Flow of a nowhere vanishing complete vector field Let X be a nowhere vanishing complete vector field on a manifold M, $\gamma: \mathbb{R} \to M$ be its flow with $\gamma(0)=p \in M$ and suppose it is not periodic. If $\gamma(\mathbb{R})$ is closed, is $\gamma$ a proper map?
The only counter-example I can think of is a curve looping back to itself such that $\lim_{t\to + \infty} \gamma(t) = \gamma(\hat{t})$ for some $\hat{t} \in \mathbb{R}$ but I don't know if this can arise from a nowhere vanishing complete vector field.
 A: I assume that your vector field is at least continuous.
I will answer your second question. Such a situation cannot arise.
Assume that $\gamma(t)$ converges to some point $p\in M$. Then, $X(\gamma(t))$ converges to $X(p)$, so $\gamma'(t)$ converges to some vector $a$. Using local coordinates, you can assume that $M=\mathbb{R}^n$, for simplicity.
Since $\gamma(t)$ converges, $\gamma'(t)$ is integrable, so that its limit (since it does exist) is zero. Thus, $X(p)=0$.
In general, non-properness means that there is a sequence $t_n$ tending to infinity such that $\gamma(t_n)$ converges to some point $p$.
A: Assume the vector field $X$ to be of class $C^1$.  As hinted by M. Dus, to answer the first question it suffices to exclude the case that there is $t_n \to \infty$ (say) such that $\gamma(t_n) \to \gamma(\tau) (=: p) $.  Take a closed flow box $U$ of $p$, with transversal $T$.  Clearly, $p$ is not isolated in $T \cap \gamma(\mathbb{R})$. Since the flow maps are homeomorphisms, the closed set $T \cap \gamma(\mathbb{R})$ contains no isolated points, hence is perfect. Therefore $T \cap \gamma(\mathbb{R})$ has cardinality $\mathfrak{c}$, which contradicts the second countability of the real line. 
