 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-3/16h/e/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

I = ω

κ_{i} = ℵ_{i}

, and

λ_{i} = ℵ_{i + 1}

\sum_{i < ω} ℵ_{i} = ℵ_{ω},

whereas every factor in the product is at most

ℵ_{ω}

. Hence

ℵ_{ω} < \prod_{i < ω} ℵ_{i + 1} \leq (ℵ_{ω})^{ℵ_{0}} \leq (ℵ_{ω})^{ℵ_{ω}} .

Now suppose

2^{ℵ_{ω}} = ℵ_{ω}

. The exponent laws for infinite cardinals would give

(ℵ_{ω})^{ℵ_{ω}} = (2^{ℵ_{ω}})^{ℵ_{ω}} = 2^{ℵ_{ω} \cdot ℵ_{ω}} = 2^{ℵ_{ω}} = ℵ_{ω},

contradicting the strict inequality. Therefore

2^{ℵ_{ω}} \neq = ℵ_{ω} .

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