An indexed family \(\ts{A_\alpha}\) of subsets of \(X\) is said to be a point-finite indexed family if each \(x \in X\) belongs to \(A_\alpha\) for only finitely many values of \(\alpha\).
Lemma (The shrinking lemma). Let \(X\) be a normal space; let \(\ts{U_1, U_2, \ldots}\) be a point-finite indexed open covering of \(X\). Then there exists an indexed open covering \(\ts{V_1, V_2, \ldots}\) of \(X\) such that \(\overline{V}_n \subset U_n\) for each \(n\).
Solution
Goal: Prove the Shrinking Lemma for countable point-finite open coverings \(\{U_n\}_{n=1}^\infty\) on a normal space \(X\).
1. Inductive construction of the family \(\{V_n\}_{n=1}^\infty\):We construct open sets \(V_k\) inductively such that for each \(k \ge 1\), \(\overline{V_k} \subseteq U_k\), and the intermediate family \(\{V_1, \dots, V_k, U_{k+1}, U_{k+2}, \dots\}\) covers \(X\).Proof:
1.1. Induction hypothesis for step \(k\):Assume \(V_1, \dots, V_{k-1}\) have been chosen such that \(\overline{V_j} \subseteq U_j\) for all \(j < k\), and the family \(\{V_1, \dots, V_{k-1}, U_k, U_{k+1}, \dots\}\) is an open cover of \(X\). (For \(k=1\), this is the original cover \(\{U_n\}_{n=1}^\infty\).)
1.2. Construction of closed set \(A_k\):Define:\[A_k = X \setminus \left( \bigcup_{j=1}^{k-1} V_j \cup \bigcup_{i=k+1}^\infty U_i \right).\]As the complement of an open set, \(A_k\) is closed in \(X\).
1.3. Containment \(A_k \subseteq U_k\):If \(x \in A_k\), then \(x \notin V_j\) for all \(j < k\) and \(x \notin U_i\) for all \(i > k\). Since \(\{V_1, \dots, V_{k-1}, U_k, U_{k+1}, \dots\}\) covers \(X\), \(x\) must belong to \(U_k\).
1.4. Application of normality:Because \(X\) is normal and \(A_k\) is closed with \(A_k \subseteq U_k\) (where \(U_k\) is open), there exists an open set \(V_k \subseteq X\) satisfying:\[A_k \subseteq V_k \subseteq \overline{V_k} \subseteq U_k.\]
1.5. Coverage of the updated family:By construction, \(V_k \cup \left( \bigcup_{j=1}^{k-1} V_j \cup \bigcup_{i=k+1}^\infty U_i \right) \supseteq A_k \cup (X \setminus A_k) = X\).Thus \(\{V_1, \dots, V_k, U_{k+1}, U_{k+2}, \dots\}\) covers \(X\).
2. Verification that \(\{V_n\}_{n=1}^\infty\) covers \(X\) via point-finiteness:Proof:
2.1. Let \(x \in X\) be an arbitrary point.
2.2. Since the original cover \(\{U_n\}_{n=1}^\infty\) is point-finite, the index set \(S_x = \{n \in \mathbb{Z}_+ \mid x \in U_n\}\) is finite and non-empty.
2.3. Let \(m = \max S_x\). By definition of \(m\), \(x \notin U_i\) for all \(i > m\).
2.4. By 1, the family \(\{V_1, \dots, V_m, U_{m+1}, U_{m+2}, \dots\}\) covers \(X\), so \(x\) belongs to this union.
2.5. Since \(x \notin \bigcup_{i=m+1}^\infty U_i\), it follows that \(x \in \bigcup_{j=1}^m V_j\).
2.6. Thus \(x \in V_j\) for some \(j \le m\), proving \(\bigcup_{n=1}^\infty V_n = X\).
3. Conclusion:The collection \(\{V_n\}_{n=1}^\infty\) is an open cover of \(X\) satisfying \(\overline{V_n} \subseteq U_n\) for every \(n \in \mathbb{Z}_+\). Q.E.D.