# $\infty$ as inaccessible cardinal and relation of inaccessible cardinal to second-order ZFC

(1) It is provable in ZF that ∞ satisfies a somewhat weaker reflection property, where the substructure (Vα, ∈, U ∩ Vα) is only required to be 'elementary' with respect to a finite set of formulas. Ultimately, the reason for this weakening is that whereas the model-theoretic satisfaction relation $\models$ can be defined, truth itself cannot, due to Tarski's undefinability theorem.

(2) Secondly, under ZFC it can be shown that κ is inaccessible if and only if (Vκ, ∈) is a model of second order ZFC. (http://en.wikipedia.org/wiki/Inaccessible_cardinal#Two_model-theoretic_characterisations_of_inaccessibility)

First of all, why is it $\infty$ instead of $\aleph_0$? Also, can anyone explain what (1) is actually saying, and why this must hold?

And can anyone show why (2) must hold?

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The $\infty$ symbol often denotes the class of ordinals in set theory. –  Asaf Karagila Feb 22 '13 at 18:25
Please don't ask two questions in the same thread. The second point must hold because it is true (if ZFC is consistent of course). One direction is simple, the other is slightly less simple. Have you tried proving this yourself, reading a book, strengthening your set theoretical basis in order to attack this on your own? Something? –  Asaf Karagila Feb 22 '13 at 19:55

## 1 Answer

For (1), regardless of what $U$ is, the structure $(V_n; \in, U \cap V_n)$ cannot be an elementary substructure of $(V_\omega; \in, U \cap V_n)$ because the former satisfies the sentence "there is a largest ordinal" and the latter does not. That is one reason that $\omega$ does not work in place of the class $\infty$ of ordinal numbers (which is more often denoted by $\text{ON}$ or $\text{Ord}$.)

To see why the "weaker reflection principle" in (1) is true, and in fact prove an intermediate version, take any (meta-) natural number $n$. Then one can show that there is an $\alpha$ that is sufficiently "closed" that $V_\alpha$ is a $\Sigma_n$-elementary substructure of $V$—that is, whenever an element of $V_\alpha$ has a $\Sigma_n$ property in $V$, it also has that property in $V_\alpha$. The proof uses the Tarski-Vaught criterion for elementarity (restricted to $\Sigma_n$ formulas) together with the replacement (or collection) axiom to collect the witnesses into a set. Then we close downward by rank and repeat the process $\omega$ many times to get the required closure point. The reason we must restrict to $\Sigma_n$ formulas for some $n$ is that $\Sigma_n$-definability is definable whereas $\Sigma_\omega$ (first-order) definability is not.

Note that $\Sigma_{n_1}$ elementarity with respect to a class $U$ that is definable by a $\Sigma_{n_2}$ formula will follow from $\Sigma_{n_1+n_2+1}$-elementarity without respect to $U$.

For (2), a hint is that if $\kappa$ is not regular then you can let $U$ be a cofinal function $\alpha \to \kappa$ for some $\alpha < \kappa$.

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