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

Mathematics
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A 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
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@cinar Can we use Venn Diagrams to prove it? @sauravshakya I think for question 1 and 2, they are about the topic ''sets''.

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Other answers:

@Rolypoly no we cant use venn diagrams
\(\setminus\) is not division !!!!!!!!!!!!!!!!!!!!!!!!!!
\[ \large A\setminus(B\setminus A)=A\cap(B\cap A^c)^c=A\cap(B^c\cup A)=A \] 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 \(A\subseteq B\).
\[A-(B-A)=A \cap B \]
\[B-(B-A)=A \cap B\]
this is a true statement, I just dont know how to prove it..
Let \(A=\{a,b,c,d,e\}\) and \(B=\{a,i,u,e,o\}\) then \[ \large A\setminus(B\setminus A)=A\setminus\{i,u,o\}=A \] and \[ \large B\setminus(B\setminus A)=B\setminus\{i,u,o\}=\emptyset \]
IT IS NOT TRUE !!!!!!!!!!!!!
sorry there is a typo the question is Prove A\(A\B)=B\(B\A)
\[ \large A\setminus(A\setminus B)=A\cap(A\cap B^c)^c=A\cap(A^c\cup B) \] \[ \large =(A\cap A^c)\cup(A\cap B)=A\cap B \]

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