Let A, B and C three sets. Prove that if $B-C$ $\subseteq$ $A^c$then $A \cap B \subseteq C$
Im trying to prove this with sheer logic and not making use of De Morgans laws etc.
Let $y \in (B-C\space$ $\subseteq$ $A^c)$
$y \in B, y\notin C \subseteq A^c$ —–> $A^c$ is equivalent to $B \cup C$ in the given problem.
$y \in B, y\notin C \subseteq y \in B$ or $y\in C$
Thus I proved that $B \subseteq C$ since $B$is in $B$ or $C$. (or at least I think)
Next I have to prove that $A$ is also a $\subseteq C$ and am unsure how to proceed. If I manage to do this I think I will have proved $A \cap B \subseteq C$.
Feel free to show how you would do it, if I am completely wrong.
Thanks!
Best Answer
You have to prove that $A\cap B$ is a subset of $C$. At this level, things do not require immense bursts of creativity, or a stream of great ideas. Usually just verifying the definitions is fairly straightforward and short.
So you need to show that if $x\in A\cap B$ then $x\in C$. If $x\in A\cap B$ then $x\in A$ and $x\in B$. Now comes the point to use the assumption, $B-C\subseteq A^c$. The assumption, when unraveling to definitions tells us that if $x\in B$ and $x\notin C$, then $x\notin A$.
Now our assumption is that $x\in B$, since $x\in A$ it is impossible that $x\notin C$, because then $x\notin A$. Therefore $x\in C$, as wanted.