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Are-A-B-A-B-and-A-B-completely-equivalent-Simplify-A-B-A-B-to-A-B-using-set-operations-and-their-properties-




Question Number 1698 by Rasheed Ahmad last updated on 01/Sep/15
•Are A∪B=A∩B  and  A=B   completely equivalent?  •Simplify A∪B=A∩B to A=B  using set operations and their  properties.
$$\bullet{Are}\:\boldsymbol{\mathrm{A}}\cup\boldsymbol{\mathrm{B}}=\boldsymbol{\mathrm{A}}\cap\boldsymbol{\mathrm{B}}\:\:{and}\:\:\boldsymbol{\mathrm{A}}=\boldsymbol{\mathrm{B}}\: \\ $$$${completely}\:{equivalent}? \\ $$$$\bullet{Simplify}\:\boldsymbol{\mathrm{A}}\cup\boldsymbol{\mathrm{B}}=\boldsymbol{\mathrm{A}}\cap\boldsymbol{\mathrm{B}}\:{to}\:\boldsymbol{\mathrm{A}}=\boldsymbol{\mathrm{B}} \\ $$$${using}\:{set}\:{operations}\:{and}\:{their} \\ $$$${properties}. \\ $$
Answered by 123456 last updated on 01/Sep/15
if A∪B=A∩B them A=B  we have that A∪B=A∩B them ∀x,x∈A∪B,x∈A∩B  them supose that A≠B, without loss  of generality suppose tbat ∣A∣>∣B∣  x∈A,x∉B  x∈A∪B,x∉A∩B (contradition)  hence A=B  and if A=B, the  A∪B=A∪A=A  A∩B=A∩A=A  A∪B=A∩B  so  A∪B=A∩B⇔A=B
$$\mathrm{if}\:\mathrm{A}\cup\mathrm{B}=\mathrm{A}\cap\mathrm{B}\:\mathrm{them}\:\mathrm{A}=\mathrm{B} \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{that}\:\mathrm{A}\cup\mathrm{B}=\mathrm{A}\cap\mathrm{B}\:\mathrm{them}\:\forall{x},{x}\in\mathrm{A}\cup\mathrm{B},{x}\in\mathrm{A}\cap\mathrm{B} \\ $$$$\mathrm{them}\:\mathrm{supose}\:\mathrm{that}\:\mathrm{A}\neq\mathrm{B},\:\mathrm{without}\:\mathrm{loss} \\ $$$$\mathrm{of}\:\mathrm{generality}\:\mathrm{suppose}\:\mathrm{tbat}\:\mid\mathrm{A}\mid>\mid\mathrm{B}\mid \\ $$$${x}\in\mathrm{A},{x}\notin\mathrm{B} \\ $$$${x}\in\mathrm{A}\cup\mathrm{B},{x}\notin\mathrm{A}\cap\mathrm{B}\:\left(\mathrm{contradition}\right) \\ $$$$\mathrm{hence}\:\mathrm{A}=\mathrm{B} \\ $$$$\mathrm{and}\:\mathrm{if}\:\mathrm{A}=\mathrm{B},\:\mathrm{the} \\ $$$$\mathrm{A}\cup\mathrm{B}=\mathrm{A}\cup\mathrm{A}=\mathrm{A} \\ $$$$\mathrm{A}\cap\mathrm{B}=\mathrm{A}\cap\mathrm{A}=\mathrm{A} \\ $$$$\mathrm{A}\cup\mathrm{B}=\mathrm{A}\cap\mathrm{B} \\ $$$$\mathrm{so} \\ $$$$\mathrm{A}\cup\mathrm{B}=\mathrm{A}\cap\mathrm{B}\Leftrightarrow\mathrm{A}=\mathrm{B} \\ $$
Answered by 123456 last updated on 01/Sep/15
A∪B=A∩B  (A∪B)∪A=(A∩B)∪A  A∪B=A  (A∪B)∪B=(A∩B)∪B  A∪B=B  A=B
$$\mathrm{A}\cup\mathrm{B}=\mathrm{A}\cap\mathrm{B} \\ $$$$\left(\mathrm{A}\cup\mathrm{B}\right)\cup\mathrm{A}=\left(\mathrm{A}\cap\mathrm{B}\right)\cup\mathrm{A} \\ $$$$\mathrm{A}\cup\mathrm{B}=\mathrm{A} \\ $$$$\left(\mathrm{A}\cup\mathrm{B}\right)\cup\mathrm{B}=\left(\mathrm{A}\cap\mathrm{B}\right)\cup\mathrm{B} \\ $$$$\mathrm{A}\cup\mathrm{B}=\mathrm{B} \\ $$$$\mathrm{A}=\mathrm{B} \\ $$
Commented by 123456 last updated on 01/Sep/15
(A∪B)∩A=(A∩B)∩A  A=A∩B  (A∪B)∩B=(A∩B)∩B  B=A∩B
$$\left(\mathrm{A}\cup\mathrm{B}\right)\cap\mathrm{A}=\left(\mathrm{A}\cap\mathrm{B}\right)\cap\mathrm{A} \\ $$$$\mathrm{A}=\mathrm{A}\cap\mathrm{B} \\ $$$$\left(\mathrm{A}\cup\mathrm{B}\right)\cap\mathrm{B}=\left(\mathrm{A}\cap\mathrm{B}\right)\cap\mathrm{B} \\ $$$$\mathrm{B}=\mathrm{A}\cap\mathrm{B} \\ $$
Commented by Rasheed Ahmad last updated on 01/Sep/15
G^(oo) DD_(eductio) N _!^!
$$\mathrm{G}^{{oo}} \mathrm{DD}_{\mathrm{eductio}} \mathrm{N}\:_{!} ^{!} \\ $$
Commented by 123456 last updated on 02/Sep/15
also we can extend it to  ∪_(i=1) ^n A_i =∩_(i=1) ^n A_i ⇔A_i =A_j ,i∈{1,...,n},j∈{1,...,n}  n∈N^∗
$$\mathrm{also}\:\mathrm{we}\:\mathrm{can}\:\mathrm{extend}\:\mathrm{it}\:\mathrm{to} \\ $$$$\underset{{i}=\mathrm{1}} {\overset{{n}} {\cup}}\mathrm{A}_{{i}} =\underset{{i}=\mathrm{1}} {\overset{{n}} {\cap}}\mathrm{A}_{{i}} \Leftrightarrow\mathrm{A}_{{i}} =\mathrm{A}_{{j}} ,{i}\in\left\{\mathrm{1},…,{n}\right\},{j}\in\left\{\mathrm{1},…,{n}\right\} \\ $$$${n}\in\mathbb{N}^{\ast} \\ $$
Commented by Rasheed Soomro last updated on 03/Sep/15
Generalization! V. Good!
$${Generalization}!\:{V}.\:{Good}! \\ $$

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