Communicated by the Editors

In this note we show by a simple direct proof that Folkman’s necessary and sufficient condition for an infinite family of sets with finitely many infinite members to have a transversal implies Woodall’s condition. A short proof of Folkman’s theorem results by combining with Woodall’s proof. Several authors, Brualdi and Scrimger [ 11, Folkman [2] and Woodall [5], have given necessary and sufficient conditions for an infinite family of sets with finitely many infinite members to have a transversal, generalising the well-known theorems of Hall [ 3] and Jung and Rado [4 1. In each case the proof of necessity is straightforward. Woodall’s proof of his theorem is one page long, and he also establishes simply the equivalence between his conditions and those of Brualdi and Scrimger. On the other hand the proof of Folkman’s theorem in 121 is quite intricate. In this note we show by a simple direct proof that Folkman’s condition implies Woodall’s condition. A short proof of Folkman’s theorem results by combining with Woodall’s proof. Throughout this note A = (Aili E 1) is a family of subsets of a set E such that A i is finite for all i E I, with lb0 finite. For J E I we abbreviate lJ iEJ A i to A(J). THEOREM 1 (J. Folkman 121). A has a transversal if and only if (1.1) for every non-negative integer n and pair J, J ’ of sets such that J ’ G J C I, A(J)\P(J’) is finite and (A(J)\P(J’)) < IJ\J ’ )- n there is a finite set J ” 5 J ’ such that all finite sets L with J ” c L c J ’ satisfy IA(L)l 2 ILI + n. Let X be a subset of E. If the subfamily (,4,/i E I,) has a transversa