Discussion
Let F be the class of subsets of R defined by F = {A | either A is countable or Ac is countable}. Then, which of the following is not true ?
Answer: C
The statement that is not true is \([0, \infty) \in \mathbf{F}\).
F is the countable-cocountable class. The interval \([0, \infty)\) is uncountable, and its complement \((-\infty, 0)\) is uncountable too, so neither condition holds and \([0, \infty) \notin \mathbf{F}\).
The other statements are true. R belongs to F because its complement is empty, hence countable. F is closed under countable unions: if all \(B_{n}\) are countable the union is countable, and if some \(B_{m}\) has countable complement then the complement of the union lies inside \(B_{m}^{c}\) and is countable. Finally \(Q^{c}\), the irrationals, has complement Q, which is countable, so \(Q^{c} \in \mathbf{F}\). In fact F is a sigma-algebra.
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