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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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