Show that a regular Lindelöf space is metrizable if it is locally metrizable. [Hint: A closed subspace of a Lindelöf space is Lindelöf.] Regularity is essential; where do you use it in the proof?
Solution
Let \(x\in X\). Choose a metrizable neighborhood \(N_x\) of \(x\), and choose an open set \(U_x\) with \[ x\in U_x\subset N_x. \] Regularity is used now: choose an open \(V_x\) such that \[ x\in V_x\subset\overline{V_x}\subset U_x. \] The subspace \(\overline{V_x}\) is closed in Lindelöf \(X\), hence Lindelöf. It is also a subspace of the metrizable space \(N_x\), so it is metrizable. Every Lindelöf metric space is second countable; therefore \(\overline{V_x}\), and hence its open subspace \(V_x\), has a countable basis.
The family \(\{V_x:x\in X\}\) is an open cover of Lindelöf \(X\). Choose a countable subcover \[ V_{x_1},V_{x_2},\dots. \] For each \(i\), let \(\mathcal B_i\) be a countable basis for \(V_{x_i}\). Since the \(V_{x_i}\) are open in \(X\), \[ \mathcal B=\bigcup_{i=1}^\infty\mathcal B_i \] is a countable basis for \(X\). The space \(X\) is regular and second countable, so the Urysohn metrization theorem implies \(X\) is metrizable.
The essential use of regularity was the shrinking \[ x\in V_x\subset\overline{V_x}\subset U_x, \] which makes \(\overline{V_x}\) a closed Lindelöf subspace while keeping it inside a metrizable neighborhood.