 Solution

ID: past-exam-of-the-analytic-topology-and-axiomatic-set-theory-course-of-the-university-of-oxford/2010/b/5/d/ii/solution

Put

κ = ℵ_{ω}

ℵ_{n} < ℵ_{n + 1}

κ = \sum_{n < ω} ℵ_{n} < \prod_{n < ω} ℵ_{n + 1} \leq κ^{ℵ_{0}} .

The sum is

κ

because the countable union has cardinality at most

κ

and its summands are cofinal in

κ

. Thus the power is strictly larger than

κ

, a fact that cannot be replaced by the regular-cardinal calculation of the preceding part.

For an upper bound, use

κ \leq 2^{κ}

κ^{ℵ_{0}} \leq (2^{κ})^{ℵ_{0}} = 2^{κ \cdot ℵ_{0}} = 2^{κ} = κ^{+},

where the last step is GCH at

κ

. There is no cardinal strictly between

κ

and its successor, so

ℵ_{ω}^{ℵ_{0}} = ℵ_{ω + 1} .

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