Is the complement of the ends of a manifold bounded? Let $M$ be a connected manifold with precisely $k$ ends $\epsilon_1,...,\epsilon_k$. Choose a collection $(U_i)_{i=1}^k$ of pairwise disjoint open $\epsilon_i$-neighborhoods. Then I wonder how to prove that $N := M \setminus (\bigcup_{i=1}^k U_i)$ is a bounded subset of M, i.e, a set with compact closure.
Thoughts I had so far: It is clear that each component of $N$ is bounded by itself. However, it is quite possible that $N$ consists of infinitley many such components, so an argument on finite union doesnt work. Does anybody have an idea ?
Edit: As established in the comments, the problem lies in the fact that there at least two definitions for ends of manifolds that are non-equivalent in general: 
I used a version of Siebemanns definition (1965), where an end of $X$ is an equivalence class of sequences $U_1 \supset U_2 \supset U_3...$ of connected open subsets with compact boundary, so that $\bigcap_{i=1}^\infty cl(U_i) = \emptyset$. Two such sequences $(U_i)$ and $(V_i)$ are equivalent if for each $n \in \mathbb N$ there exists $j \in \mathbb N$, so that $U_i \supset V_j$ and $V_i \supset U_j$. The set of corresponding ends is denoted by $\mathcal E_1$. 
If $X$ admits a compact exhaustion $(K_i)_{i=1}^\infty$ (for example if $X$ is a manifold),  i.e. a sequence of compact subsets $K_1 \subset K_2 \subset ...$ with $\bigcup_{i=1}^\infty int(K_i) = X$, then one can define another set of ends $\mathcal E_2$ as the inverse limit of the system  $\pi_0(X\setminus K_1) \leftarrow \pi_0(X \setminus K_2) \leftarrow \pi_0(X \setminus K_2)....$. 
Now my original question can be formulated as follows:
If $X$ is a connected manifold, is there a bijection of sets $f: \mathcal E_1 \to \mathcal E_2$ ?
 A: Assume that $M$ is a noncompact manifold. Then there exists a bijection
between both sets of ends.
Start with $e\in\lim_{\longleftarrow}\pi_0(M\setminus K)$. For each exhausting sequence $K_1\subset K_2\subset\dots$ of compact subsets with $\bigcup_iK_i=M$, let $U_i$ be the connected component of $M\setminus K$ indexed by $e$. Then $\partial U_i=cl(U_i)\cap K_i$ is compact. One checks that different compact exhaustions of $M$ give equivalent sequences.
In the opposite direction, let $(U_i)_i$ be a Siebenmann end, and let $K_i$ be a compact exhaustion as above. For each $i$ there exists $j$ with $\partial U_i\subset K_j$, and we regard $V_i=U_i\setminus K_j$. If $V_i$ is connected, then $V_i$ defines a point $e(K_j)\in\pi_0(M\setminus K_j)$. If not, we find $k>i$ with $U_k\cap K_j=\emptyset$ because $\bigcap_kcl(U_k)=\emptyset$. Then $U_k$ is contained in one of the unbounded connected components of $V_i$, which we pick as $e(K_j)$. Because the sequence $K_i$ is an exhaustion, this extends to the whole inverse system.
Back to the original question. If there are only finitely many ends, choose one neighbourhood $U_i$ for each end and take $N$ as in the question. In particular, there exist compact sets $K_i$ such that $U_i$ contains one connected component of $M\setminus K_i$. Put $K=K_1\cup\dots\cup K_k$, then $K$ is compact and $N\subset K$.
