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

• B
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 ∩ Bc) = A ∩ (A ∩ Bc)c = A ∩ (Ac ∪ B) = (A ∩ Ac) ∪ (A ∩ B) = ∅ ∪ (A ∩ B) = A ∩ B

What did I miss?

## Answers and Replies

fresh_42
Mentor
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 ∩ Bc) = A ∩ (A ∩ Bc)c = A ∩ (Ac ∪ B) = (A ∩ Ac) ∪ (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.

• VladZH
PeroK
Science Advisor
Homework Helper
Gold Member
2020 Award
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 ∩ Bc) = A ∩ (A ∩ Bc)c = A ∩ (Ac ∪ B) = (A ∩ Ac) ∪ (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##.

• VladZH
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 ∩ Bc) = A ∩ (A ∩ Bc)c = A ∩ (Ac ∪ B) = (A ∩ Ac) ∪ (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
Science Advisor
Gold Member
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.