Prove that if $B−C \subseteq A'$ then $A \cap B \subseteq C$.

Is it perfectly reasonable to show that $A \cap B \not\subset C$, (assuming $B−C \subseteq A'$ holds) leads to a contradiction ?

Suppose (hoping fervently to eventually reach a contradiction) that $A \cap B \not\subset C$. Then $\exists x \in A \cap B$ such that $x \not\in C$ meaning $x \in A$ and $x \in B$ and $x \not\in C$. Thus $x \in A$ and $x \in B−C$ and as $B−C \subsetneq A'$ we conclude with saying $x \in A$ and $x \not\in A$ which clearly no $x$ satisfies and so contradicts the fact $\exists x \in A \cap B$ such that $x \not\in C$.

  • $\begingroup$ B' is everything not in B so B complement and yh i think it does work lol $\endgroup$ – Namch96 Nov 24 '14 at 16:55

A direct proof is:

$B-C\subseteq A'\implies B\cap C'\subseteq A'\implies A'\cap B\cap C'=B\cap C'\implies$ $(A'\cap B\cap C')\oplus(B\cap C')=\emptyset\implies (A'\oplus U)\cap(B\cap C')=\emptyset \implies$ $A\cap B\cap C'=\emptyset\implies A\cap B\subseteq C$ $\quad\square$


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