Show that if \(X\) has a countable basis, a collection \(\mathcal{A}\) of subsets of \(X\) is countably locally finite if and only if it is countable.
Solution
1. Backward direction (\(\impliedby\)): Countable implies countably locally finite.Proof:
1.1. Suppose \(\mathcal{A}\) is countable, so \(\mathcal{A} = \{A_k\}_{k=1}^\infty\) (or finite).
1.2. Decompose \(\mathcal{A} = \bigcup_{k=1}^\infty \mathcal{A}_k\) where each \(\mathcal{A}_k = \{A_k\}\).
1.3. A collection consisting of a single set \(\{A_k\}\) is locally finite in any space: each point \(x \in X\) has the open neighborhood \(X\) itself, which meets only the one member \(A_k\).
1.4. Thus \(\mathcal{A}\) is a countable union of locally finite collections, hence countably locally finite.
2. Lemma: In a second-countable space, any locally finite collection is countable.Proof:
2.1. Let \(\mathcal{B}\) be a locally finite collection of non-empty subsets of \(X\).
2.2. By local finiteness, for each \(x \in X\), there exists an open neighborhood \(U_x\) of \(x\) that intersects only finitely many elements of \(\mathcal{B}\).
2.3. The family \(\mathcal{U} = \{U_x\}_{x \in X}\) forms an open covering of \(X\).
2.4. Since \(X\) is second-countable, \(X\) is Lindelöf, so there exists a countable subcover \(\{U_{x_k}\}_{k=1}^\infty\) of \(X\).
2.5. For each \(k \ge 1\), the collection \(\mathcal{B}_k = \{B \in \mathcal{B} \mid B \cap U_{x_k} \neq \varnothing\}\) is finite.
2.6. Because \(\bigcup_{k=1}^\infty U_{x_k} = X\), every non-empty set \(B \in \mathcal{B}\) intersects some \(U_{x_k}\), which means \(\mathcal{B} = \bigcup_{k=1}^\infty \mathcal{B}_k\).
2.7. As a countable union of finite sets, \(\mathcal{B}\) is at most countable.
3. Forward direction (\(\implies\)): Countably locally finite implies countable.Proof:
3.1. By definition, \(\mathcal{A} = \bigcup_{n=1}^\infty \mathcal{A}_n\) where each \(\mathcal{A}_n\) is a locally finite collection.
3.2. By Lemma 2, each subcollection \(\mathcal{A}_n\) is at most countable.
3.3. As a countable union of countable collections, \(\mathcal{A}\) is countable.
4. Q.E.D.Steps 1 and 3 give the two implications.