Remark 4.7.3.23. Let $\lambda $ be an uncountable cardinal. We say that $\lambda $ is weakly inaccessible if it is regular and satisfies the following:
- $(\ast ')$
For every cardinal $\kappa < \lambda $, we also have $\kappa ^{+} < \lambda $. That is, $\lambda $ is not a successor cardinal.
It follows from Proposition 4.7.3.22 that every strongly inaccessible cardinal is weakly inaccessible. Moreover, the converse follows from the generalized continuum hypothesis Remark 4.7.2.14.