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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?

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Your proof is correct.

Also,

$$A \cup B = A \implies B \subseteq A$$

$$A \cap B = A \implies A \subseteq B$$

Hence $A=B$.

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Your proof is correct. Judging from your way of presentation, it would be better to call it a contraposition proof.

A direct proof is as follows. Note that $A = B$ iff $A \subset B$ and $B \subset A$. Note that $A = (A \cap B) \cup (A \cap B^{c})$ with the two sets on the right-hand side disjoint. So $A = A \cap B$ by assumption iff $A \cap B^{c} = \varnothing$, and iff $A \subset B$. Note that $A \cup B \supset A, B$ by definition. So $A \cup B = A$ by assumption iff $A \supset B$.

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$B=(A\cup B)\cap B = A\cap B=A$ since $B\subseteq A\cup B$ and $A\cup B=A=A\cap B$.

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This proof seems overly complicated, I propose $$ A=A\cap B\subset X\subset A\cup B=A\mbox{ for } X=A,B $$ then $A=A\cap B=A\cup B=B$.

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$\boxed{A\subset B}$

$x\in A \underbrace{\implies}_{A\cap B=A} x\in A\cap B \implies \left\lbrace \begin{array}lx\in A\\x\in B\implies A\subset B\end{array}\right.$

$\boxed{B\subset A}$

$x\in B\implies x\in A\cup B\underbrace{\implies}_{A\cup B=A} x\in A\implies B\subset A$

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$$A = A \cup B = A \cup (A^C \cap B) = (A \cap B) \cup (A^C \cap B) =B$$

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Part 1: claim $A\subseteq B$.

Let $x\in A$. Claim $x\in B$.

Since $A\cap B=A$, $x\in A\cap B$. By the definition of intersection of two sets, we know that $x\in A$ and $x\in B$. Hence we have shown that $x\in B$. Since $x$ is arbitrary, for any $x$, if $x\in A$, then $x\in B$. Therefore, $A\subseteq B$.

Part 2: claim $B\subseteq A$.

Let $x\in B$. Claim $x\in A$.

$x\in B$ implies that $x\in A$ or $x\in B$. By the definition of union of two sets, we know that $x\in A\cup B$. Since $A\cup B=A$, we know that $x\in A$. Since $x$ is arbitrary, for any $x$, if $x\in B$, then $x\in A$. Therefore, $B\subseteq A$.

Combine the two parts, we know that $A=B$.

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