D Pálvölgyi, H Peters, D Vermeulen - International Journal of Game …, 2018 - Springer
… is finite, we may assume without loss of generality that there is a set L with \(L = L(v^k)\) for
all k. Then, since \((v^k)_{k=1}^\infty \) converges to v, it follows from Theorem 3.1 that \(v \in L\)…