Minimal bound on the lifetime of a maximal integral curve Let $M$ be a smooth manifold and $X: M\to TM$ a smooth vector field. Then one of the fundamental theorem of differential geometry tell us there is a smooth map
$$
\alpha: \mathcal{D}(X)\to M
$$
called the flow of $X$ where $\mathcal{D}(X)$ is the domain of the flow. Moreover We have
$$
\mathcal{D}(X) = \bigcup_{t\in\mathbb{R}}\{t\}\times\mathcal{D}_t(X)\subset \mathbb{R}\times M
$$
where $\mathcal{D}_t(X)$ is the set of points $p\in M$ such the maximal integral curve of $X$ through $p$ is defined at least in the interval $[0, t]$.
Is there an $\epsilon>0$ such that for every $\left|t\right| < \epsilon$ we have $\mathcal{D}_t(X) = M$? In words, is there a minimal bound for the lifetime of a maximal integral curve?
(The negation would be for every $\epsilon > 0$ there exists a maximal integral curve $c: I\to M$ with domain strictly included
in $]-\epsilon, \epsilon[$.)
I guess that if the negation holds, one can give an example with $M = \mathbb{R}^n$ or an open subset of it.
 A: The answer to your question is no, as the counterexample in the comments shows. In fact, such an $\varepsilon$ never exists unless the vector field is complete, meaning that every integral curve exists for all time. The following is a paraphrase of Lemma 9.15 in my Introduction to Smooth Manifolds (2nd ed.):

Let $X$ be a smooth vector field on a smooth manifold. If there exists $\varepsilon>0$ such that every maximal integral curve is defined at least on the domain $(-\varepsilon,\varepsilon)$, then $X$ is complete.

(By the way, your statement in parentheses is the negation of the preceding statement, not the converse.)
A: HINT:
If the maximal interval starting at $x$ is $(a,b)$ (call the integral curve $c_x(\cdot)$ ),  then the maximal interval starting at $c_x(t_0)$ is $(a-t_0, b-t_0)$. That should tell you something about the possible domains. 
$\bf{Added:}$ For any compact subset $K$ of $X$, you can guarantee an $\epsilon$ for all $x$ in $K$. This implies: if for some $x$ the maximal interval $(a_x, b_x)$ is finite on the right ( $b_x < \infty$), then $\lim_{t\nearrow b_x} c_x(t) = \infty$, in the sense that it leaves any compact subset of $X$ for $t$ close enough to $b_x$.  
