I'm am trying to comprehend the proof on page 22 of Thomas Jech's Set theory and work out my first ever proof by transfinite induction. The problem is this:
Le $G$ be a function (on $V$), then there can only be one $\alpha$-sequence satisfying $$(\forall\xi <\alpha)\; a_\xi = G\big(\langle a_\eta : \eta<\xi\rangle \big ) \tag{$\star$}\label{moi}.$$
(I actually have no idea what $V$ is supposed to be, here.)
The book says that if $\langle a_\xi : \xi< \alpha\rangle$ and $\langle b_\xi : \xi< \alpha\rangle$ are two $\alpha$-sequences satisfying \eqref{moi}, then one shows $a_\xi = b_\xi$ by induction on $\xi$.
Now for a basis I say that $\langle a_\eta:\eta<0\rangle=\emptyset=\langle b_\eta:\eta<0\rangle$ and so $a_0=G(\emptyset )= b_0$.
Question: Is this valid? I don't know wether or not $\emptyset$ is in the domain of $G$. Especially sice $G$ is so generally (vaguely) defined.
Now if we assume that $a_\xi=b_\xi$ for all $\xi\leq \zeta$, for some $\zeta$, then $$ \langle a_\eta :\eta <\zeta+1\rangle=\langle a_\eta:\eta\leq \zeta\rangle=\langle b_\eta:\eta\leq \zeta \rangle=\langle b_\eta: \eta< \zeta+1\rangle\implies a_{\zeta+1}=b_{\zeta+1}.$$ Question: Is this correct, at all?
Lastly, if we have $a_\xi=b_\xi $ for all $\xi<\zeta$, for some $\zeta$$$\langle a_\eta:\eta<\zeta\rangle=\langle b_\eta:\eta<\zeta\rangle\implies a_\zeta=b_\zeta.$$
Is this a valid proof?
As Asaf said in the comments, $V$ is the class of all sets, so in particular $\varnothing$ is in the domain of $G$. Your argument is correct, but you don’t actually need to split it into an initial case, a successor case, and a limit case. Suppose that $\eta<\alpha$, and $a_\xi=b_\xi$ for each $\xi<\eta$. Then
$$a_\eta=G\big(\langle a_\xi:\xi<\eta\rangle\big)=G\big(\langle b_\xi:\xi<\eta\rangle\big)=b_\eta\;.\tag{1}$$
The hypothesis that $a_\xi=b_\xi$ for each $\xi<\eta$ is vacuously true for $\eta=0$, so the initial case comes for free, so to speak, and the calculation $(1)$ is valid for both limit and successor $\eta$.