Give an example showing that a Hausdorff space with a countable basis need not be metrizable.
Solution
Goal: Provide an explicit example of a topological space that is Hausdorff (\(T_2\)) and second-countable, but fails to be regular (\(T_3\)), hence cannot be metrizable.
1. Definition of the space \(\mathbb{R}_K\) (the \(K\)-topology on \(\mathbb{R}\)):Let \(K = \{\frac{1}{n} : n \in \mathbb{Z}_+\} = \{1, \frac{1}{2}, \frac{1}{3}, \dots\} \subset \mathbb{R}\). The space \(\mathbb{R}_K\) is \(\mathbb{R}\) equipped with the topology generated by the basis:\[\mathcal{B} = \{(a, b) : a < b\} \cup \{(a, b) \setminus K : a < b\}.\]
2. \(\mathbb{R}_K\) is Hausdorff:Proof: The \(K\)-topology is strictly finer than the standard Euclidean topology on \(\mathbb{R}\). Since \(\mathbb{R}\) with the standard topology is Hausdorff, any finer topology is also Hausdorff.
3. \(\mathbb{R}_K\) has a countable basis (second-countable):Proof:
3.1. Define \(\mathcal{B}_\mathbb{Q} = \{(p, q) : p, q \in \mathbb{Q}, p < q\} \cup \{(p, q) \setminus K : p, q \in \mathbb{Q}, p < q\}\).
3.2. Since \(\mathbb{Q} \times \mathbb{Q}\) is countable, \(\mathcal{B}_\mathbb{Q}\) is a countable collection of open sets.
3.3. For any \(x \in (a, b)\), choose rationals \(p, q \in \mathbb{Q}\) with \(a < p < x < q < b\); then \(x \in (p, q) \subseteq (a, b)\).
3.4. Similarly, for any \(x \in (a, b) \setminus K\), choose rationals \(p, q\) with \(a < p < x < q < b\); then \(x \in (p, q) \setminus K \subseteq (a, b) \setminus K\).
3.5. Thus \(\mathcal{B}_\mathbb{Q}\) is a countable basis for the topology of \(\mathbb{R}_K\).
4. \(\mathbb{R}_K\) is not regular:Proof:
4.1. The set \(K\) is closed in \(\mathbb{R}_K\) because \(\mathbb{R} \setminus K = (-\infty, 0) \cup ((0, 2) \setminus K) \cup (1, \infty)\) is an open set in \(\mathbb{R}_K\).
4.2. The point \(0\) does not belong to \(K\) (\(0 \notin K\)).
4.3. Suppose there exist disjoint open sets \(U, V \subseteq \mathbb{R}_K\) such that \(0 \in U\) and \(K \subseteq V\).
4.4. Since \(0 \in U\) and \(0 \notin K\), there exists a basic open neighborhood \((-\varepsilon, \varepsilon) \setminus K \subseteq U\) for some \(\varepsilon > 0\).
4.5. Since \(K \subseteq V\), for every \(n \in \mathbb{Z}_+\), \(\frac{1}{n} \in V\), so there exists an open interval \((a_n, b_n) \subseteq V\) containing \(\frac{1}{n}\).
4.6. Choose \(N \in \mathbb{Z}_+\) such that \(\frac{1}{N} < \varepsilon\).
4.7. The interval \((a_N, b_N)\) contains points of \((-\varepsilon, \varepsilon)\) not in \(K\) (since \(K\) is discrete in \((0, \infty)\)).
4.8. Thus \(((-\varepsilon, \varepsilon) \setminus K) \cap (a_N, b_N) \neq \emptyset\), which implies \(U \cap V \neq \emptyset\), contradicting disjointness.
4.9. Hence the closed set \(K\) and the point \(0\) cannot be separated by disjoint open sets, so \(\mathbb{R}_K\) is not regular.
5. Conclusion: \(\mathbb{R}_K\) is not metrizable.Proof: Every metrizable space is regular (\(T_3\)). Since \(\mathbb{R}_K\) is not regular by 4, it cannot be metrizable. Q.E.D.