Transfinite diameter

Remarks

The maximum exists because $K^n$ is compact and the product is continuous. The sequence $(\delta_n(K))$ is nonincreasing, so the limit exists. To see this, drop one point from an optimal $(n+1)$-tuple in each of the $n + 1$ possible ways and multiply the resulting $n + 1$ inequalities. The number $\delta_2(K)$ is the diameter of $K$.

Example: let $K$ be the unit circle and let $\omega = e^{2 \pi i / n}$. The square of the Vandermonde determinant of $1, \omega, \dots, \omega^{n-1}$ has absolute value $n^n$, so these points give the product $n^{n/2}$. By Hadamard's inequality no $n$ points of the closed unit disc give more, so $\delta_n(K) = n^{1/(n-1)}$, which tends to $1$.

Non-example: the limit of $\delta_n(K)$ is not the diameter in general. For the unit circle $\delta_2 = 2$, while $\delta(K) = 1$.

A finite set $K$ has $\delta(K) = 0$, since for $n$ larger than the number of points two of the $z_i$ coincide.