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We give a simpler proof that every compact metric space is supercompact. In 1969 De Groot [5] introduced the notion of supercompactness (see definition below). All supercompact spaces are compact by the Alexander subbase lemma; the converse of this was recently shown false by Bell [1]. This leaves the question of what hypotheses on a space, beyond compactness, enable one to conclude that it is supercompact. Strok and Szyman'ski [7] have shown, in particular, that all compact metric spaces are supercompact; their proof, however, is very complicated. The purpose of this note is to offer a simpler proof. We have been advised that our proof is similar to one of van Douwen [4], though ours is somewhat simpler. The author has recently shown [6] that all compact groups are supercompact. A collection e is said to be linked if whenever A, B E X, A n B #0. 9 is binary if for every linked e, c , n e #0. A space X is said to be supercompact if X has a binary closed subbase. Fix a compact metric space X and let { Cn: n E o} be a closed base for X. We shall construct a sequence { 6n: n E o} of finite families of closed sets such that for each n E w, (1) U 6m = (2) U m An is binary. By (1), S = U,, ?,, is a subbase for X; since X is compact, (2) implies that S is binary. We proceed by induction on n. Set

A simpler proof that compact metric spaces are supercompact