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

  • #1
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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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  • #2
fresh_42
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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.
 
  • #3
PeroK
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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##.
 
  • #4
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.
 
  • #5
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Thank you, guys. Seems like I confused with the formultaion
 
  • #6
WWGD
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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.
 

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