Can $\{B(x_0,k)\}_k$ exhaust one complete Riemannian manifold?

Question

Recall first the basic knowledga about the geodesic theory.

Let $(M, g)$ be a riemannian manifold of dimension $m,$ with metric $$ g(x)=\sum g_{j k}(x) d x_{j} \otimes d x_{k}, \quad 1 \leqslant j, k \leqslant m $$ The length of a path $\gamma:[a, b] \longrightarrow M$ is by definition $$ \ell(\gamma)=\int_{a}^{b}\left|\gamma^{\prime}(t)\right|_{g} d t=\int_{a}^{b}\left(\sum_{j, k} g_{j k}(\gamma(t)) \gamma_{j}^{\prime}(t) \gamma_{k}^{\prime}(t)\right)^{1 / 2} d t $$ The geodesic distance of two points $x, y \in M$ is $\delta(x, y)=\inf _{\gamma} \ell(\gamma) \quad$ over paths $\gamma$ with $\gamma(a)=x, \quad \gamma(b)=y$ if $x, y$ are in the same connected component of $M, \delta(x, y)=+\infty$ otherwise.

Of course, $$ \text {all closed geodesic balls } \overline{B\left(x_{0}, r\right)} \text { are compact,} $$ by the Hopf-Rinow.

Question:Can $\{B(x_0,k)\}_k$ exhaust one complete Riemannian manifold? i.e.,Does $\bigcup_k B(x_0,k)=X$ and $\overline{B\left(x_{0}, k\right)}\subset {B\left(x_{0}, k+1\right)}$ hold?

Answer 1

Yes, this is true if the manifold is connected.

The first thing to notice is that a manifold is locally homeomorphic to a euclidean ball and hence locally path connected. Now for connected topological spaces locally path connected implies path connected (see for example here) and so connected manifolds are path connected.

In particular, on a connected Riemannian manifold $M$ the distance between any two points is finite, and so $\cup_{k\in\mathbb N} B(x_0,k)=M$ for all $x_0\in M$.

Furthermore $ \overline {B(x_0,k)}\subseteq B(x_0,k+1)$ for all $k\in \mathbb N$:

$(M,\delta$) is a metric space and in any metric space $(X,d)$ it holds that $$\overline{B(x_0,r)}\subseteq \{x\in X:d(x,r)\leq r\}$$ for all $r>0$, which follows from the continuity of the metric. Actually on a Riemannian manifold even equality holds, but this does not matter here.