Here is my attempt at proving the theorem:
Proof. Suppose $A \subseteq B \setminus C$. Let $x$ be an arbitrary element of $A$. We can conclude that $x \in B \setminus C$, since $A$ is a subset of $B \setminus C$. It follows from $x \in B \setminus C$, that $x \in B$ and $x \notin C$. We have shown that for any $x \in A$, $x \notin C$. Thus, if $A \subseteq B \setminus C$ then $A$ and $C$ are disjoint.
As far as I can tell, my logic is correct, but I still feel like I’ve done something wrong. I feel unsure if what I’ve demonstrated is logically sufficient to conclude $A$ and $C$ are disjoint. Any criticism would be welcome. Thanks!
Best Answer
Your proof is correct.
Alternate:
If possible suppose that $x\in A\cap C$ then $x\in A$ and $x\in C$. Since $x\in A\subset B\setminus C$ so $x\notin C$ $-$ a contradiction that $x\in C$. Hence, $A\cap C=\emptyset$.