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

• B

## Main Question or Discussion Point

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?

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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.

PeroK
Homework Helper
Gold Member
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 $A, B$.

Math_QED
Homework Helper
2019 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?
You missed nothing. This is correct.

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

WWGD