Is it possible to give an explicit characterization of compact subsets of $[0,\infty)$. Is it true that given any compact subset $K \subseteq [0,\infty)$ then $[0,\infty) \setminus K$ has only one path connected component? Why?
I'm trying to show formally that $[0,\infty)$ has only one end, so we take a compact subset $C$ of $[0,\infty)$ and let $r: [0,\infty) \rightarrow [0,\infty)$ be a continuous map. Then there must exist $L>0$ such that $r([L,\infty))$ is contained in some component of $[0,\infty) \setminus C$.
Why there is only one equivalence class though? it seems to me that the complement of a compact set in $[0,\infty)$ has two connected path components.
From page $6$ of http://arxiv.org/pdf/0806.3771.pdf
Definition A proper ray is a map $[0,\infty) \rightarrow X$ such that the preimage of a compact set is compact again. Two proper rays $c_{0},c_{1}: [0,\infty) \rightarrow X$ converge to the same end if for every compact subset $C \subseteq X$ there is $R>0$ such that $c_{0}([R,\infty))$ and $c_{1}([R,\infty))$ lie in the same component of $X \setminus C$. This defines an equivalence relation on the set of proper rays. The set of equivalence classes is the set of ends. The number of ends of $X$ is the cardinality of this set.
Any compact subset $S$ of $[0,\infty)$ is contained in an interval $[0,x]$ for some $x$. Then $r^{-1}([0,x])$ is compact so it fails to intersect $[L,\infty)$ for some $L$. Thus, $r([L,\infty)) \subset (x,\infty) \subset [0,\infty) \setminus S$.
So, given two proper rays $c_0, c_1$ and some compact $C \subset [0,\infty)$, we have by the above argument $L_0, L_1$ s.t. $c_0([L_0,\infty)$ and $c_1([L_1,\infty))$ lie in the same component of $[0,\infty) \setminus C$. Let $R = \max(L_0,L_1)$.