On a compact metric space $(\Omega, d)$ the Wasserstein-1 distance is given by $$ W_1(\mu, \nu) = \sup_{\varphi \in \text{Lip}(\Omega)} ~ \int \varphi(x) d\mu(x) - \int \varphi(x) d \nu(x), $$ where $\text{Lip}(\Omega)$ denotes the $1$-Lipschitz functions on the metric space.
For the discrete metric $d(x, y) = 1_{x \neq y}$, it can be shown that $W_1(\mu, \nu) = \frac{1}{2} |\mu - \nu|_{TV}$.
My question is the following: is the above supremum actually attained if we use the discrete metric?
In my optimal transport reference, this is proven using the Arzelà-Ascoli theorem, which uses equicontinuity of the dual variable due to the Lipschitz constraint. Is there a reason this argument could fail for the discrete metric?
I'm not sure if the same argument goes through, but you can directly show the supremum is attained:
With the discrete metric, any function $\varphi$ with $|\varphi(x)|\le \frac 12$ for all $x \in \Omega$ is $1$-Lipschitz. Let $P \sqcup N = \Omega$ be the Hahn decomposition of $\Omega$ for the signed measure $\mu - \nu$ and define $$\varphi(x) = \begin{cases} \frac 12 & x \in P \\ - \frac 12 & x \in N \end{cases}.$$ This $\varphi$ is $1$-Lipschitz with the discrete metric, and $$\int \varphi(x)d(\mu-\nu)(x) = \frac 12 |\mu - \nu|_{TV}$$ so this $\varphi$ attains the supremum.