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 $f: \mathcal E\_1 \to \mathcal E\_2$ ?*



Is the complement of the ends of a manifold bounded?
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 $N\subset K$.


