Give an example of a collection of sets \(\mathcal{A}\) that is not locally finite, such that the collection \(\mathcal{B} = \ts{\overline{A} \mid A \in \mathcal{A}}\) is locally finite.
Solution
1. Definition of local finiteness:A collection \(\mathcal{F}\) of subsets of \(X\) is locally finite if every point \(x \in X\) has an open neighborhood that intersects only finitely many members of \(\mathcal{F}\).
2. Construction of the example:Proof:
2.1. Let \(X = \mathbb{R}\) equipped with the standard Euclidean topology.
2.2. For each integer \(n \ge 2\), define the subset:\[A_n = (0, 1) \setminus \left\{\frac{1}{n}\right\}.\]
2.3. Define the collection \(\mathcal{A} = \{A_n \mid n \ge 2\}\).
2.4. The sets \(\{A_n\}_{n=2}^\infty\) are pairwise distinct because \(\frac{1}{m} \in A_n\) for all \(m \neq n\), while \(\frac{1}{n} \notin A_n\).
3. \(\mathcal{A}\) is not locally finite:Proof:
3.1. Consider the point \(x = \frac{1}{2} \in \mathbb{R}\).
3.2. Let \(U\) be an arbitrary open neighborhood of \(x = \frac{1}{2}\) in \(\mathbb{R}\).
3.3. There exists \(\varepsilon > 0\) such that \((\frac{1}{2} - \varepsilon, \frac{1}{2} + \varepsilon) \subseteq U \cap (0, 1)\).
3.4. For every \(n \ge 3\), the interval \((\frac{1}{2} - \varepsilon, \frac{1}{2} + \varepsilon)\) contains points other than \(\frac{1}{n}\), so \(U \cap A_n \neq \varnothing\).
3.5. Thus \(U\) intersects infinitely many distinct elements \(A_n \in \mathcal{A}\).
3.6. Hence \(\mathcal{A}\) is not locally finite at \(\frac{1}{2}\).
4. \(\mathcal{B} = \{\overline{A} \mid A \in \mathcal{A}\}\) is locally finite:Proof:
4.1. For every \(n \ge 2\), the topological closure of \(A_n = (0, 1) \setminus \{1/n\}\) in the Euclidean metric is:\[\overline{A}_n = [0, 1].\]
4.2. As a set of subsets of \(\mathbb{R}\), the collection of closures collapses to a single set:\[\mathcal{B} = \{ \overline{A}_n \mid n \ge 2 \} = \{ [0, 1] \}.\]
4.3. Since \(\mathcal{B}\) is a finite collection (consisting of exactly \(1\) set), every open neighborhood of any point \(x \in \mathbb{R}\) intersects at most \(1\) member of \(\mathcal{B}\).
4.4. Therefore \(\mathcal{B}\) is locally finite.
5. Q.E.D.By steps 3 and 4, \(\mathcal{A} = \{(0, 1) \setminus \{1/n\} \mid n \ge 2\}\) is not locally finite and \(\mathcal{B} = \{[0, 1]\}\) is locally finite.