# Proof that $A \cap B$ and $A \setminus B$ are disjoint.

I am trying to prove that $A \cap B$ and $A \setminus B$ are disjoint. Here is what I've done so far.

Is there anything that's wrong in my proof, and is there anything that can make it better?

Proof: $A \cap B$ and $A \setminus B$ are disjoint if $(A \cap B) \cap (A \setminus B) = \emptyset$.

First, let $x \in (A \cap B) \cap (A \setminus B)$. Then $x \in (A \cap B)$ and $x \in (A \setminus B)$. Then this means $x \in A$ and $x \in B$, and $x \in A$ and $x \notin B$. Thus, these two sets must be disjoint and therefore $x \in \emptyset$. Hence, $(A \cap B) \cap (A \setminus B) \subseteq \emptyset$.

Conversely, since the empty set is always a subset of any nonempty set, $\emptyset \subseteq (A \cap B) \cap (A \setminus B)$.

Therefore $(A \cap B) \cap (A \setminus B) = \emptyset$.

• "Conversely, since an empty set is always a subset of any set" nonempty or otherwise. It would also help to say just before "thus" the line "but $x\in B$ and $x\notin B$ cannot occur simultaneously" as well as the phrase before "first" "Suppose for contradictory purposes that there is some $x\in\dots$" Otherwise, your proof is essentially correct. – JMoravitz Jun 20 '16 at 19:02

There is nothing wrong in the proof. However, to make it better you might skip some superfluous details. For example, the following is enough:

Suppose towards a contradiction that $(A \cap B) \cap (A \setminus B) \neq \emptyset$. Then there is an $x \in (A \cap B) \cap (A \setminus B)$, which implies $x \in B$ and $x \notin B$. A contradiction.

Other way $$(A\cap B)\cap(A-B)=(A\cap B)\cap(A\cap B\,')=A\cap(B\cap B\,')=A\cap\phi=\phi$$

Let $x \in (A \cap B) \cap (A \setminus B)$. Then $x \in (A \cap B)$ and $x \in (A \setminus B)$. After this, it is better to say:
This means $(x \in A\land x \in B)\land (x \in A\land x \notin B)$. So $x\in B\land x \notin B$, which is contradiction.

The proof is correct, but it is a tad verbose.
If you are going to write a proof by contradiction, I recommend you say so up front.

You just need to show that $x$ cannot be in both $B$ and $B'.$

Alternatively, show $(A \cap B) \cap (A \cap B') = \emptyset$

$(A \cap A) \cap (B \cap B') = \emptyset$