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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!

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  • $\begingroup$ I think that your proof is perfectly right. $\endgroup$ – GReyes Feb 17 at 7:16
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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, $A\cap C=\emptyset$.

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Your proof is fine.

Option:

Let $A,B,C$ be subsets of a set $X$.

$A \cap C \subset (B$ \ $C) \cap C =$

$ (B \cap C^c) \cap C = B\cap (C^c \cap C)=$

$ B \cap \emptyset =\emptyset.$

Note : $C^c = X$ \ $C$.

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A more "algebraic alternative": $B\setminus C= B\cap C^\complement$. So given that $A \subseteq B \setminus C$ we conclude that

$$A \cap C \subseteq (B\cap C^\complement) \cap C = B \cap (C \cap C^\complement)= B \cap \emptyset = \emptyset$$

so that $A \cap C=\emptyset$. But elementwise reasoning is fine as well.

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