Show the intersection of $A$ and $B$ and the relative complement of $B$ in $A$ are disjoint and their union is $A$ I am trying to solve this problem:

Show the sets $A\cap B$ and $A\setminus B$ are disjoint and their union is $A$.

So, I can intuitively understand why the first statement is true;

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*The intersection of $A$ and $B$ contains elements that are in both $A$ and $B$ {$x$: $x \in A$ and $x\in B$} - The Venn Diagram of such a set would have only the overlapping portion of $A$ and $B$ shaded in.

*The Relative complement of $B$ in A contains the elements of $A$ that are not elements of $B$ {$x \in A$: $x \notin B$} - The Venn diagram of such a set would have only the $A$ circle (excluding the overlapping portion) shaded in.

*So the intersection of these sets would be the set of elements that are (in both $A$ and $B$) and (in $A$ but not $B$) which means if $x \in$ $A \cap B$ then $x \notin A \setminus B$ necessarily since $A \setminus B$ contains no elements of the set B by definition - inspecting the corresponding Venn diagrams we see that they have no shaded portions in common.

*My question for this portion is: how would I go about proving this statement using set theory notation? Would I start by assuming that $x \in A \cap B$ then go to show that this assumption leads to $x \notin A \setminus B$? Which would then mean the intersection was $\emptyset$?

For the second part of the question I, again, have an intuitive understanding of why this is true:

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*For one, inspecting the Venn diagrams we can see that combining the two diagrams leads to the entire circle for $A$ being shaded in which is interpreted as the set $A$.

*Symbolically: $(A \cap B)$ $\cup$ ($A \setminus B$) = {$x$: [$x \in A$ and $x \in B$] or [$x \in A$: $x \notin B$]} which allows us to select points from either set and thus reconstruct the set $A$ by combining the points only in A with the points in both $A$ and $B$.

*Similar to the first point above, I'm not sure how to formally prove this statement with notation. I tried negating both statements as well to see if that led to anything but it did not help to illuminate a way forward.

Any help is greatly appreciated!
 A: Your understanding of the definitions of $A \cap B$ and $A \setminus B$ is perfect. You just have to translate your (correct) informal reasoning with Venn diagrams into a formal argument.

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*To show that $A \cap B$ and $A \setminus B$ are disjoint it is enough to prove that $x \in A \cap B$ implies $x \notin A \setminus B$.
Indeed, this means that it is impossible to have an element of both $A \cap B$ and $A \setminus B$.
Let us prove this implication formally: if $x \in A \cap B$, then $x \in B$ and hence $x \notin A \setminus B$ (by definition of $A \setminus B$).


*To show that $A = (A \cap B) \cup (A \setminus B)$, it is enough to show that $A \subseteq (A \cap B) \cup (A \setminus B)$ and $(A \cap B) \cup (A \setminus B) \subseteq A$.



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*Let us show that $(A \cap B) \cup (A \setminus B) \subseteq A$. If $x \in (A \cap B) \cup (A \setminus B)$, then either $x \in A \cap B$ or $x \in A \setminus B$; in both cases $x \in A$.


*Let us show that $A \subseteq (A \cap B) \cup (A \setminus B)$.
If $x \in A$ then there are two possibilities, either $x \in B$ and then $x \in A \cap B$, or $x \notin B$ and then $x \in A \setminus B$; in both cases, $x \in (A \cap B) \cup (A \setminus B)$.
A: My approach to this problem would be to proceed by contradiction.

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*Assume $A \cap B$ and $A \setminus B$ are not disjoint. There exists $x \in (A \cap B) \cap (A \setminus B)$. Therefore, $x \in B$ because $x \in A \cap B$ and $x \not \in B$ because $x \in A \setminus B$, contradiction. So, $A \cap B$ and $A \setminus B$ are disjoint.


*Assume $(A \cap B) \cup (A \setminus B) \neq A$. There exists $x \in A \setminus ((A \cap B) \cup (A \setminus B))$. Therefore, $x \not \in B$ because $x \not \in (A \cap B)$ and $x \in B$ because $x \not \in (A \setminus B)$, contradiction. So, $(A \cap B) \cup (A \setminus B)=A$.
Hope this helps!
