Proposition 4.7.3.22. Let $\lambda $ be an uncountable cardinal. Then $\lambda $ is strongly inaccessible if and only if it is regular and satisfies the following additional condition:
For every cardinal $\kappa < \lambda $, we also have $2^{\kappa } < \lambda $ (see Remark 4.7.2.14).
Proof. Assume first that $\lambda $ is strongly inaccessible. Proposition 4.7.3.20 implies that $\lambda $ is regular. To verify $(\ast )$, we must show that if $S$ is a $\lambda $-small set, then the collection $P(S)$ of subsets of $S$ is also $\lambda $-small. This is clear, since $P(S) \simeq P( \amalg _{s \in S} \{ s\} ) \simeq \prod _{s \in S} P( \{ s\} )$ can be decomposed as a $\lambda $-small product of two-element sets.
We now prove the converse. Assume that $\lambda $ is regular and satisfies condition $(\ast )$; we wish to show that it is strongly inaccessible. Let $\{ S_ i \} _{i \in I}$ be a $\lambda $-small collection of $\lambda $-small sets; we wish to show that the product $\prod _{i \in I} S_ i$ is also $\lambda $-small. For each index $i$, the construction $s \mapsto \{ s\} $ determines an injective function from $S_ i$ to $P(S_ i)$. Taking the product over $i$, we obtain an injection
It will therefore suffice to show that the set $P( \amalg _{i \in I} S_ i )$ is $\lambda $-small. Using condition $(\ast )$, we are reduced to showing that $\amalg _{i \in I} S_ i$ is $\lambda $-small, which follows from the regularity of $\lambda $. $\square$