The Baire category theorem implies that \(\mathbb{R}\) cannot be written as a countable union of closed subsets having empty interiors. Show this fails if the sets are not required to be closed.
Solution
1. Construction of the partition:1. Let \(\mathbb{Q}\) be the set of rational numbers in \(\mathbb{R}\). Since \(\mathbb{Q}\) is countably infinite, choose an enumeration \(\mathbb{Q} = \{q_1, q_2, q_3, \dots\}\).2. Let \(A_0 = \mathbb{R} \setminus \mathbb{Q}\) be the set of all irrational numbers.3. For each \(n \in \mathbb{Z}_+\), let \(A_n = \{q_n\}\).4. Then \(\bigcup_{n=0}^\infty A_n = (\mathbb{R} \setminus \mathbb{Q}) \cup \bigcup_{n=1}^\infty \{q_n\} = (\mathbb{R} \setminus \mathbb{Q}) \cup \mathbb{Q} = \mathbb{R}\).
2. Each singleton \(A_n = \{q_n\}\) has empty interior for \(n \ge 1\):Proof: Every non-empty open subset of \(\mathbb{R}\) contains an open interval \((a, b)\) with \(a < b\), which contains uncountably many points. Since \(A_n\) contains exactly one point, \(A_n\) contains no open interval. Thus \(\operatorname{Int}(A_n) = \emptyset\).
3. The set of irrationals \(A_0 = \mathbb{R} \setminus \mathbb{Q}\) has empty interior:Proof: If \(\operatorname{Int}(\mathbb{R} \setminus \mathbb{Q}) \neq \emptyset\), it would contain a non-empty open interval \((a, b)\) with \(a < b\). By the density of \(\mathbb{Q}\) in \(\mathbb{R}\), there exists a rational number \(q \in (a, b) \subset \mathbb{R} \setminus \mathbb{Q}\), which contradicts \(q \in \mathbb{Q}\). Thus \(\operatorname{Int}(A_0) = \emptyset\).
4. \(A_0\) is not closed in \(\mathbb{R}\):Proof: By the density of the irrationals in \(\mathbb{R}\), the closure of \(A_0 = \mathbb{R} \setminus \mathbb{Q}\) is \(\overline{\mathbb{R} \setminus \mathbb{Q}} = \mathbb{R} \neq A_0\).
5. Q.E.D.By steps 1, 2, and 3, \(\mathbb{R} = \bigcup_{n=0}^\infty A_n\) is a countable union of subsets with empty interior, and by step 4 the set \(A_0\) is not closed.