Question:
Show that if $ \ A \cup B = A$ and $ \ A\cap B = A$ then $ \ A = B$
My attempt:
Proof by contradiction:
Assume $ \ A \cup B = A$ and $ \ A\cap B = A$ and $ \ A \neq B$
Case 1: $ \exists \ x \in A, x\notin B$
If $ x \in A \implies x\in A \cap B \implies x \in A \ and \ x\in B \implies x\in B$, a contradiction.
Case 2: $ \exists \ x \in B, x\notin A$
If $ \ x \in B \implies x\in A \cup B \implies x\in A$, since $ \ A \cup B = A$. Contradiction.
Is this approach correct? Could someone please show me how to do a direct proof?
how to do a direct proof
You could use this answer to Let $A$, $B$ and $C$ be sets. If $A \cup B = A \cup C$ and $A \cap B = A \cap C$, then show that $B = C$ for the case $\,C=A\,$. $\endgroup$