# Real analysis chapter 0

1. Aug 28, 2008

1. The problem statement, all variables and given/known data

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

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

2. Relevant 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

3. 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.

2. Aug 28, 2008

### konthelion

A\(A\ B) = B

For A\B, Let $$x \in A \wedge x \notin B$$
Then, A \ (A \ B) implies that

$$x \in A \wedge (x \notin A \vee x \in B)$$ by DeMorgan's Law

Then, by associative law,

$$(x \in A \wedge x \notin A) \vee ( x \in A \wedge x \in B) \implies ( x \in A \vee x \in B)$$

So, $$A \cup B = B$$ if $$A \subset B$$ or if A is the empty set.

Last edited: Aug 28, 2008
3. Aug 29, 2008

### HallsofIvy

Staff Emeritus
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. Aug 29, 2008

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. Aug 31, 2008

### boombaby

A\B=A$$\cap$$B$$^{c}$$ will help
Well, Konthelion made a mistake: that should be "so, A$$\cap$$B=B , blah blah blah"

Last edited: Aug 31, 2008