Open balls under arc length and under chord length I am currently stuck on showing that, for an open ball on $S^1$ of radius $\epsilon$ such that the ball is defined under the arc-length metric between two points, there exists a $\delta$ such that an open ball of radius $\delta$ under the chord length metric (or Euclidean distance) between two points is contained inside.
In fact, I have tried to let $\delta = \sqrt{2-2cos(\epsilon)}$ by the law of cosine and instead of getting $B_\delta \subseteq B_\epsilon$, I got the other way round. It seems I have messed up some logic inside it...
edit:
I have shown that
$d_a(x,y)<\epsilon \Rightarrow \sqrt{2-2cos(d_a(x,y))} <\delta $
where $\sqrt{2-2cos(d_2(x,y))} = d_2(x,y)$ such that any element in the ball under $d_a$ is in $d_2$, which brings up that $B_\epsilon \subseteq B_\delta$. $d_a$ represents the arc metric, $d_2$ the chord metric.
update: I have tried $\delta = 2sin(\epsilon/2)$ and $d_2(x,y)<\delta \Rightarrow 2sin^{-1}(d_2(x,y)/2) < \epsilon$, where $2sin^{-1}(d_2(x,y)/2) = d_a(x,y)$. Is this a strong enough argument to prove this statement?
 A: Note that for $0\leq \theta \leq \pi$ (the possible values of $d_a(x,y)$)
we have $\sqrt{2-2\cos(\theta)}=2\sin(\theta/2).$ Then
$$ d_2(x,y) = \sqrt{2-2\cos(d_a(x,y))}=2\sin\left(\frac12 d_a(x,y)\right). \tag1 $$
So if you set $\delta = \sqrt{2-2\cos(\epsilon )}=2\sin(\epsilon /2),$ it is possible to show that
$$ d_a(x,y) < \epsilon \iff d_2(x,y) = \sqrt{2-2\cos(d_a(x,y))} < \delta$$
and
$$d_a(x,y) < \epsilon \iff d_2(x,y) = 2\sin\left(\frac12 d_a(x,y)\right) < \delta.$$
Your first attempt just proved one of those double implications in one direction, and it happened not to be the direction you really needed.
From Equation $(1)$ it is evident that
$d_a(x,y) = 2\sin^{-1}\left(\frac12 d_2(x,y)\right)$ (exactly as you found),
so your second attempt looks fine to me.
Again you could prove the implication in either direction, but this time you did it in the direction you need.
I often like to remind people that you don't need to use the largest possible value of $\delta$; a smaller value will do fine as far as the rigor of the proof is concerned. But in this case there seems to be no difficulty in dealing with the values of $\delta$ you chose for each $\epsilon.$ The arc sine function actually makes this choice of $\delta$ as easy as anything I could imagine.
