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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?

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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.

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    $\begingroup$ On the other hand, it is very easy to give examples of $\varphi$ that force any reflecting point to be a limit ordinal, or a cardinal, or a beth fixed point... $\endgroup$ – Andrés E. Caicedo Jun 14 '14 at 16:18
  • $\begingroup$ I've added some details on my question. I believe that clarifies why neither the answer nor the comment is what I'm looking for. $\endgroup$ – Arthur Jun 15 '14 at 5:54
  • $\begingroup$ @Arthur: Oh, that is quite simple. By insisting to take the next one he assures we go up. Imagine that we picked a bunch of formulas which all reflect at $Z(\alpha)$, then by not going one ordinal up, we might not move from $\alpha$ at all. $\endgroup$ – Asaf Karagila Jun 15 '14 at 6:53
  • $\begingroup$ @AsafKaragila He could do that by defining $\beta_0 =\alpha+1$ instead. In the proof it is phrased as if he's specifically looking for a limit $\beta $, and I'm wondering why. $\endgroup$ – Arthur Jun 15 '14 at 7:30
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    $\begingroup$ @Arthur: I'd say nicer, it's rarely stronger. But yes, that's the case. Because, as remarked, there are plenty of things which get reflected from a certain point onwards (think of any statement of the form "there is a set whose properties depend only on its elements" (e.g. empty, infinite ordinal, uncountable ordinal, $\aleph$ fixed point, $\beth$ fixed point, an $\aleph$ and $\beth$ simultaneous fixed point, inaccessible cardinal, and so on and so on), such statement once true in $Z(\alpha)$ will stay true in $Z(\beta)$ for all $\beta>\alpha$. $\endgroup$ – Asaf Karagila Jun 15 '14 at 7:58

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