Zermelo-Fraenkel Set Theory

CHAPTER 6. CARDINALS 45
Lemma 6.18 If the cardinal p is infinite, then
1. cf(p) is infinite,
2. cf(p) p,
3. cf(p) is regular.
Proof. 1. Obvious.
2. If p = |I|, then p = ∑ i∈I |{i}| = ∑ i∈I 1.
3. Assume that p = ∑ i∈I p i, |I| = cf(p), p i < p, and I = ⋃ j∈J I j, |J|, |I j | < |I|. Then
p = ∑ ( ∑ )
j∈J i∈I j
p i and (by definition of cf(p)) for all j, ∑ i∈I j
p i < p, contradiction. □
Lemma 6.19 1. cf(1) is undefined, cf(2) = cf(3) = cf(4) = . . . = 2,
2. ℵ 0 is regular,
3. ℵ α+1 is regular.
Proof. 2. Any finite union of finite sets is finite.
3. If ℵ α+1 = ∑ i∈I p i with |I| and all p i smaller than ℵ α+1 , then these cardinals are all
ℵ α , and ℵ α+1 = ∑ i∈I p i ∑ i∈I ℵ α = |I| · ℵ α ℵ 2 α = ℵ α , a contradiction.
□
What about the cofinality of limit alephs (i.e., ℵ α where α is a limit)?
Definition 6.20 Regular limit alephs are called (weakly) inaccessible.
Limit alephs that “you can point at” (cf. Exercise 112) always turn out to be singular.
In fact, ZFC (if consistent) is unable to prove that weak inaccessibles exist:
Theorem 6.21 If ZFC is consistent, then so is ZFC + “no inaccessibles exist”.
However, by Gödel’s 2nd incompletability theorem, there is no (sensible) consistency proof
for ZFC + “some inaccessible exists”, even assuming the consistency of ZFC.
Lemma 6.22 (AC)
1. ℵ cf(ℵα)
α > ℵ α ,
2. cf(2 ℵα ) > ℵ α .
(For α = 0, this is the content of 6.16.)
The following lemma is needed here and there in the exercises below. As an example,
cf. the proof of 6.37 (p. 50) where it is used.
Lemma 6.23 If κ and λ are initials such that λ < cf(κ), then κ λ = ⋃ α