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. 



Can a complete manifold have an uncountable number of ends?
The answer is yes. Consider a hyperbolic pair of pants where all three boundary circles are of the same size. Glue countably many of them together such each new one is glued to the existing manifold along exactly one boundary circle. Then the manifold looks like the boundary of a fattened tree.


Fix $p$ in one of the pairs of pants $Y\_0$. For all other pairs of pants $Y$, pick a point $q\in Y$, then the minimal geodesic joining $p$ and $q$ will leave $Y$ through one circle. Label the two remaining boundary circles with $0$ and $1$. Then all glueing circles get exactly one label, except for those bounding $Y\_0$.


For each sequence $(a\_n)\_n\in\{0,1\}^{\mathbb N}$, we construct a ray from $p$ leaving $Y\_0$ through the same boundary component. Whenever the ray enters the $n$-th pair of pants $Y$ along its way, we let it leave through the circle labelled $a\_n$. This way, we construct an uncountable number of rays that are not pairwise cofinal.


