Does every metric space have a minimax radius? Let $(X,d)$ be a metric space with $\mathcal{P}(X)$ the space of probability measures on its Borel $\sigma$-algebra. For $\mu,\nu\in\mathcal{P}(X)$, define
\begin{align*}
d(\mu,\nu)&:=\int_X\int_Xd(x,y)\ \mbox{d}\mu(x)\ \mbox{d}\nu(y),
\\A(\mu)&:=\inf_{\nu\in\mathcal{P}(X)}d(\mu,\nu),
\\B(\nu)&:=\sup_{\mu\in\mathcal{P}(X)}d(\mu,\nu),
\\\alpha&:=\sup_{\mu\in\mathcal{P}(X)}A(\mu),
\\\beta&:=\inf_{\nu\in\mathcal{P}(X)}B(\nu).
\end{align*}
Then does $\alpha=\beta$ always hold? If not, does it at least hold for $X$ separable?
This is a follow up question of Does every connected compact metric space have a unique always attainable average distance? There it was shown in the answer that $\alpha=\beta$ for $X$ compact by a minimax theorem. This got me curious whether I could find a non-compact metric space for which $\alpha<\beta$.
Note that $0\leq\alpha\leq\beta<\infty$ for $X$ bounded. For $X$ unbounded, consider $x\in X$ and a sequence $x_1,x_2,\ldots\in X$ with $d(x,x_i)>2^i$. Take $\mu=\sum_{i\geq1}2^{-i}\delta_{x_i}$, such that $d(\mu,\nu)=\infty$ for all $\nu\in\mathcal{P}(X)$. It follows that $\alpha=\beta=\infty$. For $X$ totally bounded, its completion is compact, and thus $\alpha=\beta$. So any counterexample has to be bounded, but not totally bounded.
The reason I suspect the separable case to be easier is that then the space is strongly Lindelöf, which makes measures and their support behave much more intuitively. For example, a measure may have empty support, but only if $X$ is non-separable. However, I was not even able to come up with a counterexample for non-separable spaces. Part of this is due to the fact that it is very difficult to construct these pathological measures.
 A: Not an answer, but too long for a comment.
There is a large literature on this sort of problem in which "unique always attainable average distance" of a metric space is referred to as its "rendezvous number". A proof of the existence of a rendezvous number for every connected compact metric spaces was first published in this paper by Oliver Gross in 1961.  In this paper, the common value of your $\ \alpha\ $ and $\ \beta\ $, when they're equal, is called a "weak rendezvous number". It's not difficult to construct examples which have a weak rendezvous number but no rendezvous number, and the paper gives one.
One possible approach to answering your question might be to start from an optimisation problem with a duality gap.
A: Nice question!
No, not even for separable metric spaces; in fact, the below counterexample is on a countable set of points.
The construction is as follows. We start with one point (the top, in the picture). At any given stage (there are $4$ or $5$ depicted below, depending on how you count), we proceed to the next stage by adding/appending two points at distance $2/3$ (say) from each other, with one point being at distance $1/2$ from all the previous points, and the other at distance $1$ from all the previous points.
So, in the picture below, red denotes distance $1/2$, green denotes distance $2/3$, and blue denotes distance $1$. We chose these numbers so that, defining the distance between a point and itself to be $0$, we (don't have to worry about the triangle inequality and thus) obtain a metric space.

Now, we claim $\alpha = 1/2$ and $\beta = 1$.
Indeed, for any given $\mu \in \mathcal{P}(X)$ and $\epsilon > 0$, we can choose finite $X' \subseteq X$ so that $\mu(X') \ge 1-\epsilon$ and let $\nu$ be the delta mass at any point that has distance $1/2$ to every point of $X'$, to see that $\alpha \le (1-\epsilon)\cdot \frac{1}{2}+\epsilon \cdot 1 = \frac{1}{2}+\frac{\epsilon}{2}$.
And similarly, given any $\nu \in \mathcal{P}(X)$ and $\epsilon > 0$, we can choose finite $X' \subseteq X$ so that $\nu(X') \ge 1-\epsilon$ and let $\mu$ be the delta mass at any point that has distance $1$ to every point of $X'$, to see that $\beta \ge (1-\epsilon)\cdot 1 + \epsilon \cdot 0 = 1-\epsilon$.
