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    Countable basis versus metrizability for locally compact Hausdorff spaces

    E-RX228

    Area
    Topology
    Topics
    Compactness, Metrizability
    Status
    draft

    Let \(X\) be a locally compact Hausdorff space. Is it true that if \(X\) has a countable basis, then \(X\) is metrizable? Is it true that if \(X\) is metrizable, then \(X\) has a countable basis?

    Solution

    1. Answer to Question 1: Yes, second countable implies metrizable:

    1.1. A locally compact Hausdorff space is regular (\(T_3\)): for any point \(x \in X\) and any closed set \(C \subset X\) not containing \(x\), local compactness and the Hausdorff property imply that \(x\) and \(C\) can be separated by disjoint open sets.

    Munkres Theorem 32.2 (locally compact Hausdorff spaces are completely regular).

    1.2. By Urysohn’s Metrization Theorem, every regular space with a countable basis (second-countable) is metrizable.

    Urysohn’s Metrization Theorem.

    1.3. Thus, if \(X\) is a locally compact Hausdorff space with a countable basis, \(X\) is metrizable.

    1 and 2.

    2. Answer to Question 2: No, metrizability does not imply second countability:

    2.1. Let \(X\) be any uncountable set (such as \(\mathbb{R}\)) equipped with the discrete metric \(d(x, y) = 1\) for \(x \neq y\) and \(d(x, x) = 0\).

    definition of the discrete metric space.

    2.2. \(X\) is a metric space, hence metrizable and Hausdorff.

    metric spaces are metrizable and Hausdorff.

    2.3. \(X\) is locally compact: for each \(x \in X\), the singleton \(\{x\} = B(x, 1/2)\) is an open neighborhood of \(x\) which is finite, hence compact.

    finite subsets of a topological space are compact.

    2.4. In the discrete topology, every singleton \(\{x\}\) is an open set.Any topological basis \(\mathcal{B}\) for \(X\) must contain a subset of \(\{x\}\) around \(x\), which forces \(\{x\} \in \mathcal{B}\) for every \(x \in X\).

    definition of a topological basis.

    2.5. Since \(X\) is uncountable, the cardinality of any basis is \(|\mathcal{B}| \ge |X| > \aleph_0\).Thus \(X\) does not have a countable basis.

    uncountability of \(X\).

    3. Conclusion:

    1. Yes: every second-countable locally compact Hausdorff space is metrizable.

    2. No: an uncountable discrete space is a metrizable, locally compact Hausdorff space without a countable basis.

    1 and 2.

    Q.E.D.