I'm reading for an exam on set theory using Kunen's first edition, and we've more or less been told that the reflection theorem will come up.
The proof constructively yields an ordinal $\beta>\alpha$, where $\alpha $ is given. I can see from the construction that $\beta $ turns out to be a limit ordinal, however, I'm struggling to see exactly why it needs to be one.
Some extra details: The reflection theorem says that if you have a class $\bf Z$ that is built up by a hierarchy of sets $Z(\alpha)$ indexed by the ordinals, a finite collection $\phi_1, \ldots, \phi_n$ of formulas, and an ordinal $\alpha$, then there is some $\beta > \alpha$ such that all the formulas are absolute for $Z(\beta), \bf Z$.
The proof uses the finite (and subformula-closed) collection of sentences $\phi_1, \ldots, \phi_n$ and defines a class function $\mathbf F_i:\mathbf{ON}\to \mathbf{ON}$ for each sentence. If $\phi_i$ is of the form $\exists x\phi_j(x, y_1, \ldots, y_m)$, then $\mathbf F_i(\xi)$ is the first ordinal such that $\exists x \in Z(\mathbf{F}_i(\xi))\phi_j(x, y_1, \ldots, y_m)$ with all the $y_j \in Z(\xi)$ (if such an $x$ exists in $\bf Z$). Otherwise $\mathbf F_i(\xi) = 0$.
The proof then starts with a $\beta_0 = \alpha$, and recursively defines $\beta_{p + 1}$ for $p \in \omega$ to be the max of $$ \beta_p +1 \qquad \mathbf F_1(\beta_p) \qquad \ldots \qquad \mathbf F_n(\beta_p) $$ and then $\beta = \sup\{\beta_p \mid p \in \omega\}$.
I'm wondering about why he includes $\beta_p + 1$ in that list. Surely, if there were a formula $\phi_i$ in our collection that forced the $\beta$ to be a limit ordinal (or something stronger, like a cardinal, a regular cardinal, a limit cardinal or a $\beth$-fixed point), then the corresponding $\mathbf F_i$ would take care of that, wouldn't it?
It doesn't have to be.
If $\varphi$ is the statement $\exists x\forall y(y\notin x)$, then $V\models\varphi$ and for every $\alpha>0$, $V_\alpha\models\varphi$. Limit or not.