A \(G_\delta\) set in a space \(X\) is a set \(A\) that equals a countable intersection of open sets of \(X\). Show that in a first-countable \(T_1\) space, every one-point set is a \(G_\delta\) set.
There is a familiar space in which every one-point set is a \(G_\delta\) set, which nevertheless does not satisfy the first countability axiom. What is it?
Solution
Goal: Prove that one-point sets in first-countable \(T_1\) spaces are \(G_\delta\) sets, and identify a familiar non-first-countable space with \(G_\delta\) singletons.
1. Part (a): Singletons are \(G_\delta\) in first-countable \(T_1\) spaces.Proof:
1.1. Let \(x \in X\). Since \(X\) is first-countable, there exists a countable local neighborhood basis \(\{B_n\}_{n=1}^\infty\) of open neighborhoods at \(x\).
1.2. We claim that \(\{x\} = \bigcap_{n=1}^\infty B_n\).
1.3. Since \(x \in B_n\) for every \(n \ge 1\), \(\{x\} \subseteq \bigcap_{n=1}^\infty B_n\).
1.4. For any point \(y \in X\) with \(y \neq x\), the \(T_1\) separation axiom implies that the complement \(U = X \setminus \{y\}\) is an open neighborhood of \(x\).
1.5. By definition of a local neighborhood basis, there exists some \(n_0 \ge 1\) such that \(B_{n_0} \subseteq U = X \setminus \{y\}\).
1.6. Thus \(y \notin B_{n_0}\), which implies \(y \notin \bigcap_{n=1}^\infty B_n\).
1.7. Therefore, \(\bigcap_{n=1}^\infty B_n = \{x\}\).
1.8. Because each \(B_n\) is open, \(\{x\}\) is a countable intersection of open sets, hence a \(G_\delta\) set.
2. Part (b): Space where singletons are \(G_\delta\) but first-countability fails.A familiar example is the product space \(\mathbb{R}^\omega = \prod_{n=1}^\infty \mathbb{R}\) equipped with the box topology.Proof:
2.1. Singletons are \(G_\delta\):Let \(\mathbf{x} = (x_n)_{n=1}^\infty \in \mathbb{R}^\omega\). For each integer \(k \ge 1\), define the box-open set:\[U_k = \prod_{n=1}^\infty \left(x_n - \frac{1}{k}, x_n + \frac{1}{k}\right).\]Then:\[\bigcap_{k=1}^\infty U_k = \prod_{n=1}^\infty \bigcap_{k=1}^\infty \left(x_n - \frac{1}{k}, x_n + \frac{1}{k}\right) = \prod_{n=1}^\infty \{x_n\} = \{\mathbf{x}\}.\]Thus every one-point set in \(\mathbb{R}^\omega_{\text{box}}\) is a \(G_\delta\) set.
2.2. Failure of first countability:As shown in §30 Example 2, the origin \(\mathbf{0} \in \mathbb{R}^\omega\) has no countable local neighborhood basis in the box topology (given any countable family \(\{B_m\}_{m=1}^\infty\) of box-open neighborhoods of \(\mathbf{0}\) with \(B_m = \prod_{n=1}^\infty (-\varepsilon_{m, n}, \varepsilon_{m, n})\), the diagonal open neighborhood \(W = \prod_{n=1}^\infty (-\frac{1}{2}\varepsilon_{n, n}, \frac{1}{2}\varepsilon_{n, n})\) contains no \(B_m\)).
2.3. Hence \(\mathbb{R}^\omega\) in the box topology is not first-countable.
3. Conclusion:Singletons are \(G_\delta\) in any first-countable \(T_1\) space, and \(\mathbb{R}^\omega\) with the box topology is a non-first-countable space with \(G_\delta\) singletons. Q.E.D.