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 $X \setminus B \subseteq X \setminus A$.
$\endgroup$ 22 Answers
$\begingroup$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. $a \notin B$.
Then $a \in X$ and $a \notin B$ 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 $x \notin A$.
In particular, this contradicts our assumption that $a \in A$. Hence $a \in B$ necessarily.
Then $A \subseteq B$ as required.
$\endgroup$ 2 $\begingroup$To show $A \subseteq B$, we need to show that for any $x \in A \Rightarrow x \in B$.
We proceed using the contrapositive approach i.e. we instead prove $x \not \in B \Rightarrow x \not \in A$
Suppose $x \not \in B$
By definition, $X\setminus B=\{x \in X, x \not\in B\}$
So, $x \in X\setminus B$
Since $X\setminus B \subseteq X\setminus A$, we have $x \in X\setminus A$.
By definition, $X\setminus A=\{x \in X, x \not\in A\}$
So, $x \not \in A$.
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