A.2 Cardinal arithmetic

Lemma A.6 mathlib

Let $# μ$ be a strong limit cardinal. Then there are precisely $# μ$ -many subsets of $μ$ of size strictly less than $cof (ord (# μ))$ .

Proof ▶

Endow $μ$ with its initial well-ordering. Each such subset is bounded in $μ$ with respect to this well-ordering as its size is less than $cof (ord (# μ))$ . So it suffices to prove there are precisely $# μ$ -many bounded subsets of $μ$ . □

\begin{aligned} # {s : Set μ ∣ \exists ν : μ, \forall x \in s, x < ν} & = # ⋃_{ν : μ} {s : Set μ ∣ \forall x \in s, x < ν} \\ \leq \sum_{ν : μ} # {s : Set μ ∣ \forall x \in s, x < ν} \\ = \sum_{ν : μ} # {s : Set {x : μ ∣ x < ν}} \\ = \sum_{ν : μ} \underset{< μ}{\underset{⏟}{2^{# {x : μ ∣ x < ν}}}} \\ \leq μ \end{aligned}