A space \(X\) is said to be perfectly normal if \(X\) is normal and if every closed set in \(X\) is a \(G_\delta\) set in \(X\).
Show that every metrizable space is perfectly normal.
Show that a perfectly normal space is completely normal. For this reason the condition of perfect normality is sometimes called the “\(T_6\) axiom.” [Hint: Let \(A\) and \(B\) be separated sets in \(X\). Choose continuous functions \(f, g: X \to [0, 1]\) that vanish precisely on \(\overline{A}\) and \(\overline{B}\), respectively. Consider the function \(f - g\).]
There is a familiar space that is completely normal but not perfectly normal. What is it?
Solution
Goal: Characterize perfectly normal spaces (\(T_6\)), prove that metrizability implies perfect normality, perfect normality implies complete normality (\(T_5\)), and identify the ordinal space \([0, \Omega]\) as a completely normal non-perfectly normal space.
1. Part (a): Every metrizable space is perfectly normal.Proof:
1.1. Every metric space \((X, d)\) is \(T_4\) (normal).
1.2. Let \(A \subseteq X\) be an arbitrary closed set.
1.3. For each \(n \in \mathbb{Z}_+\), define \(U_n = \{x \in X : d(x, A) < \frac{1}{n}\}\).
1.4. The distance function \(x \mapsto d(x, A)\) is continuous, so each \(U_n\) is open in \(X\).
1.5. Since \(A\) is closed, \(x \in A \iff d(x, A) = 0 \iff \forall n \ge 1, d(x, A) < \frac{1}{n} \iff x \in \bigcap_{n=1}^\infty U_n\).
1.6. Thus \(A = \bigcap_{n=1}^\infty U_n\) is a countable intersection of open sets (\(G_\delta\)).
1.7. Hence \(X\) is perfectly normal.
2. Part (b): Perfectly normal spaces are completely normal (\(T_5\)).Proof:
2.1. Let \(A, B \subseteq X\) be separated sets, so \(\overline{A} \cap B = \emptyset\) and \(A \cap \overline{B} = \emptyset\).
2.2. Since \(X\) is perfectly normal, the closed sets \(\overline{A}\) and \(\overline{B}\) are closed \(G_\delta\) sets in \(X\).
2.3. By Urysohn’s theorem for \(G_\delta\) zero sets, there exist continuous functions \(f, g: X \to [0, 1]\) vanishing precisely on \(\overline{A}\) and \(\overline{B}\):\[f^{-1}(\{0\}) = \overline{A}, \qquad g^{-1}(\{0\}) = \overline{B}.\]
2.4. Define \(h: X \to \mathbb{R}\) by \(h(x) = f(x) - g(x)\). Since \(f\) and \(g\) are continuous, \(h\) is continuous.
2.5. Define \(U = h^{-1}((-\infty, 0)) = \{x \in X : f(x) < g(x)\}\) and \(V = h^{-1}((0, \infty)) = \{x \in X : f(x) > g(x)\}\).
2.6. Since \((-\infty, 0)\) and \((0, \infty)\) are disjoint open intervals in \(\mathbb{R}\), \(U\) and \(V\) are disjoint open subsets of \(X\).
2.7. For any \(a \in A \subseteq \overline{A}\), \(f(a) = 0\). Since \(a \notin \overline{B}\), \(g(a) > 0\), so \(f(a) < g(a) \implies a \in U\). Thus \(A \subseteq U\).
2.8. For any \(b \in B \subseteq \overline{B}\), \(g(b) = 0\). Since \(b \notin \overline{A}\), \(f(b) > 0\), so \(f(b) > g(b) \implies b \in V\). Thus \(B \subseteq V\).
2.9. Therefore separated sets are separated by disjoint open sets, so \(X\) is completely normal.
3. Part (c): The space \(\overline{S}_\Omega = [0, \Omega]\) is completely normal but not perfectly normal.Proof:
3.1. Every well-ordered set in the order topology is completely normal (Munkres Theorem 32.4).
3.2. The singleton \(\{\Omega\}\) is closed in \(\overline{S}_\Omega\).
3.3. Suppose \(\{\Omega\} = \bigcap_{n=1}^\infty U_n\) for some collection of open sets \(U_n \subseteq \overline{S}_\Omega\).
3.4. For each \(n\), \(\Omega \in U_n\), so there exists \(\alpha_n < \Omega\) such that \((\alpha_n, \Omega] \subseteq U_n\).
3.5. The countable set \(\{\alpha_n : n \in \mathbb{Z}_+\} \subset S_\Omega\) has an upper bound \(\alpha = \sup_{n} \alpha_n < \Omega\) in \(S_\Omega\).
3.6. Then the uncountably infinite set \((\alpha, \Omega) \subset \bigcap_{n=1}^\infty (\alpha_n, \Omega] \subseteq \bigcap_{n=1}^\infty U_n = \{\Omega\}\), a contradiction.
3.7. Thus \(\{\Omega\}\) is not a \(G_\delta\) set, so \(\overline{S}_\Omega\) is not perfectly normal. Q.E.D.