Size and rank bounds below an inaccessible

We first justify the small-union estimate used here. For every infinite cardinal theta, $\mathrm{\mid }\theta \times \theta \mathrm{\mid }=\theta$: by induction on infinite cardinals order pairs by their maximum coordinate, then lexicographically. An initial segment ending at coordinates below $\gamma +1<\theta$ has size at most $\mathrm{\mid }(\gamma +1)\times (\gamma +1)\mathrm{\mid }<\theta$, by the induction hypothesis at $\mathrm{\mid }\gamma +1\mathrm{\mid }$ (or by finite counting). This well-order has type at most theta, since otherwise its first theta elements would be a proper initial segment of size theta. The reverse bound uses the injection $\xi \mapsto (\xi ,0)$. Consequently for fewer than kappa sets of size below kappa, regularity bounds the set of their cardinalities and the index cardinal below a common infinite cardinal $\eta <\kappa$. Such an eta exists because a strong-limit cardinal is a limit cardinal: if kappa were the successor of eta, Cantor diagonalization would give $2^{\eta }\ge \kappa$. AC selects injections of the sets into eta, embedding their disjoint union into $\eta \times \eta$. Its size is therefore below kappa.

Induct on $\alpha <\kappa$. The empty V_0 is small. At a successor, $\mathrm{\mid }{V}_{\alpha +1}\mathrm{\mid }=2^{\mathrm{\mid }{V}_{\alpha }\mathrm{\mid }}<\kappa$ by strong limit. At a nonzero limit alpha there are fewer than kappa earlier levels, so step 1.1 bounds their union below kappa. If $x\in V_{\kappa }$, it is a subset of some earlier V level, hence has size below kappa.

If $Y\subseteq V_{\kappa }$ and $\mathrm{\mid }Y\mathrm{\mid }<\kappa$, Replacement collects the ranks of its members; regularity bounds their supremum plus one below kappa. Thus the supremum of their ranks plus one, namely rank(Y), is below kappa, so $Y\in V_{\kappa }$. For empty Y the rank is zero.

For $\mu ,\nu <\kappa$, graphs inject the set of functions $\nu \to \mu$ into $\mathcal{P}(\nu \times \mu )$. Step 1.1 bounds the product size by some infinite $\eta <\kappa$, so ${\mu }^{\nu }\le 2^{\eta }<\kappa$. This also covers finite cardinals; more explicitly ${\mu }^0=1$, $0^{\nu }=0$ for positive nu and $1^{\nu }=1$.

Let C be the set of infinite strong-limit cardinals below kappa. It is unbounded: above a given bound start with an infinite cardinal ${\theta }_0<\kappa$ larger than it, put ${\theta }_{n+1}=2^{{\theta }_n}$ and $\delta ={\sup }_n{\theta }_n<\kappa$ by regularity. Cantor diagonalization makes the sequence strictly increasing. Its supremum is a cardinal: a bijection of delta with a smaller ordinal would inject a larger theta_n into a smaller cardinal. For every cardinal $\mu <\delta$ some theta_n exceeds mu, hence $2^{\mu }\le {\theta }_{n+1}<\delta$. Thus delta is in C. If delta<kappa is a nonzero limit accumulation point of C, it is similarly a cardinal, and for every cardinal mu<delta there is $\rho \in C\cap \delta$ above mu, giving $2^{\mu }<\rho <\delta$. Thus delta is in C, proving closure. No enumeration choices are needed in this last uniquely defined iteration.