Can a complete manifold have an uncountable number of ends? 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. 
 A: Let me rephrase S. Goette's example in a slightly different way. Consider the complement of the triadic Cantor set in the complex plane. This is a Riemann surface that can be uniformized: there is a complete Riemannian metric of constant minus one curvature. The ends of this surface are in one-to-one correspondence with the points of the Cantor set. 
You can see the decomposition into pairs of pants by drawing nested circles around the intervals that appear in the construction of the Cantor set. 
In some sense, uncountably many ends is the generic case. Let us restrict ourselves to planar surfaces, namely complements of compact sets in the Riemann sphere. The set of compact subsets in a compact set has a natural topology, the Vietoris topology (which coincides with the topology given by the Hausdorff metric). So we can say that two open subsets are close if their complement are close with respect to this topology. It happens that generically a compact subset of the sphere has uncountably many connected components, hence its complement has uncountably many ends.
A: 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.
