Show that every regular Lindelöf space is normal.
Solution
Let \(A,B\subset X\) be disjoint closed sets. Since \(X\) is regular, for each \(a\in A\) choose an open \(U_a\ni a\) with \[ \overline{U_a}\cap B=\varnothing. \] Because closed subspaces of Lindelöf spaces are Lindelöf, \(A\) is Lindelöf. Hence choose a countable family \(U_1,U_2,\dots\) covering \(A\). Similarly choose open sets \(V_1,V_2,\dots\) covering \(B\) with \[ \overline{V_n}\cap A=\varnothing. \]
Define \[ U=\bigcup_{n\ge1}\left(U_n\setminus\bigcup_{i=1}^n\overline{V_i}\right), \] \[ V=\bigcup_{n\ge1}\left(V_n\setminus\bigcup_{i=1}^n\overline{U_i}\right). \] Each is open. Every \(a\in A\) belongs to some \(U_n\), and \(a\notin\overline{V_i}\) for all \(i\), so \(A\subset U\). Similarly \(B\subset V\).
They are disjoint. If \(x\) belonged to both, choose \(m,n\) such that \[ x\in U_m\setminus\bigcup_{i\le m}\overline{V_i} \] and \[ x\in V_n\setminus\bigcup_{i\le n}\overline{U_i}. \] If \(m\le n\), then \(x\in U_m\subset\overline{U_m}\), contradicting the second membership; if \(n\le m\), then \(x\in V_n\subset\overline{V_n}\), contradicting the first. Thus \(X\) is normal.