Are segment domains closed? Let $M$ be a complete Riemannian manifold. Its segment domain is defined by:
$$
\mathbf{seg}(p)= \{v\in T_pM: \exp_p(tv):[0,1] \to M \textit{ is a segment} \ \ \} 
$$ (Note: "segment" has many different terminology like "minimizing curve" and "minimizing geodesic segment")
My question is: How to show $\mathbf{seg}(p)$ is closed in $T_pM$?

Remember that a curve is a segment if its length is equal to the distance between the endpoints; so 
$$ \mathbf{seg}(p) = \{ v \in T_p M : d(p, \exp_p v) - |v| = 0 \} = f^{-1}(0)$$
where $f : T_p M \to \mathbb R$ is defined by $$f(v) = d(p, \exp_p v) - |v|.$$
Since the exponential map, the distance function and the norm are all continuous, $f$ is continuous; so the segment domain is the pre-image by a continuous map of a closed set, and is thus closed.
