Show that every locally compact Hausdorff space is completely regular.
Solution
Let \(X\) be locally compact Hausdorff. Its one-point compactification \(X^*=X\cup\{\infty\}\) is compact Hausdorff, hence normal.
Let \(A\subset X\) be closed and \(x\in X\setminus A\). The set \[ A^*=A\cup\{\infty\} \] is closed in \(X^*\), because its complement \(X\setminus A\) is open in \(X\), hence open in \(X^*\) away from the point at infinity. By Urysohn’s lemma there is a continuous \[ F:X^*\to[0,1] \] with \(F(x)=0\) and \(F(A^*)=\{1\}\). Restricting \(F\) to \(X\) gives a continuous function separating \(x\) from \(A\). Therefore \(X\) is completely regular.