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Real Analysis

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Answer #1

be two sets. let A and ret P be and any set containing both p. Then A CP and BEP A xt AUB let =31 EA oss => XEP a E B T. ASPYE (AUB) I CAMB) > YE (A1B) (BIA) let be any sets let 2. A, B und WE CAIB) U(BIA) Then (at A and x&B) on WEB and a&A) → XE (A

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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 thalet 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 @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 ②Here 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

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