Let $M$ be a complete and noncompact Riemannian manifold. Fix a point $p$ in $M$. Let $\gamma$: $[0, L]\rightarrow M$ (parametrized by its arc length) be a geodesic starting from $p$. Denote by $d(\cdot, \cdot)$ the distance function on $M$ induced by the Riemannian metric. $\gamma$ is minimal if $d(\gamma(s_1), \gamma(s_2))=|s_1-s_2|$. A ray is a minimal geodesic defined on $[0, +\infty)$. Since $M$ is noncompact, there exists at least one ray from $p$.

Two rays $\gamma_1, \gamma_2$ from the same point $p$ are called cofinal if for any $r\geq0$ and all $s>r$, $\gamma_1(s)$ and $\gamma_2(s)$ lie in the same component of $M\backslash B(0, r)$, where $B(0, r)=\{x\in M| d(p, x)<r\}$. An equivalence class of cofinal rays is called an end of $M$. My question is:

Can $M$ have uncountably many ends? It seems that the answer is no. But I am not very sure and I can not find a convictive proof.

countablyinfinitely many ends. $\endgroup$ – HJRW Dec 18 '15 at 20:352more comments