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

The family is a subset of all complex sequences, so its cardinality is at mostFor each $z\in \mathbb{C}$, the sequence $(z,0,0,\dots )$ 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 $\mathfrak{c}$. ThusThe 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.