When Does A \ (A \ B) Equal B in Set Theory?

In summary: Anyway, assuming Konthelion's assumption is true, then x is in A\capB and x is not in A\(A\B) because if x is in A\(A\B) then x is in A and x is not in A\B, and if x is not in A\(A\B) then x is in B. So, x is in A\capB and x is not in A\(A\B).
  • #1
icantadd
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Homework Statement



where A and B are sets and ' \ ' means set difference

under what conditions does A \ (A \ B) = B

Homework Equations



S\A = {X | X in S and X not in A}
B < A (subset) means if x is in B then x is in A


The Attempt at a Solution


I get that if B is a subset of A then A \ (A \ B) = B. But I don't know how to prove it. Here's my attempt thus far

assume x is in A \ (A \ B)
to be proved x is in B

from assumption x is in A and x is not in (A \ B) , then
x is in A and not { L | L in A and L not in B}
because B is a subset of A
x is in A and . . .

. . .
It's like I know that A \ B contains nothing from B but because A includes B, subtracting it from A will give back B. I don't know how to phrase it right for making a mathematical argument. Any guidance will be strongly appreciated.
 
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  • #2
A\(A\ B) = B

For A\B, Let [tex] x \in A \wedge x \notin B [/tex]
Then, A \ (A \ B) implies that

[tex] x \in A \wedge (x \notin A \vee x \in B) [/tex] by DeMorgan's Law

Then, by associative law,

[tex] (x \in A \wedge x \notin A) \vee ( x \in A \wedge x \in B) \implies ( x \in A \vee x \in B) [/tex]

So, [tex]A \cup B = B[/tex] if [tex] A \subset B[/tex] or if A is the empty set.
 
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  • #3
Assuming B is a subset of A:

If x is in A\(A\B) then x is in A and x is not in A\B. x not in A\B means either x is not in A (which is impossible) or x is in B. Therefore x in A\(A\B) implies x is in B.
If x is in B, since B is a subest of A, x is in A. Since x is in B, it is NOT in A\B and therefore is in A\(A\B).

Those two prove that, if B is a subset of A, A\(A\B)= B.

But you also have to prove the other way: if A\(A\B)= B, then B is a subset of A.

if x is in B, then x is NOT in A\B but is in A\(A\B) because A\(A\B)= B. Therefore x is in A and so B is a subset of A.
 
  • #4
Thank you for your help. I actually got it just after posting here, when I read the chapter and found that if B is a subset of A, A\B can be rewritten as B* or B's Complement in A. Then my modeling of the problem became more crystaline, and I went from there, only slightly different than yours HallsOfIvy

For Konthelion, I have not seen the step you made to get to "x in A and (x not in A or x in B)". Can you explain why you were able to do this?
 
  • #5
A\B=A[tex]\cap[/tex]B[tex]^{c}[/tex] will help o:)
Well, Konthelion made a mistake: that should be "so, A[tex]\cap[/tex]B=B , blah blah blah"
 
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FAQ: When Does A \ (A \ B) Equal B in Set Theory?

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