Components in the complement of compact subset in manifold Let $M$ be a (smooth) connected manifold and $K \subset M$ a compact subset. Then $M \setminus K$ consists of a number of components. Let $(U_j)_{j \in \mathcal J}$ be the collection of bounded components in $M \setminus K$, i.e, the components with compact closure in $M$. I wonder if its true that the set $U:=\bigcup_{j \in \mathcal J} U_j$ is again a bounded subset of $M$. 
Some thoughts I have made so far: It is clear that $M \setminus K$ may have an infinite, even uncountable number of bounded components, so one can't argue that $U$ is bounded because it is the finite union of bounded subsets. Also, the fact that $K$ is compact obviously plays a big role. I attempted to argue by contradiction: Suppose that $U$ was unbounded. Then $U$ cannot be contained in any compact connected subset of $M$. Somehow, I feel this should also imply that $K$ cannot be contained in any compact connected subset of $M$, which is a condradiction. But I am not sure how to make that last step.
 A: Warning: This answer is not complete, as it contains a claim that has yet to be proved.
Wlog we may assume that $M$ is not compact.
Let $K \subset M$ be a given compact subset. Then, there exists a compact, connected set $L \subset M$, so that $K \subset int(L)$. For example, one could pick a complete Riemannian metric on $M$ and choose $L$ to be the closure of the ball $B(x,r)$ for some point $x \in M$ and $r > 0$ large. Let us look at the closed set $M \setminus int(L)$ and let $U$ be component of $M \setminus K$. We make the following 
Claim: If $\partial U \cap (M \setminus int(L)) = \emptyset$, then either $U \cap (M \setminus int(L)) = \emptyset$ or $U$ is unbounded.
Let $U_i$ be a bounded component of $M \setminus K$ with $U \cap (M\setminus int(L)) \neq \emptyset$. By the above claim, it follows that also $\partial U \cap (M\setminus int(L)) \neq \emptyset$. However, as $M$ was connected, we must have $\partial U_i \subseteq \partial K \subset int(L)$, where we arrive at a contradiction. Hence the union $U = \bigcup_{j \in \mathcal J} U_j$ of all bounded components in $M \setminus K$ is contained in $int(L)$. In particular, $\overline{U} \subset L$, so $U$ is bounded.
A: A very similar question recently appeared on MO; the only difference is that the OP studies $K\cup\bigcup_j{U_j}$, which I will call $X$.
Since the duplicate received a complete answer, I sketch the argument here:

*

*Show that $x$ is not in $X$ iff there is a trajectory in $M$ starting at $x$ that "heads off to infinity".

*Show that (1) implies $X$ is closed.

*Embed your manifold in $\mathbb{R}^d$.  Such a strong use of the manifold nature of $M$ is is necessary, because there is a counterexample for (metrizable) CW complexes: Choose $p\in S^1$ and let $M=(S^1\times\mathbb{N})/{\sim}$, where $\sim$ identifies all the points $\{(p,n):n\in\mathbb{N}\}$.  (This is almost, but not quite, the Hawaiian earring.)  Choose $K$ to be any "leaf"; say, $S^1\times\{0\}$.  Then each connected component of $M\setminus K$ is $(S^1\setminus\{p\})\times\{n\}$, which is relatively compact.  But the union of all such sets with $L$ is $M$, which is not compact.

*Expand $K$ to a relative (closed) ball $B$.  (This enlarges $X$, and compactness propagates to subspaces.)

*Note that each point in $\partial B$ intersects exactly one connected component of $M\setminus B$ in an open set, and that $\partial B$ is compact. Conclude.

