Prove if $A \cup C \subseteq A \cap C$ then $A=C$
I'm attempting this proof and am having second thoughts that I am right.
Attempt(Contrapositive):
Assume $A \neq C$
Assume WLOG $x \in A$ and $x \notin C$
Then $x \in A \cup C$ but $x \notin A \cap C$ thus $A \cup C \not\subset A \cap C$
I felt like this was the correct way to prove this but I think I was supposed to assume $x \notin A \cup C \implies x \notin A \cap C$
Are both ways correct?
Best Answer
Your first proof checks out! For the contrapositive, $A\cup C \not\subseteq A\cap C$ is the conclusion you have to make, so you do not need to make any assumptions about it. Note you can also show this directly by showing $A \subseteq C$ and $C \subseteq A$, assuming $A \cup C \subseteq A \cap C$.
Let $x \in A$. By the hypothesis, $$x \in A \cup C \subseteq A \cap C \implies x \in A \cap C \implies x \in C.$$ Since $x \in A$ was chosen arbitrarily, you have $A \subseteq C$. Similarly, for any $x \in C$, the same argument holds that $$x \in C \implies x \in A \cup C \implies x \in A \cap C \implies C \subseteq A.$$