I do have a question about a fragment of a proof of a theorem in iterated forcing.
It is the one that Jech invokes in the following form as theorem 16.30 in the 'Seth Theory' book and as Part II, Thm 2.7 in the 'Multiple Forcing' book, and Baumgartner refers to it a Theorem 2.2. in his chapter on Iterated forcing:
Let $\kappa $ be regular uncountable. Let $(P_\lambda: \lambda < \alpha)$ be an iteration of forcing notions with direct limit $P_\alpha$, and assume that the set of stages $\delta<\alpha $ where $P_\delta$ is the direct limit of the previous forcings is stationary in $\alpha$ (assuming $cf(\alpha) = \kappa$. If all $P_\alpha$ satisfy the $\kappa$-c.c., then so does $P_\kappa$.
Both Baumgartner and Jech prove this theorem by starting with stating a tiny thing that: If $A = \{p_\xi: \xi < \kappa\}$ and $cf(\alpha) \neq \kappa$, then there is an $\beta < \alpha$ and a subset $B \subseteq A$ with $|B| = \kappa$ such that for all $p \in B$, $support(p) \subseteq \beta$.
They do not comment on this argument.
I tried to prove it for myself and I must admit I struggle.
For any $p \in A$ define $\gamma_p$ to be the minimal ordinal below $\alpha$ such that $$support(p) \subseteq \gamma_p$$
(so we identify entire $A$ with the set of ordinals – each smaller than $\alpha$).
I can see the trivial implication in the case that $cf(\alpha) > \kappa$, because then the set $\{\gamma_p: p \in A\}$ has to be bounded in $\alpha$ (cause otherwise we would have a set of size $\kappa$ ordinals smaller than $\alpha$ that would be cofinal in $\alpha$).
But then suppose $cf(\alpha) < \kappa$. Even if I try to remove the set of $\gamma_p$'s giving me a cofinal family, I cannot guarantee that I can somehow carve out a bounded set of size $\kappa$.
It seems that what we are trying to prove is actually equivalent to the following:Let $A = \{\gamma_\delta: \delta < \kappa\}$ be set of ordinals, each smaller than $ \alpha$. If $cf(\alpha) < \kappa$, then there is a subset $B$ of $A$ of size $\kappa$ that is bounded in $\alpha$.
If, for each $\eta < \alpha$ I try to define, say $C_\eta = \{\gamma_\delta \in A: \gamma_\delta \geq \eta\}$, then what we are asked to prove is: there is an $\eta < \alpha$ such that $|A \setminus C_\eta| = \kappa$.
But why would that be true?
Can I please ask you for a hint or reference to a place where this statement has an actual proof?
Edit: Thanks to Asaf I think I got it now, but since I do not necessarily agree it is absolutely trivial, I am going to write down the argument here, in case anybody find it puzzling in the future – it is probably a standard exercise, but e.g. I never had it at my set theory courses nor is it included in the Hungarian book on the classical problems in set theory.
Trivial claim: If $C \subseteq \sf{Ord}$ is of size $\delta$, then it has a subset of order-type $\delta$. Indeed, just define a function $f: \delta \rightarrow C$ by $f(0) := \min(C)$, and $f(\beta) := \min(C \setminus f[\beta])$ (where $f[\beta]$ denotes the image of $[0, \beta)$ under $f$).
And now the main claim:
Let $\alpha$ be an arbitrary limit ordinal, and let $\delta$ be an ordinal such that $cf(\alpha) < cf(\delta)$. Then any subset $X \subseteq \alpha$ with $\sf{ot}(X) = \delta$ is bounded in $\alpha$.
For the proof suppose otherwise, that there is an unbounded $X = \{x_\eta: \eta < \delta\} \subseteq \alpha$ of type $\delta > cf(\alpha)$.
By definition of $cf(\alpha)$ there is an unbounded subset $C = \{c_\xi: \xi < cf(\alpha)\} \subseteq \alpha$ (assume the enumeration is increasing of course).
But since $X$ is unbounded as well, for any $\xi < cf(\alpha)$ there is an $\eta < \delta$ with $c_\xi < x_\eta$.
Define $g: cf(\alpha) \rightarrow \delta$ by $g(\xi) = min\{\eta < \delta: c_\xi < x_\eta \: \& \: \forall \zeta < \xi \: g(\zeta) < \eta \}$. Then the set $Y:=\{x_{g(\xi)}: \xi < cf(\alpha)\} \subseteq X$ has to be unbounded in $\alpha$ (since $C$ was), so $\sup(Y) = \alpha$. But $cf(\alpha) < cf(\delta) \leq \delta$, so any subset $Y \subseteq X$ of order type $cf(\alpha)$ has to be bounded in X, a fortiori in $\alpha$, i.e. $\sup(Y) < \sup(X) = \alpha$, which is a contradiction.
This allows to prove the lemma needed for the iterated forcing theorem I asked about: if $A \subseteq \alpha$ is a set of of size $\delta = cf(\kappa)$ of sequences $p \in I^\alpha$ with bounded support, for each $p \in A$ define $\gamma_p = \sup\{\gamma < \alpha: p(\gamma) \neq \dot{1} \}$. Then $A' = \{\gamma_p: p \in A\}$ is a subset of $\alpha$ of size $\leq \delta$. W.l.o.g. assume $|A'|=\delta$. $A'$ has a subset $B$ of the order type $\delta$. By the argument above (trivially, since $\delta$ is a cofinality, $cf(\delta) = \delta$), the set $B \subseteq A'$ has to be bounded in $\alpha$ by some $\beta < \alpha$, and this is the ordinal we were looking for.
Best Answer
What you really want to prove here is the following:
This is trivial. If it's not bounded, then the cofinality of $\alpha$ is $\kappa$.
Now if you have any subset of size $\kappa$, it has a subset of type $\kappa$ as well.
Now look at your conditions. Each one is finite, since the iteration is a finite support one (according to Jech's Multiple Forcing reference). Replace each condition with its maximal nontrivial coordinate. Either there is a coordinate containing $\kappa$ of them, or there is a set of size $\kappa$ of different maximal coordinates. Apply the above, and move on.