Question

Answer 1

Solution © let A and B be two sets. Now to prove AUB is the smallest set containing A and B For this it is enough to show that for any set s which contains both A and B have At AUB CS we let s be any set containing A and B Now AUB = fx xe A or xe B] let XE AUB TEA or YEB Y YES because Ass and BCS AUBC S Hence AUB is smallest set containing A and B- Next we have to prove НОВ is the largest set contained in both A and B For this it is enough to show that for any set T wil which is contained in A and B T CAOB let T be any set contained in A and B Now AnB = {2 |IE A and Xe B] we have

let xet xed and me because TCA and TCB B = x^ 10 8 TC AAB T Thus AAB is the largest set contained in both A and B. Solution @ let Alb and c any sets. To prove (A\B) u (B1A) = (AUB)| (ANB) let TE ABU BA -> XEAB or xeble + YEA but x 6B or YEB but x¢ A TJ T ㅠ * xe PUB but a¢ AMB XE (AUB)) CAOB) Therefore (AlBlu (BIA) < (AUB) (A0B) — Now let x6(AUB) | Anb) ƏYE AUB but & € AOB XEA or YEB but X&A or x & B OF XEB but 20 B ł YEA but 24 B

Answer 2

Solution © let A and B be two sets. Now to prove AUB is the smallest set containing A and B For this it is enough to show that for any set s which contains both A and B have At AUB CS we let s be any set containing A and B Now AUB = fx xe A or xe B] let XE AUB TEA or YEB Y YES because Ass and BCS AUBC S Hence AUB is smallest set containing A and B- Next we have to prove НОВ is the largest set contained in both A and B For this it is enough to show that for any set T wil which is contained in A and B T CAOB let T be any set contained in A and B Now AnB = {2 |IE A and Xe B] we have

let xet xed and me because TCA and TCB B = x^ 10 8 TC AAB T Thus AAB is the largest set contained in both A and B. Solution @ let Alb and c any sets. To prove (A\B) u (B1A) = (AUB)| (ANB) let TE ABU BA -> XEAB or xeble + YEA but x 6B or YEB but x¢ A TJ T ㅠ * xe PUB but a¢ AMB XE (AUB)) CAOB) Therefore (AlBlu (BIA) < (AUB) (A0B) — Now let x6(AUB) | Anb) ƏYE AUB but & € AOB XEA or YEB but X&A or x & B OF XEB but 20 B ł YEA but 24 B

T αε ΑΒ or T XEBLA de AB) U (BIA) - (AUB) 1 (A1B) (A\B) U (BIA) From A and (* we get (Alb) U (BIA) - (AUB) | ANB) solution ② let let A,B and C be any sets. ACB then to prove cb c cla let XE CL B = TEC but x&B XEC but 16A (because ACB henee YEA is not possible) Xocla [cB C clA

Here converse is not true Counter example : let A= {4}, B={13,6-41,2,3] Here clb = 2133 c/A={1,2,3} = db c dla but A&B Thank you.