Let $A, B, C$, and $D$ be sets with $C \subseteq A$ and $D \subseteq B$. Prove that $D-A \subseteq B-C$.

Question

I know that in order to prove that $D-A \subseteq B-C$. I need to show that if ex $ x \in (D-A)$ then $ x \in (B-C)$. Here are a few things I know. If $ x \in A$ then $ x \in C$ and also if $ x \in B$ then $ x \in D$. I'm not sure how to approach the prove what method I should use to prove it. Can anyone please explain to me how to approach the problem? Thank you.

Answer 1

The following statement:

I know. If $x∈A$ then $ x∈C$ and also if $x∈B$ then $x∈D.$

Should have been:

I know. If $x∈C$ then $x∈A$ and also if $x∈D$ then $x∈B.$

Back to $$ D-A \subseteq B-C$$

If $ x $ is in $D-A$,then $x$ is in $D$ and not in $A$.

Thus $x$ is in $B$ and not in$ C$.

That is, $x$ is in $B-C.$

Answer 2

Your way to write things is wrong: you have to prove that for all $x\in D\setminus A$, one has $x\in A\setminus C$.

So take any such $x$: it belongs to $D$, but not to $A$. As it belongs to $D$, it belongs to $B$ by hypothesis. And as it is not in $A$, it can't be in $C$ since $A$ contains $C$. Thus, $x$ lies in $B\setminus C$.