Show that the three sets $A\setminus B$, $A\cap B$, and $B\setminus A$ are disjoint and that their union is $A\cup B$

Question

To prove they are disjoint, suppose for the sake of contradiction that they aren't i.e., $\exists x: x\in (A\setminus B) \cap (A\cap B) \cap (B\setminus A) \iff x \in A\setminus B$ and $x\in A\cap B$ and $x\in B\setminus A \iff (x\in A \land x \notin B) \text{ and } (x\in A \land x\in B)\text{ and }(x \in B \land x\notin A)$ so $x\in A$ and $x \notin A$ and $x \in B$ and $x \notin B$ which is a contradiction. Thus by the definition of the empty set, $(A\setminus B) \cap (A\cap B) \cap (B\setminus A)=\emptyset$.

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For the second one let $x\in (A\setminus B) \cup (A\cap B) \cup (B\setminus A) \iff x\in A\setminus B$ or $x \in A\cap B$ or $x \in B\setminus A \iff (x\in A \text{ and } x\notin B)$ or $(x\in A \text{ and } x\in B)$ or $(x \in B \text{ and } x\notin A)$.

If $x\notin A$ then $x\notin B$ or $x \in B$ or $x \in A \implies x\in A$ or $x\in B \implies x\in A\cup B$.

If $x\in A$ then $(x\in A \text{ and } x\notin B)$ or $(x\in A \text{ and } x\in B)$ or $x\in B \implies x\in A$ or $x\in B\implies x\in A\cup B$.

In both cases $x\in A\cup B$ thus $ (A\setminus B) \cup (A\cap B) \cup (B\setminus A) \subseteq A\cup B$.

Now suppose $x \in A\cup B$. Then $x\in A$ or $x\in B$ and I'm not sure how to continue here.

My questions are: is my first proof correct? Is the second proof up to that point correct? If so, how could I continue in order to show $ A\cup B \subseteq (A\setminus B) \cup (A\cap B) \cup (B\setminus A)$?

Edit: The first proof should be proving the sets are pairwise disjoint. Here is a post where this gets answered for one of these cases and the rest are similar. Proof that $A \cap B$ and $A \setminus B$ are disjoint.

Answer 1

According to the properties involved, one has that: \begin{align*} (A\backslash B)\cap(A\cap B) & = (A\cap B^{c})\cap(A\cap B)\\\\ & = A\cap(B^{c}\cap(A\cap B))\\\\ & = A\cap(B^{c}\cap(B\cap A))\\\\ & = A\cap ((B^{c}\cap B)\cap A)\\\\ & = A\cap (\varnothing\cap A)\\\\ & = A\cap\varnothing = \varnothing \end{align*}

Similarly, we prove that $A\cap B$ and $B\backslash A$ are disjoint. As to the third case, we may proceed as follows: \begin{align*} (A\backslash B)\cap(B\backslash A) & = (A\cap B^{c})\cap(B\cap A^{c})\\\\ & = A\cap(B^{c}\cap(B\cap A^{c}))\\\\ & = A\cap((B^{c}\cap B)\cap A^{c})\\\\ & = A\cap(\varnothing\cap A^{c})\\\\ & = A\cap\varnothing = \varnothing \end{align*}

As to the union, let us do it in two steps. First, it results that \begin{align*} (A\backslash B)\cup(A\cap B) & = (A\cap B^{c})\cup(A\cap B)\\\\ & = A\cap(B^{c}\cup B) = A \end{align*}

And finally we obtain the desired claim: \begin{align*} (A\backslash B)\cup(A\cap B)\cup(B\backslash A) & = A\cup(B\backslash A)\\\\ & = A\cup(B\cap A^{c})\\\\ & = (A\cup B)\cap(A\cup A^{c})\\\\ & = A\cup B \end{align*}

Hopefully this helps!

Answer 2

1. You need to prove that the three sets are pairwise disjoint. You can obviously have $A \cap B \cap C = \emptyset$ without any two being disjoint.
2. You make unnecessarily heavy work of the first direction. Delete “let”. Then just note that what is on the right after the second biconditional is trivially the inclusive disjunction of $x \in A$ and $x \in B$.
3. And you need to do no more work for the other direction, because note you have proved biconditionals. So you can equally go in the opposite direction from the last to the first and you are done!