- #1
VladZH
- 56
- 1
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?
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?