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2 for the case of bipartite graphs. , n } and the Ai are subsets of E ; assume, as may be done without loss of generality, that I n E = 0; and consider the bipartite graph G = (N, F), where N = I u Eand Let I ‘ G I , E’ G E and suppose that S = I’ u E’ is a set separating I and E in C. Then A(I \ 1’) n (E \ E’) = 0 and so A(I \ 1’) G E’. Hence, by %, I S I= 11’1 + IE’I3 11’1 + J A ( I \ ri)i 3 1 ’1 + IT\ rq = n. T h u s no set separating I and E contains fewer than n nodes of G. e. I possesses a transversal.

We might therefore expect some theorems involving partial transversals to be symmetric with respect t o these two collections of objects. 6 below are results of this kind. 1. The lack of notational symmetry, on the other hand, can be corrected by a mere change of notation (and terminology). We recall, then, that we are dealing with two classes of objects: the sets Ai, i ~ of1 the family 41 = ( A i : ig I), and the elements of the ground set E. Now whereas the elements are (of course) distinct, this is not in general true of the sets.

3 every subset of an admissible set is itself admissible. Equally, every restriction of an admissible mapping is again admissible. The subsets A, B of X, Y respectively are said to be linked (in symbols: A tf B) if there exists an admissible bijection of A into B. ) If A. B are singletons, say A = {x}, B = : y > , then the relation {x} ++ { y ) is written as x - y . This is consistent with the usage of the symbol <+ introduced earlier. By convention, 0 ++ 0. A subset A of X is said to be totally admissible (or, more briefly, a totalset) if A t--) Y.

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