Proving set implication

Question

Can someone give me a starting point to prove if ${\bar A} \subset B$, then $(C - B) \cup A = A$.

Answer 1

If $C-B$ is empty then you are done. That is the trivial case. If $C-B$ is non-empty then there are elements in $C$ which are not in $B$, and consequently not in $\bar{A}$. But if an element is not in $\bar{A}$, necessarily it is in $A$. So $(C-B)\subseteq A$. Hence $(C-B)\bigcup A = A$.

Answer 2

Let $x \in C\setminus B$. Then $x \not \in B$. If $\overline A \subset B$ so if $x\in \overline A$ then $x \in B$ but $x \not\in B$ so $x \not \in \overline A$. So $x \in A$.

So $(C\setminus B)\subset A$ and $(C\setminus B)\cup A = A$.

Answer 3

Can someone give me a starting point to prove if ${\bar A} \subset B$, then $(C \smallsetminus B) \cup A = A$.

$\overline A\subset B~$ means that if any element is not in $A$ then it is in $B$, and that implies that it is not in $C\smallsetminus B$.

Thus by contraposition, ...