closed geodesics must intersect in a positive curvature 2-dim manifold Let $M$ be a 2-dimensional Riemannian manifold of positive curvature and $A, B$ two closed geodesics. Show that $A$ and $B$ must intersect.
I know that this is a Frankel-Hadamard type of proof and I also looked at this post here. But I don't know how to hit the problem without the assumption of $M$ being compact and connected.
I think that what I can do for now is to assume $A$ and $B$ don't intersect. Then if I can find a shortest geodesic c from $A$ to B, and show that this geodesic must be perpendicular to each closed geodesic, and then find a parallel field along $c$ that is tangent to $A$ and $B$ at the endpoints of $c$, using the second variation formula to get a shorter curve from $A$ to $B$ I can get it done. But I don't really know how to define the geodesic and the parallel field.
I'd appreciate any help.
 A: EDIT: In my original answer, I assumed that $\Gamma$ was a family of geodesics, but it can be any family of curves such that $\Gamma(s,\cdot) = \gamma$. The answer has been edited to reflect this.
Here is a sketch of one way to see this:
First, recall that if a curve $\gamma: [0,1] \rightarrow M$ is a constant speed geodesic whose length is the distance from $A$ to $B$, it minimizes the energy functional
$$ E[c] \int_{t=0}^{t=1} g(c',c')\, dt, $$
where $c: [0,1] \rightarrow M$ is any smooth curve such that $c(0) \in A$ and $c(1) \in B$.
Let $\Gamma: (-\delta,\delta)\times [0,1] \rightarrow M$ be a variation of $\gamma$, where $\Gamma(s,0) \in A$ and $\Gamma(s,1) \in B$. Let $S = \partial_s\Gamma$ and $T = \partial_t\Gamma$. The energy   of each curve $\Gamma(s,\cdot)$ is $$ E[\Gamma(s,\cdot)] = \int_{t=0}^{t=1} |T|^2\,dt. $$
Since $\Gamma(0,\cdot) = \gamma$ minimizes energy,
\begin{align*}
0 &= \frac{1}{2}\left.\frac{d}{ds}E[\Gamma(s,\cdot)]\right|_{s=0}\\
&= \int_{t=0}^{t=1} g(T,\nabla_ST)\,dt\\
&= \int_{t=0}^{t=1} g(T,\nabla_TS)\,dt\\
&= g(T(0,1),S(0,1)) - g(T(0,0),S(0,0)) - \int_{t=0}^{t=1} g(\nabla_TT, S)\,dt\\
&= g(T(0,1),S(0,1)) - g(T(0,0),S(0,0))
\end{align*}
Since this holds for any variation $\Gamma$ such that $S(0,0)$ is tangent to $A$ and $S(0,1)$ is tangent to $B = S(0,0)$, it follows that $\gamma$ is orthogonal to both $A$ and $B$.
