Question on minimizing property of geodesics I'm trying to understand the third chapter of this book which eventually leads up to the Hopf-Rinow theorem: https://www.math.tecnico.ulisboa.pt/~gcardoso/GeoRiem/nata_textb.pdf
On page 108, I'm rather confused by the definition and utility of the concept of "totally normal neighbourhood" (Seemed to be called a convex neighbourhood in other literature". It seems to be used in the proof for theorem 4.6, which asserts that any length minimizing curve between two arbitrary points $p$ and $q$ on a riemannian manifold must be a geodesic. 
It says on the next page that:
"Notice that, given any two points p, q in a totally normal neighborhood
V, there exists a geodesic c: I →M connecting p to q; if γ: J→M is
any other piecewise differentiable curve connecting p to q, then l(γ)≥l(c),
and l(γ)=l(c) if and only if γ is a reparametrization of c. The proof of
Theorem 4.6 is now an immediate consequence of the following observation:
if c: I→Mis a piecewise differentiable curve connecting p to q such that
l(c)≤l(γ) for any curve γ: J→M connecting p to q, we see that c must be
a reparametrized geodesic in each totally normal neighborhood it intersects."
I don't understand why the 'totally normalness' is necessary. I want to argue this way: suppose that such a length minimizing curve is not a geodesic, there lies on the curve some point $r_1$ such that $\frac{D\dot c(t)}{dt}\neq 0$. For this offending point $r_1$, construct a normal neighbourhood centered about $r_1$, call this $S_\epsilon$. take any other point on the curve within $S_\epsilon$, say $r_2$. Then we can draw a geodesic connecting $r_1$ to $r_2$ which would reduce the overall length of the call.
Hence in summary: I argue that for a global curve to be minimized, I need each small part to be locally a geodesic, which in turns makes the curve a geodesic, for if not, if can perturb the local sections so slightly that I further minimize the length of the curve.
Where does this concept of totally normal come in?
 A: Example
I think the following illustrates why neighbourhoods need to be totally normal. It is based on an example from Pressley (2010), Section 9.4.
Let $M$ be the Euclidean plane with the origin removed. For any $p \in M$ let $U = \mathbb{R}^2 \setminus \{-p\}$. Let $V = M$. Then $\exp_p : U \to M$ is a diffeomorphism from $U$ onto $V$. Moreover $U$ contains the origin and $V$ is open. Therefore $V$ is a normal neighbourhood.
However $V$ cannot be a totally normal neighbourhood as it cannot be contained within a ball of finite radius.
Now for geodesics and shortest paths: given $p \in M$ let $q = -p$. There is no shortest path in $M$ from $p$ to $q$. It “should” be the line segment from $p$ to $q$ but the origin has been deleted. Indeed consider a path $\gamma$ from $p$ to $q$. With one important exception, local shortening can be applied (as proposed in the question) to repeatedly reduce the length of $\gamma$. The exception, however, is that local shortening cannot pass through the origin.
Intuition
Some further thoughts on what is going on intuitively.  When we “locally shorten” the path from $x$ to $y$, what we are doing is getting a tangent vector at $x$ that generates a geodesic that passes through $y$.
We know that we can do this at any $x$, for the normal neighbourhood at $x$.  But the normal neighbourhood may stop short of reaching $y$.
With a total normal neighbourhood, we can guarantee that a geodesic starting from $x$ can reach $y$.
Reference
Pressley, Andrew, Elementary Differential Geometry: Second Edition, Springer Undergraduate Mathematics Series. London: Springer-Verlag. (2010). DOI 10.1007/978-1-84882-891-9_9
