$\alpha \cap \beta = \varnothing$: An Analysis on Set Disjointness

Question: Prove: $A\cap B$ and $A\setminus B$ are disjoint and that $A=(A\cap B)\cup (A\setminus B)$.

In the above question I would like to know how to come to the desired conclusion. What is the idea here?

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For the first:

Suppose $x \in A\cap B$ and $x \in A\backslash B$. The first statement implies $x\in B$, while the second implies $x\not\in B$ which is a contradiction.

For the second:

Suppose $x\in (A\cap B)\cup (A\backslash B)$.Then $x\in A \cap B$ or $x\in A\backslash B$. Both cases imply, $x\in A$ and so $(A\cap B)\cup (A\backslash B)\subseteq A$.

Now suppose $x\in A$. If $x\in B$, then $x\in A\cap B$ and if $x \not\in B$ then, $x\in A\backslash B$. Both cases imply $x \in (A\cap B)\cup (A\backslash B)$ and so $A\subseteq (A\cap B)\cup (A\backslash B)$.

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Hmm.. This seems a bit weak because it isn't clear why contradiction implies disjointness. – Loie Benedicte Aug 17 '13 at 19:12
@LoieBenedicte I purposely left it there for you to fill in the rest of the details of the proof. If you want, I can refine my first proof to add more details for your better understanding. – Alraxite Aug 17 '13 at 19:16

Let $x \in A\cap B$ . Then $x \in A$ and $x \in B$. This implies that $x \not \in B^{c}$ and thus $x \not \in A -B$. For the second part, you can prove that one is the subset of the other and vice versa.

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The complement with respect to what? – Loie Benedicte Aug 17 '13 at 18:07

My approach was the following:

If $(A\cap B)\cap (A\setminus B) = \varnothing$, then if $x\in (A\cap B)$, then it must be that $x\not\in (A\setminus B)$, so what does it mean to not be in the set $A\setminus B$? Well, it means $\neg(x\in A\setminus B)$, which means $x\not\in (A\setminus B)$, which then means $x\not\in A \lor x\in B$, so if $x\in A\cap B$, then $\dots$ Hmm $\dots$ Maybe this is the wrong idea.

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First: They are asking to prove $(A \cap B) \cap (A \setminus B) = 0$. Knowing that $A \setminus B$ is equal to simply $A, x \epsilon A$ and $x$ is NOT an element of $B$, we can continue:

The intersection of A and B includes elements both in A and B. $A \setminus B$ has elements in A not in B. Thus, the intersection is null.

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