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?](https://math.stackexchange.com/questions/4468488/does-every-connected-compact-metric-space-have-a-unique-always-attainable-averag) 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.