In The Higher Infinite, Kanamori gives a proof that Vopenka's principle implies the existence of (many) extendibles. The relevant theorem is Proposition 24.15 on p.337. Working in a $V_\kappa$ which satisfies Vopenka's principle (for $\kappa$ inaccessible) he first defines the Vopenka filter $F$ over $\kappa$ as:
$X\in F$ iff there is a natural sequence of models $\langle M_\alpha:\alpha<\kappa\rangle$ such that its critical points are all in $X$.
where a natural sequence of models is a sequence of models of the form $M_\alpha = \langle V_{f(\alpha)}, \{\alpha\}, R_\alpha\rangle$ (where $f(\alpha)\leq f(\beta)$ for $\alpha<\beta$ and $\alpha<f(\alpha)$). He then shows that the set of extendible cardinals in $V_\kappa$ is in $F$.
It strikes me that there is a simpler proof of this result than the one Kanamori gives. In particular, following Kanamori we first define a function $g$ such that $g(\alpha) = \alpha$ if $\alpha$ is extendible and $g(\alpha)$ is the least $\alpha+\delta$ such that $\alpha$ is not $\delta$-extendible otherwise. We can then define a natural sequence $\langle M_\alpha:\alpha<\kappa\rangle$ such that $M_\alpha = \langle V_{lub\{g(\beta):\beta\leq\alpha\} + \omega}, \{\alpha\}, \emptyset\rangle$, and it is straightforward to show that the critical points of this sequence are extendible. What am I missing?
(Also in the proof of Proposition 24.14 why do we add $f\upharpoonleft \alpha$ to $N_\alpha$?) Thanks in advance!