TITLE: Gorelic's Forcing for large Lindelöf spaces with points $G_\delta$
QUESTION [6 upvotes]: I am trying to understand a step for proving that there exists large Hausdorff Lindelöf Spaces with points $G_\delta$ using forcing. I am following Isaac Gorelic's "The Baire Category And Forcing Large Lindelöf Spaces With Points $G_\delta$. (http://www.jstor.org/discover/10.2307/2160344?uid=3737664&uid=2&uid=4&sid=21105156482053)
When proving that $\mathbb P$ has the $\omega_2$-c.c. if $|\mathbb Q|=\omega_2$, $\mathbb Q \subset \mathbb P$, he says that by CH, by the $\Delta$-system lemma and simple counting we may assume that there are $p \neq q$ such that $\text{tp}\, I^p=\text{tp}\, I^q$, $\text{tp}\, A^p=\text{tp}\, A^q$, that by denoting $\xi'$ the unique ordinal in $A^q$ corresponding to a $\xi \in A^p$ such that te order type of $A^p\cap \xi$ is the same of $A^q \cap \xi'$ we have $\forall \xi \in A^p\, \forall \eta \in A^p\, f^p_\xi(\eta)=f_{\xi'}^q(\eta')$. Picking these $p, q$ is easy. He also states that we may assume that $\forall x \in \Delta:=A^p\cap A^q, \, f^p_\xi(x)=f_{\xi'}^q(x)$. This is also something easy to do, but I think he made a typo here. Maybe what he wanted was $\forall x, \xi \in \Delta:=A^p\cap A^q, \, f^p_\xi(x)=f_\xi^q(x)$, but I may me wrong. We will call this last supposition (*).
Then he defines a common extension $r$ of the form $r=\langle I^p\cup I^r, A^p\cup A^Q, F^r, G^r, T^p \cup T^q\rangle$. I think he made a typo here, I think $I^r$ was supposed to mean $I^q$.
Then he says how to build such $G^r$, of course, the set $F^r$ will be determined by $G^r$.
He divides the proof in three cases.
The first case states that if $\alpha \in I^p \cup I^q$, then we set $g_\alpha^r=g_\alpha^p \cup g_\alpha^q$. I think he made another typo here, I think $I^p \cup I^q$ was supposed to mean $I^p \cap I^q$. At this point, the reader must verify that this function is well defined. So let $x \in A^p \cap A^q \setminus A^\alpha$. Let $\xi \in A_\alpha$. Then $\xi \in A^p\cap A^q$ so by (*) it follows that $g_\alpha^p(x)=f_\xi^p(x)=f_\xi^q(x)=g_\alpha^q(x)$, as intended.
Question 1: is the supposition (*) really neccessary or can we show that this function is well defined only by using what Gorelic stated?
For the second case, suppose $\alpha \in I^p \setminus I^q$ (the third case is analogous). So by the Baire Category Theorem, there is an $H: A^q\rightarrow 2$ with $$h \in (U_{g_{\alpha'}}|A^q)\cap \bigcap\{U_B|A^p:B\in T^p\}\setminus\{f^q_\eta: \eta \in A_{\alpha'}\}\neq \emptyset$$
Okay, no problems with this. Then we set $g_\alpha^r=g_\alpha^p\cup h$.
Question 2: why is $g_\alpha^r$ well defined?
Edit: In order to try to show that $g_\alpha^r$ is well defined, I noticed that we may also suppose that for every $\alpha \in I_p$ and for every $x \in \Delta\setminus A_\alpha$, $g^p_\alpha(x)=g^q_{\alpha '}(x)$. So in case $x\in A^p \cap A^q \setminus A_\alpha \setminus A_{\alpha'}$, it follows that $g^p_\alpha(x)=g_{\alpha'}(x)=h(x)$. But I still don't know what to to in case $x \in A_{\alpha'}$.

REPLY [6 votes]: I find that actually this
 forcing is conceptually clear and not too technical.
Don't get bogged down with a few typos.
Fact 3 is about amalgamation of similar conditions.
Only $g_\alpha^p$-s are important for amalgamation (since the rest is preassigned in the ground model).
There are only $\omega_1$-many order types of sets $I^p$. Assume all $p \in Q$ have the same
ordertype $(I^p) < \omega_1$. Then using CH and the $\Delta$-system lemma for $\aleph_2$ countable sets $I^p$, $p \in Q$, choose $p$ and $q$ which are isomorphic structures. Namely, (1) for $\alpha \in I^p \cap I^q$, (a) type($I^p \cap\alpha $) = type ($I^q \cap\alpha$), and (b) $g_\alpha^p \cup g_{\alpha}^q$ is a function, and (2) for $\alpha \in I^p \setminus I^q$, and $\alpha' \in I^q \setminus I^p$ such that type($I^p \cap\alpha $) = type ($I^q \cap\alpha'$), $g_\alpha^p \upharpoonright (A^p \cap A^q) = g_{\alpha'}^q \upharpoonright (A^p \cap A^q)$.
Now regarding Case 2: $g_\alpha^r$ is a function, because (as you've noticed) $h\upharpoonright $ domain ($g_{\alpha'}^q$) $= g_{\alpha'}^q$, and as $A_{\alpha'} \cap A^p = \emptyset$, $x \in A_{\alpha'} \Rightarrow x\notin $ domain ($g_\alpha^p$).

REPLY [4 votes]: A shorthand for the proof of Fact 5 is to take a countable elementary submodel of the ground model.