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

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Discussion Overview

The discussion revolves around proving the identity A \ (A \ B) = B, as presented in Munkres' Topology. Participants explore definitions and logical steps involved in set operations, aiming to clarify the identity's validity.

Discussion Character

  • Technical explanation, Debate/contested, Mathematical reasoning

Main Points Raised

  • One participant presents a step-by-step proof of the identity, using definitions of set difference and complements.
  • Another participant suggests that the identity may not hold unless A is the entire space, questioning the generality of the claim.
  • A different participant proposes that the identity cannot be universally true, indicating that A \ X is always a subset of A.
  • One participant affirms that the original proof is correct, indicating no errors in the reasoning presented.
  • Another participant suggests a brute force method to show the identity by demonstrating mutual inclusion of elements between the sets.

Areas of Agreement / Disagreement

There is no consensus on the validity of the identity for all sets A and B. Some participants affirm the proof's correctness, while others challenge its general applicability.

Contextual Notes

Participants express uncertainty regarding the conditions under which the identity holds, particularly concerning the relationship between sets A and B.

VladZH
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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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VladZH said:
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.
 
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VladZH said:
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##.
 
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VladZH said:
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.
 
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Thank you, guys. Seems like I confused with the formultaion
 
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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