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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?

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2 Answers 2

up vote 6 down vote accepted

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.

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Thanks for both of your illuminating responses. Anton, that one-line demonstration is very nice! I wonder whether the no answer would still hold if I changed condition (2) to state that S minus any one of its points was still connected but did not stipulate that it should be homeomorphic to a straight line. Condition (1), of course, implies that S is arc-wise connected. It seems also that bringing Hilbert space into the picture doesn't change the situation in any way. –  Garabed Gulbenkian Mar 29 '13 at 19:35

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 $S$.

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