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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!

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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.

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  • $\begingroup$ This was my idea as well. The point is that I am not sure that the function $f_i$ is continuous. $\endgroup$
    – frank
    Feb 24 '16 at 2:53
  • $\begingroup$ It clearly depends only on the points $x_i$ and $x_{i+1}$, And its value is the arctan of some rational function of the coordinates of these points $\endgroup$ Feb 24 '16 at 3:24
  • $\begingroup$ I see. And you are claiming that these functions $f_i$ are continuous, right? $\endgroup$
    – frank
    Feb 24 '16 at 4:27
  • $\begingroup$ See additional notes. $\endgroup$ Feb 25 '16 at 2:16

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