Show that A cup B = A cap B implies A = B. | vedclass

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Let $x \in A$. Since $A \subseteq A \cup B$,we have $x \in A \cup B$. Given $A \cup B = A \cap B$,it follows that $x \in A \cap B$. By definition of i

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Let $x \in A$. Since $A \subseteq A \cup B$,we have $x \in A \cup B$.
Given $A \cup B = A \cap B$,it follows that $x \in A \cap B$.
By definition of intersection,$x \in A$ and $x \in B$.
Thus,$x \in B$,which implies $A \subseteq B$.
Similarly,let $y \in B$. Since $B \subseteq A \cup B$,we have $y \in A \cup B$.
Given $A \cup B = A \cap B$,it follows that $y \in A \cap B$.
By definition of intersection,$y \in A$ and $y \in B$.
Thus,$y \in A$,which implies $B \subseteq A$.
Since $A \subseteq B$ and $B \subseteq A$,we conclude $A = B$.

Show that $A \cup B = A \cap B$ implies $A = B$.

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