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

Question

(1) It is provable in ZF that ∞ satisfies a somewhat weaker reflection property, where the substructure (V_α, ∈, U ∩ V_α) 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 (V_κ, ∈) 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?

Answer 1

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 $\alpha < \kappa$.