Cℵ_ω — the cardinal indexed by the first limit ordinal ω
Dℵ_{ω₁} — singular only if the continuum hypothesis holds
ℵ_ω is singular because the sequence ℵ₀, ℵ₁, ℵ₂, ... is cofinal in ℵ_ω and has length ω = ℵ₀ < ℵ_ω, so cf(ℵ_ω) = ω < ℵ_ω. By contrast, ℵ₁ and ℵ₂ are successor cardinals: no countable sequence of cardinals below ℵ₁ can have supremum ℵ₁ (the union would be at most countable), so cf(ℵ₁) = ℵ₁ — regular. Option D is a misconception: cf(ℵ_{ω₁}) = ω₁ < ℵ_{ω₁}, making ℵ_{ω₁} singular in ZFC, independent of CH.
Question 2 Multiple Choice
Using König's theorem (cf(2^κ) > κ for all κ), which of the following is provably false in ZFC?
A2^{ℵ₀} = ℵ₁
B2^{ℵ₀} = ℵ₂
C2^{ℵ₀} = ℵ_ω
D2^{ℵ₀} = ℵ_{ω₁}
König's theorem gives cf(2^{ℵ₀}) > ℵ₀. So 2^{ℵ₀} cannot have cofinality ≤ ℵ₀. Since cf(ℵ_ω) = ω = ℵ₀, setting 2^{ℵ₀} = ℵ_ω would give cf(2^{ℵ₀}) = ω ≤ ℵ₀ — a direct contradiction. The other options are consistent: cf(ℵ₁) = ℵ₁ > ℵ₀, cf(ℵ₂) = ℵ₂ > ℵ₀, and cf(ℵ_{ω₁}) = ω₁ > ℵ₀. Only ℵ_ω is ruled out because it is singular with cofinality ω.
Question 3 True / False
ℵ₁ is a regular cardinal because no countable sequence of cardinals smaller than ℵ₁ can have supremum equal to ℵ₁.
TTrue
FFalse
Answer: True
A cardinal κ is regular when cf(κ) = κ. For ℵ₁: any collection of countably many cardinals each less than ℵ₁ (i.e., countable cardinals) has union at most countable. So no sequence of length ≤ ℵ₀ can be cofinal in ℵ₁, which means cf(ℵ₁) = ℵ₁ — it is regular. Every uncountable successor cardinal is regular in ZFC. Limit cardinals whose index has smaller cofinality, like ℵ_ω, are singular.
Question 4 True / False
Singular cardinals are rare and exotic: regular cardinals predominate in the hierarchy, and singular ones appear primarily at isolated points.
TTrue
FFalse
Answer: False
Singular cardinals vastly outnumber regular ones by density. Every successor cardinal (ℵ₁, ℵ₂, ℵ₃, ...) is regular, but every limit cardinal whose index has smaller cofinality is singular: ℵ_ω, ℵ_{ω+ω}, ℵ_{ω²}, ℵ_{ω_ω}, and many others are all singular. The singular cardinals are densely packed throughout the hierarchy; the regular cardinals are the exceptions, not the rule.
Question 5 Short Answer
Use the definition of singular cardinals and König's theorem to explain why 2^{ℵ₀} cannot equal ℵ_ω.
Think about your answer, then reveal below.
Model answer: ℵ_ω is singular with cf(ℵ_ω) = ω = ℵ₀. König's theorem states cf(2^κ) > κ for any cardinal κ. Setting κ = ℵ₀ gives cf(2^{ℵ₀}) > ℵ₀. But if 2^{ℵ₀} = ℵ_ω, then cf(2^{ℵ₀}) = cf(ℵ_ω) = ω = ℵ₀ ≤ ℵ₀, directly contradicting König's inequality. Therefore 2^{ℵ₀} ≠ ℵ_ω is provable in ZFC without additional axioms.
This is a clean example of how cofinality constraints propagate through cardinal arithmetic. König's theorem acts as a filter on what values cardinal exponentiation can take: the result must have cofinality greater than the base cardinal. Any singular cardinal with cofinality ≤ ℵ₀ is immediately ruled out as the value of 2^{ℵ₀}. This shows that ZFC, despite leaving the exact value of the continuum undetermined, still imposes non-trivial structural constraints on what that value can be.