Let the given sets be A,B,C, etc. For the new sets we may take A,BA',CB'A', etc. where BA' denotes the intersection of B and the complement of A and CB'A' denotes the intersection of C with the complements of B and A.
Supposing that the element k is in the given set K but not in any of the earlier given sets A, B,...,J, then k is in KJ'I'...A'. But k is not in any one of the earlier new sets IH'G'...A', HG'F'...A',...,A since it is not in any of H, G,...,A. Neither is k in any of the later new sets LK'J'...A', etc. all of which have a factor K'.
Thus k is in the one and only new set KJ'I'...A', which means that every element of the given sets is in exactly one of the new sets and this makes the new sets disjoint and makes their union the same as the union of the given sets.
(Page 7 of www.mcs.drexel.edu/~phitczen/notesfin.ps gives a typeset version of this solution. See also Exercise 2 on page 61 for an application.) |