# 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 $f: \mathcal E_1 \to \mathcal E_2$ ?

• By a bounded set, I mean a set with compact closure. This is independent of any particular metric and coincides with the usual definition of metric boundedness, provided I have chosen a complete metric on $M$ Nov 29 '15 at 9:20
• Also, it does not directly follow from the definition of an end. For example, one could take the wedge sum of infinitley many closed intervals, along with attaching a half-open interval of the form [0,1) at 0. The resulting space has one end, but one can easily find an end neighborhood with unbounded complement Nov 29 '15 at 9:24
• The infinite wedge of closed intervals is not a nice space. If you put the quotient topology on it, it is not first countable. Even if you put the "french railway metric" on it, the center does not have a compact neighbourhood. So, what would an end be in such a space? Nov 29 '15 at 11:35
• I think the problem lies in the fact that there are at least two definitions for ends out there that are non-equivalent in general. I use the definition from Siebemann's PHD thesis (1965, Princeton), where an end is, essentially, an equivalence class of nested sequences $(U_i)_{i=1}^n$ of connected open sets with compact boundary with the property that $\bigcap_{i=1}^\infty cl(U_i) = \emptyset$. Two such sequences are equivalent if they "contain one another". There is also the (more modern) definition as the limit of a particular inverse system, found on the english wikipedia Nov 29 '15 at 11:47
• @BerniWaterman: in resolving these foundational confusions you may benefit from reading arxiv.org/abs/1210.6741 especially section 3. Nov 29 '15 at 14:51

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 $N\subset K$.
• The fact that the set $V_i$ that you defined contains an unbounded component is not so trivial in my eyes. It is true, however, so your proof works fine! Nov 29 '15 at 13:58