Let \(X\) be a space that is locally \(m\)-euclidean. Show that \(X\) is Hausdorff if and only if \(X\) is completely regular.
Solution
If \(X\) is completely regular, then by definition it is \(T_1\), and complete regularity separates points from closed sets; in particular distinct points have disjoint neighborhoods. Thus \(X\) is Hausdorff.
Conversely, suppose \(X\) is Hausdorff and locally \(m\)-euclidean. Every point has a neighborhood homeomorphic to an open subset of \(\mathbb R^m\); shrinking inside such a chart gives a neighborhood whose closure in the chart is compact. Hence \(X\) is locally compact. Every locally compact Hausdorff space is completely regular, being a subspace of its compact Hausdorff one-point compactification. Therefore \[ X\text{ is Hausdorff}\quad\Longleftrightarrow\quad X\text{ is completely regular}. \]