Show that if \(X\) has a countable basis \(\ts{B_n}\), then every basis \(\mathcal{C}\) for \(X\) contains a countable basis for \(X\). [Hint: For every pair of indices \(n, m\) for which it is possible, choose \(C_{n,m} \in \mathcal{C}\) such that \(B_n \subset C_{n,m} \subset B_m\).]
Solution
Goal: Prove that if a topological space \(X\) has a countable basis \(\mathcal{B} = \{B_n\}_{n \in \mathbb{Z}_+}\), then every basis \(\mathcal{C}\) for \(X\) contains a countable subcollection \(\mathcal{C}' \subseteq \mathcal{C}\) that is also a basis for \(X\).
1. Selection of a countable subcollection \(\mathcal{C}' \subseteq \mathcal{C}\):Proof:
1.1. Define the index set of pairs:\[J = \{(n, m) \in \mathbb{Z}_+ \times \mathbb{Z}_+ \mid \exists C \in \mathcal{C} \text{ such that } B_n \subseteq C \subseteq B_m\}.\]
1.2. For each pair \((n, m) \in J\), choose one basis element \(C_{n, m} \in \mathcal{C}\) satisfying \(B_n \subseteq C_{n, m} \subseteq B_m\).
1.3. Define the subcollection:\[\mathcal{C}' = \{C_{n, m} \mid (n, m) \in J\} \subseteq \mathcal{C}.\]
1.4. Because \(J \subseteq \mathbb{Z}_+ \times \mathbb{Z}_+\), \(J\) is at most countable, and hence \(\mathcal{C}'\) is a countable collection.
2. Verification that \(\mathcal{C}'\) is a basis for \(X\):Proof:
2.1. Let \(U \subseteq X\) be an arbitrary open set, and let \(x \in U\).
2.2. Since \(\mathcal{B} = \{B_n\}\) is a basis and \(U\) is open, there exists \(m \in \mathbb{Z}_+\) such that \(x \in B_m \subseteq U\).
2.3. Since \(\mathcal{C}\) is a basis and \(B_m\) is open, there exists \(C \in \mathcal{C}\) such that \(x \in C \subseteq B_m\).
2.4. Since \(\mathcal{B}\) is a basis and \(C\) is open, there exists \(n \in \mathbb{Z}_+\) such that \(x \in B_n \subseteq C\).
2.5. Thus we have \(B_n \subseteq C \subseteq B_m\), which implies that the pair \((n, m)\) belongs to \(J\).
2.6. By definition of \(\mathcal{C}'\), there is an element \(C_{n, m} \in \mathcal{C}'\) satisfying \(B_n \subseteq C_{n, m} \subseteq B_m\).
2.7. Consequently:\[x \in B_n \subseteq C_{n, m} \subseteq B_m \subseteq U.\]
2.8. Thus \(x \in C_{n, m} \subseteq U\) with \(C_{n, m} \in \mathcal{C}'\).
3. Conclusion:\(\mathcal{C}'\) is a countable subset of \(\mathcal{C}\) that forms a basis for the topology of \(X\). Q.E.D.