 Solution

ID: past-exam-of-the-logic-and-set-theory-course-of-the-university-of-oxford/2010/b/4/e/solution

False. Take

κ = 2^{ℵ_{0}}

and

λ = ℵ_{0}

, both infinite. The explicit pairing bijection between

N^{2}

and

N

κ^{λ} = (2^{ℵ_{0}})^{ℵ_{0}} = 2^{ℵ_{0} \cdot ℵ_{0}} = 2^{ℵ_{0}} = κ .

The power-set cardinal need not be assumed well-orderable. Every bijection used in this calculation is explicitly available in ZF, so the counterexample does not use choice.

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