ℶ₁ is defined as 2^{ℵ₀}. What famous open mathematical question is precisely equivalent to asking whether ℶ₁ = ℵ₁?
ACantor's theorem: is every infinite set strictly smaller than its power set?
BThe axiom of choice: does every set have a well-ordering?
CThe continuum hypothesis: is there no cardinal strictly between ℵ₀ and 2^{ℵ₀}?
DThe axiom of regularity: do infinite sets ever contain themselves?
ℶ₁ = 2^{ℵ₀} by definition. ℵ₁ is the first uncountable cardinal — the smallest cardinal strictly larger than ℵ₀. Asking ℶ₁ = ℵ₁ is precisely asking whether 2^{ℵ₀} = ℵ₁, i.e., whether no uncountable cardinal exists below the power set of ℕ. This is the continuum hypothesis, proven independent of ZFC by Gödel and Cohen. Cantor's theorem (option A) is already proved — it shows ℶ₁ > ℵ₀, but says nothing about how much larger ℶ₁ is.
Question 2 Multiple Choice
A set theorist asks: 'What is the cardinality of P(ℝ) — the power set of the real numbers?' The correct answer in beth notation is:
Aℶ₁, because P(ℝ) is the power set of an uncountable set
Bℶ₂, because |ℝ| = ℶ₁ and P(ℝ) = 2^{ℶ₁} = ℶ₂ by definition
Cℵ₂, by the generalized continuum hypothesis
D2^{ℵ₁}, since |ℝ| = ℵ₁ by the continuum hypothesis
|ℝ| = ℶ₁ = 2^{ℵ₀} by definition. Then |P(ℝ)| = 2^{|ℝ|} = 2^{ℶ₁} = ℶ₂. Beth notation is natural here precisely because ℶ₂ is defined as one power-set application above ℶ₁. Option C would only be correct if GCH holds (ℶ₁ = ℵ₁, so ℶ₂ = ℵ₂). Option D assumes CH (|ℝ| = ℵ₁), which is independent of ZFC. The clean beth answer — ℶ₂ — holds regardless of GCH, making it more robust.
Question 3 True / False
For any ordinal α, the beth number ℶ_α is greater than or equal to the aleph number ℵ_α.
TTrue
FFalse
Answer: True
This inequality holds in ZFC without assuming GCH. The proof is by transfinite induction: ℶ₀ = ℵ₀ (equality at 0). For successor ordinals, ℶ_{α+1} = 2^{ℶ_α} ≥ 2^{ℵ_α} ≥ ℵ_{α+1} (using the inductive hypothesis and König's theorem). At limit ordinals, the sup preserves the inequality. GCH strengthens this to equality everywhere; without GCH, the beths can strictly exceed the corresponding alephs by an arbitrary amount.
Question 4 True / False
The beth and aleph sequences measure the same thing — the beth sequence is simply an alternative notation for the same aleph numbers.
TTrue
FFalse
Answer: False
They measure fundamentally different things. The aleph sequence enumerates infinite cardinals by ordinal rank: ℵ₁ is the first uncountable cardinal, ℵ₂ is the next, defined by position in the ordering of all cardinals. The beth sequence enumerates cardinals by power-set iteration: ℶ₁ is |P(ℕ)|, ℶ₂ is |P(P(ℕ))|, regardless of their aleph-rank. They start equal (ℶ₀ = ℵ₀) but may diverge at all higher indices. They are equal for all α if and only if GCH holds — which is independent of ZFC.
Question 5 Short Answer
What is the key difference between how aleph numbers and beth numbers measure the sizes of infinite sets?
Think about your answer, then reveal below.
Model answer: Aleph numbers measure infinite cardinalities by ordinal rank — ℵ₁ is simply the first uncountable cardinal, ℵ₂ the next, defined by their position in the well-ordering of all infinite cardinals. Beth numbers measure cardinalities by power-set iteration — ℶ₁ = 2^{ℵ₀} is the cardinality of P(ℕ), ℶ₂ = 2^{ℶ₁} is the cardinality of P(P(ℕ)), each step applying one more power set. The sequences coincide at ℶ₀ = ℵ₀ but whether they remain equal at higher indices is exactly what GCH asserts — and GCH is independent of ZFC.
The practical value of the distinction: the beth hierarchy is the natural yardstick for cardinalities arising from actual constructions (power sets, function spaces), while the aleph hierarchy is the complete enumeration of all infinite cardinalities. You can state |ℝ| = ℶ₁ and |P(ℝ)| = ℶ₂ with certainty. Whether these equal ℵ₁ and ℵ₂ respectively depends on an independent axiom. Beth notation lets you reason about set sizes without committing to the open question of their rank among all cardinals.