Local existence of geodesics on a Riemannian manifold In my differential geometry notes there is the following proposition:
Prop. Let $(M,g)$ be a Riemannian manifold and let $p\in M$. Then there exist an open set $U\subseteq M$, containing $p$, and $\epsilon>0$, such that $\forall q\in U,\forall v\in T_qM$, with $|v|<\epsilon$, there exists a unique geodesic $\sigma\colon (-2,2)\longrightarrow$ M , satisfying the initial conditions $\sigma(0)=q$ and $ \dot{\sigma}(0)=v$.
The result is proved by the following theorem about Differential equations.
Thm. Let $(\star)$ be the following system of differential equations 
$$\dfrac{d^2 u_k}{dt^2}=F_k(u_1,\ldots,u_m,u_1',\ldots,u_m'), \: \: k=1,\ldots,m.$$
Assume that the functions $F_k$ are smooth on some open subset $U\subseteq \mathbb{R}^{2m}$, containing the point $q_0=(u^0_1,\ldots,u^0_m,v_1^0,\ldots,v_m^0)$. Then there exist an open subset $W\ni q_0$ and $\epsilon>0$, such that for all $\bar{q}=(\bar{u}_1,\ldots,\bar{u}_m,\bar{v}_1,\ldots,\bar{v}_m)\in W$, the system of differential equations has an unique solution $u\colon (-\epsilon,\epsilon)\longrightarrow U$, satisfying the initial conditions $u(0)=(\bar{u}_1,\ldots ,\bar{u}_m)$ and $\dot{u}(0)=(\bar{v}_1,\ldots ,\bar{v}_m)$.
In the proof of the proposition is taken a chart $(U,\phi)$ containing the point $p$ and it's assumed (by the theorem above) the existence of $\epsilon_1,\epsilon_2>0$ and a geodesic $\sigma\colon (-2\epsilon_2,2\epsilon_2)\longrightarrow M$ satisfying  the conditions $\sigma(0)=q$ and $\dot{\sigma}(0)=v$, for all $q$ in some open $\widetilde{U}\subseteq U$ and for all $v\in T_qM$, with $|v|<\epsilon_1$. Now I don't understand why the number $\epsilon_1$ does not depend from the choice of the point $q\in \widetilde{U}$. Can anyone help me to understand?
 A: $\newcommand{\eps}{\varepsilon}$First, an analogy: If $U$ is an open set in some metric space and $p \in U$, then there exists $\eps > 0$ and a neighborhood $\widetilde{U}$ of $p$ such that for every $q$ in $\widetilde{U}$, the $\eps$-ball $B_{\eps}(q)$ about $q$ is contained in $U$.
The standard idiom is: Pick $\eps$ so that the $2\eps$-ball about $p$ is contained in $U$, then let $\widetilde{U}$ be the $\eps$-ball about $p$. By the triangle inequality, if $q \in \widetilde{U}$ (i.e., $d(p, q) < \eps$) and $x \in B_{\eps}(q)$ (i.e., $d(q, x) < \eps$), then
$$
d(p, x) \leq d(p, q) + d(q, x) < 2\eps,
$$
so $x \in B_{2\eps}(p) \subset U$.

In your situation, the metric space is the Riemannian manifold $M$ equipped with the infimum-of-path-lengths distance function. Using the language of the Riemannian exponential map, $U$ is the image under $\exp_{p}$ of the open $2\eps_{2}$-ball in $T_{p}M$, and $\widetilde{U}$ may be taken to be the image of some ball of sufficiently small radius that the union $\bigcup_{q \in B_{\eps}(p)} B_{\eps_{1}}(q)$ is contained in $U$.
