I am looking at Jech's proof of properties of Namba forcing. The conditions consist of perfect subtrees of $\omega_2^{<\omega}$, where perfect means every node has $\omega_2$ many (not necessarily immediate) successors. The following is a key step in showing the forcing doesn't add reals (so $\omega_1$ is preserved): suppose $T\Vdash\dot{f}:\omega\rightarrow 2$ is a function, inductively for $s\in\omega_2^{<\omega}$ construct $T_s$ and $\alpha_s\in 2$ such that:
(i) $T_\emptyset\subset T$, and if $s_1\subset s_2$ then $T_{s_1}\supset T_{s_2}$,
(ii) if $s\in\omega_2^n$ then $T_s\Vdash\dot{f}(n)=\alpha_s$,
(iii) $T_s$ has stem $t_s$, namely every node in $T_s$ is compatible with $t_s$, and if $s_1,s_2\in\omega_2^n$ are different then $t_{s_1},t_{s_2}$ are incompatible.
Jech then remarks that (iii) implies any condition $T'$ extending $\bigcup_{s\in\omega_2^n}T_s$ is compatible with one of the $T_s$. Why? This would be true if $T'$ contains one of the stem $t_s$ (e.g., if the stems have bounded height, but I can't see if that can be arranged). Is it impossible that every node in $T'$ is strictly below some $t_s$?
Other questions: what are some other references for basic properties of Namba forcing? There seems to be another variant of Namba forcing, such that above the stem every node has $\omega_2$ many immediate successors; is that equivalent to Jech's definition? Does Namba forcing preserve $\omega_3$?
Best Answer
I think you don't quite need all of the stems to have the same length. Rather, you inductively construct the stems $t_s$, so that for each $s$ all the stems $t_{s\frown i}$ have the same length $h_s$. At each step you do this just like in the base case (pick $\omega_2$ incompatible extensions of $t_s$, strengthen to decide $\dot{f}(n+1)$, and realize that there are only countably many levels to hold all your $\omega_2$ many stems).
With this additional property, let's try to prove Jech's claim. It's instructive to first look at the case $n=1$, so pick a tree $T'$ contained in $\bigcup_{i<\omega_2} T_i$. If $b_\emptyset$ is any node of height $h_\emptyset$ then it must be one of the $t_i$, since those are the only nodes in $\bigcup_i T_i$ of that height. Therefore $T'$ is compatible with that $T_i$.
More generally, let's take $T'\subseteq \bigcup_{s\in \omega_2^{n+1}} T_s$. Let $b_0$ be any node in $T'$ of height $h_\emptyset$. It must be one of the $t_i$ for $i<\omega_2$, since those are the only nodes in $\bigcup_{s\in \omega_2^{n+1}} T_s$ of that height. Call that stem $b_0=t_{i_0}$. Next, let $b_1$ be an extension of $b_0$ in $T'$ of height $h_{i_0}$. By construction, the only nodes in $\bigcup_{s\in \omega_2^{n+1}} T_s$ above $t_{i_0}$ of height $h_{i_0}$ are the stems $t_{i_0\frown i}$ for $i<\omega_2$, so $b_1$ must be one of those.
Continuing on in this way, we can show that $T'$ contains one of the $t_s$ for $s\in\omega_2^{n+1}$, and so is compatible with that $T_s$.
I don't know of a good source on Namba forcing, but Shelah has some material in Proper and Improper Forcing (specifically, chapters X and XI). In particular, he mentions your other version of Namba forcing and proves that they are not equivalent (Claim XI.4.2: neither adds generics for the other).
Regarding $\omega_3$, any forcing that preserves $\omega_1$ and makes $\omega_2$ of countable cofinality will collapse $\omega_3$. See Claim XI.4.3 of Shelah's book. The same claim also states that both versions of Namba forcing can consistently have the $\omega_4$-cc.