Here is my attempt at a solution for a proof about disjoint unions of sets $A$ and $B$. Can you please point out the mistakes? Thank you all.
Let $A$ and $B$ be any set. Prove:
- $A$ is the disjoint union of $A\setminus B$ and $A\cap B$
- $A\cup B$ is the disjoint union of $A\setminus B$, $A\cap B$, and $B\setminus A$.
Part 1
Suppose $A$ and $B$ are not disjoint. Then,
$ \{x | x\in A\setminus B\text{ and }x\in A\cap B\} $
Since $A\subseteq A\setminus B$ and $A\subseteq A\cap B$, $A\setminus B = A\cap B$. So, $A\setminus B\nsubseteq B$, or $A\setminus B\subseteq B^{C}$, and $A\cap B\subseteq B$.
Then, $B\cup B^{C} = \emptyset$, and $A$ and $B$ are disjoint.
Part 2
Suppose $A\cup B$ is not disjoint. Then,
$ \{x | x\in A\setminus B\text{ and }x\in A\cap B\text{ and }x\in B\setminus A\} $
$A\setminus B\subseteq B^{C}$, $A\cap B\subseteq A$, $A\cap B\subseteq B$, and $B\setminus A\subseteq A^{C}$. But since $B\cup B^{C} = \emptyset$ and $A\cap A^{C} = \emptyset$, $A\cup B$ is disjoint.