Let $x$ be a point of an affine manifold $(M,\nabla)$, then there exists an open neighborhood $\mathscr U$ of $x$ such that any point $p\in \mathscr U$ can be connected with $x$ by a geodesic.
The existence of a subset $\mathscr U$ of $M$ s.t. $x\in \mathscr U$ is easy to show. How could we prove that $\mathscr U$ can be open?
Many thanks!
This is true for any connection defined on $M$. Let $x\in M$, consider the exponential map $exp_x:T_xM\rightarrow M$ such that $exp_x(v)$ is the value at $1$ of the geodesic $c_v$ such that $c'_v(0)=v$. The domain of the exponential is an open subset containing $0$ and $d{exp_x}_0(v)=lim_{t\rightarrow 0}exp_x(c(t))$ where $c$ is a curve such that $c'(0)=v$, if $v\in U$, you can choose $c=c_v$. This implies that $d{exp_x}_0(v)=v$, for every $v\in U$ there exists $t_0\in R$ such that $t_0v\in U, t_0\neq 0$ we deduce that $d{exp_x}_0(t_0v)=t_0v$ and since the differential is linear that $d{exp_x}_0(v)=v$, by the local inverse theorem, you deduce that there exists a neighborhood $V$ of $0\in T_xM$ such that the restriction $exp_x$ to $V$ is a diffeomorphism onto its image. The open subset $exp_x(V)$ is the requested open subset.