Show that for a topological space \(X\), if every \(x\in X\) has a neighborhood whose closure is a regular space, then \(X\) is regular.
Solution
We use the neighborhood characterization of regularity: for every \(x\in X\) and every open neighborhood \(U\ni x\), it is enough to find an open \(V\ni x\) with \(\overline V\subset U\).
Fix \(x\in U\). By hypothesis there is a neighborhood \(N\) of \(x\) such that \(\overline N\), with the subspace topology, is regular. Choose an open set \(G\subset X\) with \[ x\in G\subset N\cap U. \] Then \(G\) is open in the subspace \(\overline N\). By regularity of \(\overline N\), there is a set \(W\), open in \(\overline N\), such that \[ x\in W, \qquad \overline W^{\,\overline N}\subset G. \] Write \(W=O\cap\overline N\) for some open \(O\subset X\), and set \(V=O\cap G\). Then \(V\) is open in \(X\), contains \(x\), and \(V\subset W\subset\overline N\). Since \(\overline N\) is closed in \(X\), \[ \overline V^{\,X}\subset \overline W^{\,\overline N}\subset G\subset U. \] Thus \(X\) is regular.