Let \(\mathcal{A}\) be the following collection of subsets of \(\mathbb{R}\):
\[ \mathcal{A} = \ts{(n, n+2) \mid n \in \mathbb{Z}}. \]
Which of the following collections refine \(\mathcal{A}\)?
\[ \begin{array}{l} \mathcal{B} = \ts{(x, x+1) \mid x \in \mathbb{R}}, \\ \mathcal{C} = \ts{(n, n+\tfrac{3}{2}) \mid n \in \mathbb{Z}}, \\ \mathcal{D} = \ts{(x, x+\tfrac{3}{2}) \mid x \in \mathbb{R}}. \end{array} \]
Solution
1. Definition of refinement:A collection \(\mathcal{R}\) of subsets of \(\mathbb{R}\) refines \(\mathcal{A}\) if for every element \(R \in \mathcal{R}\), there exists an element \(A \in \mathcal{A}\) such that \(R \subseteq A\).
2. \(\mathcal{B} = \{(x, x+1) \mid x \in \mathbb{R}\}\) refines \(\mathcal{A}\):Proof:
2.1. Let \(I = (x, x+1) \in \mathcal{B}\) for some \(x \in \mathbb{R}\).
2.2. Let \(n = \lfloor x \rfloor \in \mathbb{Z}\) be the greatest integer less than or equal to \(x\), so \(n \le x < n+1\).
2.3. Then the left endpoint satisfies \(x \ge n\), and the right endpoint satisfies \(x + 1 < (n + 1) + 1 = n + 2\).
2.4. Thus \((x, x+1) \subseteq (n, n+2) \in \mathcal{A}\).
2.5. Therefore \(\mathcal{B}\) refines \(\mathcal{A}\).
3. \(\mathcal{C} = \{(n, n + \frac{3}{2}) \mid n \in \mathbb{Z}\}\) refines \(\mathcal{A}\):Proof:
3.1. Let \(I = (n, n + \frac{3}{2}) \in \mathcal{C}\) for some \(n \in \mathbb{Z}\).
3.2. Since \(n \le n\) and \(n + \frac{3}{2} < n + 2\), we have \((n, n + \frac{3}{2}) \subset (n, n+2) \in \mathcal{A}\).
3.3. Therefore \(\mathcal{C}\) refines \(\mathcal{A}\).
4. \(\mathcal{D} = \{(x, x + \frac{3}{2}) \mid x \in \mathbb{R}\}\) does not refine \(\mathcal{A}\):Proof:
4.1. Consider the element \(J = (0.8, 2.3) \in \mathcal{D}\) obtained by taking \(x = 0.8\).
4.2. For \(J \subseteq (n, n+2)\) to hold for some \(n \in \mathbb{Z}\), we would require \(n \le 0.8\) (so \(n \le 0\)) and \(n + 2 \ge 2.3\) (so \(n \ge 0.3\)).
4.3. No integer \(n\) satisfies \(0.3 \le n \le 0\).
4.4. Testing candidates: \((0.8, 2.3) \not\subset (0, 2)\) since \(2.3 > 2\), and \((0.8, 2.3) \not\subset (1, 3)\) since \(0.8 < 1\).
4.5. Thus \((0.8, 2.3)\) is not contained in any element of \(\mathcal{A}\), so \(\mathcal{D}\) does not refine \(\mathcal{A}\).
5. Q.E.D.By steps 2, 3, and 4, \(\boxed{\mathcal{B} \text{ and } \mathcal{C} \text{ refine } \mathcal{A}; \ \mathcal{D} \text{ does not}}\).