> [!definition] > > Let $X$ be a [[n-Manifold|n-manifold]], and let $B \subset X$. $B$ is a **coordinate ball** if there exists a chart $(U, \varphi)$, $p \in X$, and $r > 0$ such that $\widehat{B} = B(\hat p, r)$, and $B$ is **regular** if there exists a coordinate ball $B'$ such that $B' \supset \ol{B}$. > [!theorem] > > Let $X$ be a $n$-manifold, then $X$ has a [[Base|base]] consisting of countably many regular coordinate balls. > > *Proof*. Let $x \in X$, then there exists a chart $(U, \varphi)$ such that $x \in U$. Since $\widehat{U}$ is [[Open Set|open]], there exists $r > 0$ such that $B(\hat x, r) \subset \widehat{U}$. Let $\{\widehat{B_{x, q_n}}\}_1^\infty$ be an enumeration of open balls of rational radius strictly less than $r$, then they are all regular. Moreover, let $V \in \cn^o(x)$ be an open [[Neighbourhood|neighbourhood]]. Assume without loss of generality that $V \subset U$, then $\widehat{V}$ is open in $\widehat{U}$, and there exists $q_n$ such that $\widehat{B_{x, q_n}} \subset \widehat{V}$. Therefore $\{\widehat{B_{x, q_n}}\}_1^\infty$ forms a countable [[Neighbourhood Base|neighbourhood base]] for $x$. > > Since $X$ is [[Second Countable|second countable]], there exists a countable dense subset. For each one of them, we can find a neighbourhood base consisting of regular coordinate balls. Combining them gives the desired base.