see that $D_a$ is dense: if $a$ is not in the domain of some condition $p\in P$, then we can extend $p$ by adding $(a,0)$ to get a new condition $p'=p\cup\{(a,0)\}$. The domain of $p'$ is still a finite subset of $\omega_2\times \omega$, since we only added one new value. Therefore $p'\in P$ and since $a\in\mathrm{dom}(p')$ also $p'\in D_a$. Hence we found an extension of $p$ that is in $D_a$.
Now, by definition of a generic filter, $G$ intersects all dense subsets of $P$ (this is essentially what "generic" means), so $G$ intersects $D_a$. Therefore $G$ contains some $p$ with $a\in\mathrm{dom}(p)$, therefore $a\in\mathrm{dom}(f)$, since $p\subset f$.
