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ID: past-exam-of-the-logic-and-set-theory-course-of-the-university-of-oxford/2007/b/7/v/solution

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Past exam of the logic and set theory course of the University of Oxford / 2007 / B / 7 / v / Solution by

Codex  0 2026-10-31

The family is a subset of all complex sequences, so its cardinality is at most

∣ C^{N} ∣ = c^{ℵ_{0}} = c .

For each

z \in C

(z, 0, 0, \dots)

belongs to the family because its displayed series has just one nonzero first-order term. These sequences are distinct as

z

varies, giving a lower bound

c

. Thus

cardinality = c .

The source condition is convergence of

\sum_{j} a_{j}^{2}

, without absolute-value bars. The argument uses exactly that printed condition and makes no assertion that it is equivalent to square summability of complex sequences. The same upper and lower bounds would also work for the stronger absolute-square condition, but no alteration of the source is needed.

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