If A and B are any two non-empty sets and A i | vedclass

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(B) Given that $A$ is a proper subset of $B$,denoted as $A \subset B$ and $A 
eq B$.Since $A \subset B$,the set $A - B = \emptyset$,which implies $n(

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$1$

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(B) Given that $A$ is a proper subset of $B$,denoted as $A \subset B$ and $A 
eq B$.Since $A \subset B$,the set $A - B = \emptyset$,which implies $n(A - B) = 0$.By definition,the symmetric difference is $A \Delta B = (A - B) \cup (B - A)$.Thus,$n(A \Delta B) = n(A - B) + n(B - A) = 0 + n(B - A) = n(B - A)$.Since $A$ is a proper subset of $B$,there must be at least one element in $B$ that is not in $A$. Therefore,$n(B - A) \geq 1$.To minimize $n(A \Delta B)$,we choose the smallest possible value for $n(B - A)$,which is $1$.Hence,the minimum value of $n(A \Delta B)$ is $1$.

If $A$ and $B$ are any two non-empty sets and $A$ is a proper subset of $B$. If $n(A) = 4$,then the minimum possible value of $n(A \Delta B)$ is (where $\Delta$ denotes the symmetric difference of set $A$ and set $B$).

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