# $\alpha$ exists so that for any points $x_n$ there is a point at average distance $\alpha$ from the $x_n$.

Let $X$ be a connected and compact metric space. Prove a real number $\alpha$ exists so that for every finite set of points $x_1,x_2,\dots, x_n\in X$ (not necessarily distinct) there exists $x\in X$ such that:

$\dfrac{d(x_1,x)+d(x_2,x)+\dots + d(x_n,x)}{n}=\alpha$

• Seems like something to do with the intermediate value theorem applied to a finite sum of metrics. – Rick Sanchez Apr 20 '16 at 2:22

1. Let us prove the statement for a fixed $n$. Denote$$f_n(x_1,x_2,\ldots,x_n,x) = \frac{d(x_1, x) + d(x_2, x) + \ldots + d(x_n, x)}{n},$$$$M(x_1,x_2,\ldots,x_n)=\max_{x\in X}f_n(x_1,x_2,\ldots,x_n, x),$$$$m(x_1,x_2,\ldots,x_n)=\min_{x\in X}f_n(x_1,x_2,\ldots,x_n,x),$$and let $I(x_1, x_2, \ldots, x_n)$ be the closed interval $[m(x_1, x_2, \ldots, x_n), M(x_1, x_2, \ldots, x_n)]$. If we prove all the intervals $I(x_1, x_2, \ldots, x_n)$, $x_i \in X$ have a common point, we are done. To prove it, it is enough to establish any two of them have a common point, by Helly's theorem combined with the finite intersection property of a family of compact sets. Arguing by contradiction suppose $I(x_1, x_2, \ldots, x_n)$ and $I(y_1, y_2, \ldots, y_n)$ do not intersect and let $a$ be a real number that separates them. Suppose without loss of generality that$$f_n(x_1, x_2, \ldots, x_n, x) < a < f_n(y_1, y_2, \ldots, y_n, y) \text{ for all }x,y \in X.$$Now, in the first inequality substitute consecutively $x = y_1$, $x = y_2$, $\ldots$ , $x = y_n$ and sum the terms. Then do the same with the second one substituting $y = x_1$, $y = x_2$, $\ldots$ , $y = x_n$. We get that one and the same sum is both less and greater than $a$, a contradiction. Thus all the intervals have a common intersection, which is a nonempty interval $I_n$.
2. We prove all $I_n$, $n \in \mathbb{N}$ have a common point. Again, it is enough to prove it for any two of them. Let us first observe that$$m \mid n \implies I_n \subset I_m.$$Indeed let $c \in I_n$ and let $x_i \in X$, $i = 1, 2, \ldots, m$. Then we get $k$ copies of each $x_i$, $i = 1, 2, \ldots, m$, where $k \cdot m = n$ and apply what $c \in I_n$ means for the new $n$ points in $X$. Using that observation, for any $m$, $n$ we have$$I_{mn} \subset I_m, \quad I_{mn} \subset I_n,$$which means $I_n$, $I_m$ have nonempty intersection. Thus, all $I_n$ have nonempty intersection and any $\alpha$ inside that intersection will do the job.