Question

ILULIITUL 10.37 Theorem. (The Generalized Distributive Laws for Sets of Sets.) Let S be a set and let be a non-empty set of sets. Then: (a) SNU =USNA: AE}. (b) Sund= {SUA:AE). Proof (a) Let = {SNA: AE }. We wish to show that S U = UB. For each 1, we have BESUS iff x S and 2 EU iff xe S and there exists AE such that EA iff there exists AE such that reS and x E A iff there exists Aed such that ceSnA iff there exists B E such that I eB iff z EU Thus the set SnUJ has the same elements as the set UB, so these two sets are equal. This proves (a). The proof of (b) is left as an exercise. Exercise 26. Prove Theorem 10.37(b). fant Show that:

Answer 1

Here we will prove 10:37(6). To show surna) = n { SUA : AEMY het BE SUCn.A) 5 BES or BENA :. BES de tored BENA Casei) Suppose BES. - BESUA, for each AEA : BE n {SUA: AE A} case ii) suppose BE OM A BEA, for each AEA :BESUA , for each AEM. -- BE 0 {SUA: AER? 5. In each Bf case BE O {SUA: ACM} is it BESUCO AI then BE O{SUA: AEA? SU COX S O{SUA: AEM? Now let xf n&SUA: AEA} : 2 E SUA, for each Atd. i 2€S or LEA for each AE . case 1) If t€S XE SU CA) case ii) If x EA for each at A. xena : DC E SUCOM) :. In each case LE SU COA) Hence of suchen) : n {SUA: Af @ AL C SUCO) - OX

i By © and ©, e sucnm = n{SUA: AE A?