The proof is taken from course Algorithm II, Princeton, coursera. In the proof of iii => i, Why/How iii implies the existence of cut (A, B)?
2026-03-26 09:26:22.1774517182
maxflow-mincut theorem, why no augmenting path implies existence of maxflow
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$A$ is being defined just as the subset of vertices satisfying the stated property. So there is nothing to prove about it: it simply exists. Now potentially it might be the empty set, but that (together with its complement $V(G)$) would still count as a cut. Having said that, it is not empty here as paths of length $0$ count, and so you will always at least have $s\in A$