Proof of the identity A\(A\B)=B

[VladZH]

I'm trying to proof an identity from Munkres' Topology

A \ ( A \ B ) = B

By definition A \ B = {x : x in A and x not in B}

A \( A \ B) = A \ (A ∩ B^c) = A ∩ (A ∩ B^c)^c = A ∩ (A^c ∪ B) = (A ∩ A^c) ∪ (A ∩ B) = ∅ ∪ (A ∩ B) = A ∩ B

What did I miss?

 

[fresh_42]

I'm trying to proof an identity from Munkres' Topology

A \ ( A \ B ) = B

By definition A \ B = {x : x in A and x not in B}

A \( A \ B) = A \ (A ∩ B^c) = A ∩ (A ∩ B^c)^c = A ∩ (A^c ∪ B) = (A ∩ A^c) ∪ (A ∩ B) = ∅ ∪ (A ∩ B) = A ∩ B

What did I miss?

Either that $A=X$ is the entire space, or you've found a typo. Just consider a point $b\in B\text{ \ }A$. It is clearly in $B$ but never in any set $A\text{ \ }C$ whatever $C$ might be; except $A=X$ of course.

 

[PeroK]

I'm trying to proof an identity from Munkres' Topology

A \ ( A \ B ) = B

By definition A \ B = {x : x in A and x not in B}

A \( A \ B) = A \ (A ∩ B^c) = A ∩ (A ∩ B^c)^c = A ∩ (A^c ∪ B) = (A ∩ A^c) ∪ (A ∩ B) = ∅ ∪ (A ∩ B) = A ∩ B

What did I miss?

Perhaps even more simply, from the definition it is clear that $A \text{ \ }X \subset A$. So, the identity as given cannot hold for all $A, B$.

 

[member 587159]

I'm trying to proof an identity from Munkres' Topology

A \ ( A \ B ) = B

By definition A \ B = {x : x in A and x not in B}

A \( A \ B) = A \ (A ∩ B^c) = A ∩ (A ∩ B^c)^c = A ∩ (A^c ∪ B) = (A ∩ A^c) ∪ (A ∩ B) = ∅ ∪ (A ∩ B) = A ∩ B

What did I miss?

You missed nothing. This is correct.

 

[VladZH]

Thank you, guys. Seems like I confused with the formultaion

 

[WWGD]

You can always resort to brute force by trying to show every element of B is a subset of A\(A\B) and viceversa. But, yes, you need to know the overall inclusion relation between A and B.