# Proof that if $A \subseteq B \setminus C$ then $A$ and $C$ are disjoint.

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!

• I think that your proof is perfectly right. – GReyes Feb 17 at 7:16

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, $$A\cap C=\emptyset$$.

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

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.