Recall that \(A\) is a “\(G_\delta\) set” in \(X\) if \(A\) is the intersection of a countable collection of open sets of \(X\).
Theorem. Let \(X\) be normal. There exists a continuous function \(f: X \to [0, 1]\) such that \(f(x) = 0\) for \(x \in A\), and \(f(x) > 0\) for \(x \notin A\), if and only if \(A\) is a closed \(G_\delta\) set in \(X\).
A function satisfying the requirements of this theorem is said to vanish precisely on \(A\).
Solution
Goal: Prove that in a normal space \(X\), a subset \(A \subseteq X\) is the zero set of a continuous function \(f: X \to [0, 1]\) if and only if \(A\) is a closed \(G_\delta\) set.
1. Direct implication (\(\implies\)): Zero sets of continuous functions are closed \(G_\delta\) sets.Proof:
1.1. Suppose \(f: X \to [0, 1]\) is continuous and \(A = \{x \in X : f(x) = 0\} = f^{-1}(\{0\})\).
1.2. Since \(\{0\}\) is closed in \([0, 1]\) and \(f\) is continuous, \(A\) is closed in \(X\).
1.3. For each \(n \in \mathbb{Z}_+\), let \(U_n = f^{-1}([0, \frac{1}{n})) = \{x \in X : f(x) < \frac{1}{n}\}\).
1.4. Since \([0, \frac{1}{n})\) is open in the subspace topology of \([0, 1]\), each \(U_n\) is open in \(X\).
1.5. \(x \in \bigcap_{n=1}^\infty U_n \iff \forall n \ge 1, 0 \le f(x) < \frac{1}{n} \iff f(x) = 0 \iff x \in A\).
1.6. Thus \(A = \bigcap_{n=1}^\infty U_n\) is a countable intersection of open sets, so \(A\) is a \(G_\delta\) set in \(X\).
2. Reverse implication (\(\impliedby\)): Constructing a continuous function vanishing on a closed \(G_\delta\) set.Proof:
2.1. Suppose \(A \subseteq X\) is closed and \(A = \bigcap_{n=1}^\infty U_n\), where each \(U_n\) is open in \(X\).
2.2. For each \(n \ge 1\), the complement \(X \setminus U_n\) is closed in \(X\) and disjoint from \(A\) (since \(A \subseteq U_n\)).
2.3. Since \(X\) is normal and \(A, X \setminus U_n\) are disjoint closed sets, Urysohn’s Lemma provides a continuous function \(f_n: X \to [0, 1]\) such that:\[f_n(x) = 0 \text{ for } x \in A, \qquad f_n(x) = 1 \text{ for } x \in X \setminus U_n.\]
2.4. Define \(f: X \to [0, 1]\) by the infinite series:\[f(x) = \sum_{n=1}^\infty \frac{1}{2^n} f_n(x).\]
2.5. Uniform convergence and continuity: For all \(x \in X\), \(|\frac{1}{2^n} f_n(x)| \le \frac{1}{2^n}\). Since \(\sum_{n=1}^\infty \frac{1}{2^n} = 1 < \infty\), the Weierstrass \(M\)-test proves that the series converges uniformly on \(X\). Because each partial sum is continuous, the limit function \(f\) is continuous.
2.6. Vanishing on \(A\): If \(x \in A\), then \(f_n(x) = 0\) for all \(n \ge 1\), so \(f(x) = \sum_{n=1}^\infty 0 = 0\).
2.7. Positivity outside \(A\): If \(x \notin A\), then \(x \notin \bigcap_{n=1}^\infty U_n\), so there exists some \(k \ge 1\) such that \(x \notin U_k\), meaning \(x \in X \setminus U_k\).
2.8. Then \(f_k(x) = 1\), and since \(f_n(x) \ge 0\) for all \(n\):\[f(x) \ge \frac{1}{2^k} f_k(x) = \frac{1}{2^k} > 0.\]
2.9. Thus \(f(x) = 0 \iff x \in A\), so \(f\) vanishes precisely on \(A\). Q.E.D.