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 2 years ago
1) Prove that if A and B are countable, then \[A \cap B\] is also countable.
2) Prove A\(A\B)=B\(B\A)
 2 years ago
1) Prove that if A and B are countable, then \[A \cap B\] is also countable. 2) Prove A\(A\B)=B\(B\A)

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sauravshakya
 2 years ago
Best ResponseYou've already chosen the best response.0A n B can never be greater than A and B ........ So 0<=A n B <=A and 0<=A n B <=B...... THus, if A and B are countable ......... A n B is also countable

sauravshakya
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1346927959015:dw

RolyPoly
 2 years ago
Best ResponseYou've already chosen the best response.0@cinar Can we use Venn Diagrams to prove it? @sauravshakya I think for question 1 and 2, they are about the topic ''sets''.

cinar
 2 years ago
Best ResponseYou've already chosen the best response.0@Rolypoly no we cant use venn diagrams

helder_edwin
 2 years ago
Best ResponseYou've already chosen the best response.0\(\setminus\) is not division !!!!!!!!!!!!!!!!!!!!!!!!!!

helder_edwin
 2 years ago
Best ResponseYou've already chosen the best response.0\[ \large A\setminus(B\setminus A)=A\cap(B\cap A^c)^c=A\cap(B^c\cup A)=A \] on the other hand \[ \large B\setminus(B\setminus A)=B\cap(B\cap A^c)^c=B\cap(B^c\cup A)= \] \[ \large = (B\cap B^c)\cup(B\cap A)=B\cap A \] they are not equal. unless \(A\subseteq B\).

cinar
 2 years ago
Best ResponseYou've already chosen the best response.0this is a true statement, I just dont know how to prove it..

helder_edwin
 2 years ago
Best ResponseYou've already chosen the best response.0Let \(A=\{a,b,c,d,e\}\) and \(B=\{a,i,u,e,o\}\) then \[ \large A\setminus(B\setminus A)=A\setminus\{i,u,o\}=A \] and \[ \large B\setminus(B\setminus A)=B\setminus\{i,u,o\}=\emptyset \]

helder_edwin
 2 years ago
Best ResponseYou've already chosen the best response.0IT IS NOT TRUE !!!!!!!!!!!!!

cinar
 2 years ago
Best ResponseYou've already chosen the best response.0sorry there is a typo the question is Prove A\(A\B)=B\(B\A)

helder_edwin
 2 years ago
Best ResponseYou've already chosen the best response.0\[ \large A\setminus(A\setminus B)=A\cap(A\cap B^c)^c=A\cap(A^c\cup B) \] \[ \large =(A\cap A^c)\cup(A\cap B)=A\cap B \]
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