Prove set Inclusivity I need help proving the following > Suppose $A,B \subseteq X$. > > Prove that if $X \setminus B \subseteq X \setminus A$, then $A \subseteq B$. I really need to see an example of how to do these types of problems in general. My attempt is below, but I would prefer to see a worked proof if possible. "$ \implies $": Suppose that there exists some $t \in B\subseteq X \implies t \in X$. Now $X \setminus B \subseteq X \setminus A$ so $t \notin A$. Therefore $A \subseteq B$. "$ \impliedby $": Suppose that there exists some $t \in A\subseteq B\implies t \in B$. Now, if $A \subset B$ then $X \setminus B$ contains no values of $A$. Therefore $X \setminus B \subseteq X \setminus A$.

Let us assume $X-B\subseteq X - A $.

Choose any $a \in A$. Then we claim that $a \in B$ as well. Suppose on the contrary this is not true. i.e. $a \
otin B$.

Then $a \in X$ and $a \
otin B$ implies $a \in X - B$.

Then since $X-B\subseteq X - A $ we also have $a \in X - A$.

This is equivalent to saying $x \in X$ and $x \
otin A$.

In particular, this contradicts our assumption that $a \in A$. Hence $a \in B$ necessarily.

Then $A \subseteq B$ as required.