Let \(X_1 \subset X_2 \subset \cdots\) be a sequence of spaces, where \(X_i\) is a closed subspace of \(X_{i+1}\) for each \(i\). Let \(X\) be the union of the \(X_i\); let us topologize \(X\) by declaring a set \(U\) to be open in \(X\) if \(U \cap X_i\) is open in \(X_i\) for each \(i\).
Show that this is a topology on \(X\) and that each space \(X_i\) is a subspace (in fact, a closed subspace) of \(X\) in this topology. This topology is called the topology coherent with the subspaces \(X_i\).
Show that \(f: X \to Y\) is continuous if \(f \mid X_i\) is continuous for each \(i\).
Show that if each space \(X_i\) is normal, then \(X\) is normal. [Hint: Given disjoint closed sets \(A\) and \(B\) in \(X\), set \(f\) equal to 0 on \(A\) and 1 on \(B\), and extend \(f\) successively to \(A \cup B \cup X_i\) for \(i = 1, 2, \ldots\).]
Solution
1. Part (a): Topology axioms and closed subspaces:
1.1. Topology axioms:- \(\emptyset \cap X_i = \emptyset\) and \(X \cap X_i = X_i\) are open in \(X_i\) for all \(i\), so \(\emptyset, X \in \mathcal{T}\).- If \(\{U_\alpha\}\) are open, \((\bigcup U_\alpha) \cap X_i = \bigcup (U_\alpha \cap X_i)\) is open in \(X_i\).- If \(U, V\) are open, \((U \cap V) \cap X_i = (U \cap X_i) \cap (V \cap X_i)\) is open in \(X_i\).Thus \(\mathcal{T}\) is a topology on \(X\).
set-theoretic distribution of intersections over unions.
1.2. Closed sets in \(X\):A subset \(F \subseteq X\) is closed in \(X\) if and only if \(F \cap X_i\) is closed in \(X_i\) for all \(i \ge 1\).
complement \(X \setminus F\) is open \(\iff (X \setminus F) \cap X_i = X_i \setminus (F \cap X_i)\) is open in \(X_i\).
1.3. \(X_k\) is a closed subspace of \(X\):For any fixed \(k \ge 1\) and all \(i \ge 1\):- If \(i \le k\), \(X_k \cap X_i = X_i\), which is closed in \(X_i\).- If \(i > k\), \(X_k\) is closed in \(X_{k+1} \subset \cdots \subset X_i\), so \(X_k \cap X_i = X_k\) is closed in \(X_i\).Thus \(X_k \cap X_i\) is closed in \(X_i\) for every \(i\), so \(X_k\) is closed in \(X\).Furthermore, the subspace topology on \(X_k\) coincides with its original topology.
transitivity of closed subspaces.
2. Part (b): Continuity criterion for coherent topology:
2.1. Let \(f: X \to Y\) be a map and \(V \subseteq Y\) an open set.Then \(f^{-1}(V) \cap X_i = (f|_{X_i})^{-1}(V)\) for all \(i \ge 1\).
definition of preimage.
2.2. By the definition of the coherent topology:\[ f^{-1}(V) \text{ is open in } X \iff \forall i \ge 1, \, (f|_{X_i})^{-1}(V) \text{ is open in } X_i. \]Thus \(f\) is continuous on \(X\) if and only if \(f|_{X_i}: X_i \to Y\) is continuous for all \(i \ge 1\).
characterization of continuity via preimages of open sets.
3. Part (c): Normality of \(X\):
3.1. Singletons \(\{x\}\) are closed in \(X\) because if \(x \in X_i\), \(\{x\}\) is closed in the \(T_1\) space \(X_i\), and \(X_i\) is closed in \(X\).
\(T_1\) property from normality of \(X_i\).
3.2. Let \(A, B \subset X\) be disjoint closed subsets of \(X\) (\(A \cap B = \emptyset\)).Construct a sequence of continuous functions \(f_n: X_n \to [0, 1]\) satisfying \(f_n|_{A \cap X_n} = 0\), \(f_n|_{B \cap X_n} = 1\), and \(f_{n+1}|_{X_n} = f_n\) by induction:- Base step (\(n = 1\)): \(A \cap X_1\) and \(B \cap X_1\) are disjoint closed sets in the normal space \(X_1\). By Urysohn’s Lemma, there exists a continuous \(f_1: X_1 \to [0, 1]\) with \(f_1(A \cap X_1) = 0\) and \(f_1(B \cap X_1) = 1\).- Inductive step: Given \(f_n: X_n \to [0, 1]\), let \(C = X_n \cup (A \cap X_{n+1}) \cup (B \cap X_{n+1})\).Define \(g: C \to [0, 1]\) by \(g = f_n\) on \(X_n\), \(g = 0\) on \(A \cap X_{n+1}\), and \(g = 1\) on \(B \cap X_{n+1}\).\(g\) is well-defined and continuous on the closed set \(C \subset X_{n+1}\) by the Pasting Lemma.Since \(X_{n+1}\) is normal, by the Tietze Extension Theorem \(g\) extends to a continuous function \(f_{n+1}: X_{n+1} \to [0, 1]\).
Tietze Extension Theorem on normal spaces.
3.3. Define \(f: X \to [0, 1]\) by \(f(x) = f_n(x)\) for \(x \in X_n\).\(f\) is well-defined since \(f_{n+1}|_{X_n} = f_n\), and by Part (b), \(f\) is continuous on \(X\).Moreover, \(f(A) = 0\) and \(f(B) = 1\).
Part (b) applied to \(f\).
3.4. The sets \(U = f^{-1}([0, 1/3))\) and \(V = f^{-1}((2/3, 1])\) are disjoint open neighborhoods of \(A\) and \(B\) in \(X\).Thus \(X\) is normal.
continuous separation of disjoint closed sets.
4. Conclusion:\((X, \mathcal{T})\) is a topological space, \(f: X \to Y\) is continuous iff \(f|_{X_i}\) is continuous for all \(i\), and \(X\) is normal. Q.E.D.
1 through 3.