The function $d\left ( x,y \right ) = min\left \{ \left | x-y \right |,1-\left | x-y \right | \right \}$ is a metric on $S^{1}$ where $S^{1}$ is the unit circle is the interval $\left [ 0,1 \right ]$ with 0~1, 1~0 and x~x $\forall x \in \left [ 0,1 \right ]$ with 0 identified with 1.
Here is what my notes remarked:
There are many possible arrangements of $x, y, z \in S^{1}$ . To make things more tractable, assume x=0; this amounts to a rotation. Thus $d\left ( x,y \right )=min\left \{ y,1-y \right \}$ and $d\left ( x,z \right )=\left \{ z,1-z \right \}$. Similarly, assume $y \leq \frac{1}{2}$; this amounts to a reflection. Thus, $d\left ( x,y \right )=y$.
Thus, transitivity becomes $y \leq min \left \{ z,1-z \right \}+min\left \{ \left | z-y \right |,1-\left | z-y \right | \right \}$
I'm having a bit of trouble in my attempt to understand the highlighted paragraph. If someone could explain to me how the rotation or reflection works out, that would be very helpful to me. Edit: I've included "algebraic topology" and "algebraic geometry" which might be too advance for what I'm asking but I'm doing so in hopes someone might have a better explanation to offer. Thanks in advance.
The main point is that the three points are arbitrary at the start, and you want to reduce to a problem where it's easier to see what you want. You know that rotating a circle doesn't change the distance between point, so rotating $x$ until it is where $0$ is lets you assume $x=0$. Similarly, if you draw a horizontal line through the circle, you note that the top semi-circle can be made to represent $[0,1/2]$ and the bottom semi-circle represents $[1/2,1]$ where you traverse things counterclockwise. But then if--after rotating $x$ to $0$, you find $y$ in the bottom, simply reflect everything across the horizontal line so that $y$ is in the top semi-circle. Again, reflections don't change distances, so the distances between the points are the same. Also, since we assume $x=0$, we note that this doesn't change that since $x$ is on the horizontal line.
This allows us to reduce the problem to a special case which is equivalent because the operations we did don't change distances.