Let \(X\) be a Hausdorff space such that each point of \(X\) has a neighborhood that is homeomorphic with an open subset of \(\mathbb{R}^m\). Show that if \(X\) is compact, then \(X\) is an \(m\)-manifold.
Solution
By hypothesis, for each \(x\in X\) there is a neighborhood \(U_x\) homeomorphic to an open subset of \(\mathbb R^m\). Choose an open set \(V_x\) with \[ x\in V_x\subset U_x. \] Each \(V_x\) is an open subspace of a Euclidean open set, hence is second countable.
Compactness gives finitely many points \(x_1,\dots,x_r\) such that \[ X=V_{x_1}\cup\cdots\cup V_{x_r}. \] For each \(i\), choose a countable basis \(\mathcal B_i\) for the open subspace \(V_{x_i}\). Since each \(V_{x_i}\) is open in \(X\), \[ \mathcal B=\bigcup_{i=1}^r\mathcal B_i \] is a countable basis for \(X\). The space is Hausdorff by assumption and locally homeomorphic to \(\mathbb R^m\) by hypothesis. Thus it satisfies the standard definition of an \(m\)-manifold.