Continuity of the Length of shortest arc containing angles

Question

Here's an interesting question! Let $x=(x_1,x_2,\dots,x_n)$ points on the unit circle of $\mathbb{R}^2$. And let $V_{\min}(x)$ be the length of the shortest arc containing all the points $x=(x_1,x_2,\dots,x_n)$. I would like to prove that $V_{\min}$ is continuous. Thanks!

Answer 1

Let's assume that the points are labeled in counterclockwise order.

Consider the function $f_{i}$ which is the length of the arc from $x_i$ to $x_{i+1}$ (indices taken mod $n$) that contains all other points (i.e., clockwise from $x_i$ to $x_{i+1}$).

Then your function $V$ is just $\min_i f_i$; since the min of two continuous functions is still continuous, you're done.

Post-comment additions: Let me be a little clearer. With $n$ points in some configuration, there's some between-adjacent-points interval of size at least $2\pi/n$. Pick the midpoint of one such interval as the origin for polar coordinates for your $n$ points. That means that the polar coordinate of $x_1$ will be at least $\pi/n$, and the coordinate of $x_n$ will be no more than $2\pi - \pi/n$. (A picture will help you here). If we consider arrangements of points near the starting configuration --- say, $x_i$ moves no more than $\frac{\pi}{2n}$ -- then it's pretty evident that the polar coordinate of $x_i$ varies continuously as a function of the positions of the points (a delta-epsilon argument with $\delta = \epsilon/2$ should suffice). Hence the functions $f_i$ do so.