A question about closed curves Does three dimensional Euclidean space contain unbounded closed curves that do not cross themselves?
It does not seem possible to find examples of such anmd yet it is not clear just what is standing in
the way. More precisely, let n be any positive integer not less than 3 and let E(n) be n-dimensional
Euclidean space. Does there exist a subset S of E(n) with the following properties that is not compact?
(1)S is closed, connected and locally connected. (2) If any point of S is removed, the resulting space
is still connected and is homeomorphic to a straight line. (3) If any distinct pair of points of S are
removed, the resulting space is no longer connected and has two components. Finally can S be a
non-compact subset of a separable and infinite dimensional Hilbert space?
 A: Note that $S$ is a connected 1-dimensional manifold.
Since it is not compact we get that $S$ is homeomorphic to $\mathbb R$, 
a contradiction.
A: Let $S$ be as assumed.
Choose distinct $p,q \in S$, and consider the parametrizations
$$ \psi : (0,1) \to S \setminus \{q\} $$
and
$$ \phi : (0,1) \to S \setminus \{p\}. $$
Let $p$ correspond to time $t_p \in (0,1)$ under $\psi$.
Claim: for $t < t_p$, $\phi^{-1}(\psi(t))$ is a monotone function of $t$, and similarly for $t > t_p$.
proof: clearly $\phi^{-1}(\psi(t))$ is defined and continuous for $t \in (0,t_p)$.  Moreover, it is injective as a map into $(0,1)$, by hypothesis on $\phi$ and $\psi$.  Thus, it is monotone.  Ditto $t \in (t_p,1)$.
Next claim: as $t \to t_p^-$, $\phi^{-1}(\psi(t)) \to 0$ (after reflecting $\phi$ if necessary), and as $t \to t_p^+$, $\phi^{-1}(\psi(t)) \to 1$.
proof:  as $\phi^{-1}(\psi(t))$ is monotone for $t < t_p$, it has a limit in $[0,1]$ as $t \to t_p^-$.  If this limit lies in $(0,1)$, we obtain a contradiction that $p \notin \text{im }\phi$ (since $\psi(t) \to p$ as $t \to t_p^-$ and $\phi$ is continuous as a map into $\mathbb{R}^3$).  Thus $\phi^{-1}(\psi(t)) \to 0$ or $1$ as $t \to t_p^-$, and reflecting $\phi$ if necessary we may assume it is 0.  Similar reasoning shows $\phi^{-1}(\psi(t)) \to 0$ or $1$ as $t \to t_p^+$, and it cannot approach the same limit as when $t \to t_p^-$ by injectivity/connectedness/continuity/etc.  The claim follows.
Note that it follows (by connectedness and continuity) that $\phi^{-1}(\psi(t))$ maps $(t_p - \epsilon, t_p) \cup (t_p, t_p + \epsilon)$ onto (a set containing) $(0,\delta) \cup (1 - \delta, 1)$ for some $\epsilon, \delta > 0$.  Hence for $t$ near 0 or 1, $\phi(t)$ is near $p$.  But clearly for $t$ bounded away from 0 and 1, $\phi(t)$ is bounded.  Thus $\phi$ is a bounded map, contradicting the unboundedness of $S$.
