Discussion
Let \(\mathbf{B}\) be the \(\sigma\)-field of subsets of R generated by \(\{(a, b) \mid a, b \in R\}\). Which of the following is not true ?
Answer: C
The statement that is not true is all uncountable subsets of R belong to \(\mathbf{B}\).
\(\mathbf{B}\) is the Borel \(\sigma\)-field. It has the cardinality of the continuum, while there are \(2^{\mathfrak{c}}\) subsets of R, so most subsets, and in particular many uncountable ones, are not Borel. (A Vitali set is even non-Lebesgue-measurable.)
The others are true. Singletons are closed, \(\{a\} = \bigcap (a - 1/n, a + 1/n)\), so countable sets are countable unions of Borel sets. Closed intervals are complements of open sets, and \([a, b) = \{a\} \cup (a, b)\).
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