Coordinate Balls form a Basis for Manifolds

Last updated Nov 1, 2022

Let $M$ be a Topological Manifold of Manifold Dimension $n$. Then $$\mathcal{B} := {B \subset M : B \text{ is a coordinate ball}}$$ is a Topological Basis for $M$.

We will carry over the defining Topological Basis from $\mathbb{R}^{n}$. Suppose $U \subset M$ is Open. Let $(V, \varphi)$ be any Coordinate Chart of $M$. Because $\varphi$ is a Homeomorphism and A Function is a Homeomorphism iff it is a Bijective Continuous Open Map, $\varphi(V \cap U)$ is Open in $\mathbb{R}^{n}$. Thus it can be written as a Set Union of Open Balls (by definition of Metric Topology). Thus $V \cap U$ can be written as a Set Union of Coordinate Balls. Since $(V, \varphi)$ was arbitrary, $U$ can be written as a Set Union of Coordinate Balls (from potentially multiple different Coordinate Charts). Since $U$ was arbitrary, we see that Coordinate Balls form a Topological Basis for $M$. $\blacksquare$