$\alpha \cap \beta = \varnothing$: 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?
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 $A\subseteq (A\cap B)\cup (A\backslash B)$.
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.