A.2 Cardinal arithmetic
Let \( \# \mu \) be a strong limit cardinal. Then there are precisely \( \# \mu \)-many subsets of \( \mu \) of size strictly less than \( \operatorname {\mathsf{cof}}(\operatorname {\mathsf{ord}}(\# \mu )) \).
Proof
Endow \( \mu \) with its initial well-ordering. Each such subset is bounded in \( \mu \) with respect to this well-ordering as its size is less than \( \operatorname {\mathsf{cof}}(\operatorname {\mathsf{ord}}(\# \mu )) \). So it suffices to prove there are precisely \( \# \mu \)-many bounded subsets of \( \mu \).
\begin{align*} \# \{ s : \operatorname {\mathsf{Set}}\mu \mid \exists \nu : \mu ,\, \forall x \in s,\, x {\lt} \nu \} & = \# \bigcup _{\nu : \mu } \{ s : \operatorname {\mathsf{Set}}\mu \mid \forall x \in s,\, x {\lt} \nu \} \\ & \leq \sum _{\nu : \mu } \# \{ s : \operatorname {\mathsf{Set}}\mu \mid \forall x \in s,\, x {\lt} \nu \} \\ & = \sum _{\nu : \mu } \# \{ s : \operatorname {\mathsf{Set}}\{ x : \mu \mid x {\lt} \nu \} \} \\ & = \sum _{\nu : \mu } \underbrace{2^{\# \{ x : \mu \mid x {\lt} \nu \} }}_{{\lt}\mu } \\ & \leq \mu \end{align*}