TITLE: Generic filters of inverse limits
QUESTION [7 upvotes]: Maybe this doubt is silly, but I do not understand the final step of the proof of Lemma 5.2 in Hamkins' paper Fragile measurability, Journal of Symbolic Logic 59 (1994) 262-282.
There, $\mathbb P_\lambda$ is the $\lambda$-th step of an iterated forcing construction which is the inverse limit of the previous steps, and $G$ is a $V$-generic filter. More specifically, $\mathbb P_\lambda$ is the canonical Easton support iteration which forces the GCH below $\lambda$, and $\lambda$ is a strong limit cardinal of countable cofinality.
We have a condition $r\in\mathbb P_\lambda$ (in the proof it is called $j(p)$) such that for all $\beta<\lambda$, $r(\beta)\in G(\beta)$. My doubt is why this implies that $r\in G$.
I can prove that $r|_\beta\in G_\beta=i_{\beta\lambda}^{-1}(G)$ for all $\beta<\lambda$, and maybe the conclusion is trivial, but I do not see it. In fact, I know very few about generic filters on inverse limits and their relation with previous steps.

REPLY [7 votes]: We can justify this with a general fact about forcing.
Fact: Let $G$ be $\mathbb P$-generic, where $\mathbb P$ is separative.  If $X$ is a subset of $\mathbb P$ in the ground model, $X \subseteq G$, and $m = \inf X$, then $m \in G$.
Proof:  Consider the set $\{ p \in \mathbb P : p \leq m$ or $(\exists x \in X)p \perp x \}$.  Using separativity we show this is a dense set.  Since $X \subseteq G$, this implies $m \in G$.
Now I claim that in this situation $r = \inf_{\beta<\lambda} r \restriction \beta$.  Certainly $r \leq r \restriction \beta$ for all $\beta$.  If $q$ also has this property, then $q \leq r$ by the definition of the ordering at limit stages of iterations.  ($q \leq r$ just means $q \restriction \beta \leq r \restriction \beta$ for all $\beta < \lambda$.)