Question

If $A,B$ are two sets, prove
$A\cup B=(A-B)\cup (B-A)\cup (A\cap B)$
Hence or otherwise prove
$n(A\cup B)=n\left(A\right)+n\left(B\right)-n(A\cap B)$
where $n\left(A\right)$ denotes the number of elements in $A$.

Answer 1

We know that $n\left(A\right)=n(A-B)+n(A\cap B)$
or $n(A-B)=n\left(A\right)-n(A\cap B)$ ..$\left(1\right)$
Also $n(B-A)=n\left(B\right)-n(A\cap B)$ ..$\left(2\right)$
$n(A\cap B)=n(A\cap B)$ ..$\left(3\right)$
Adding $\left(1\right)$, $\left(2\right)$ and $\left(3\right)$, we get
$n(A-B)+n(B-A)+n(A\cap B)$
$=n\left(A\right)+n\left(B\right)-n(A\cap B)$ ..$\left(4\right)$
But $A-B$, $B-A$ and $A\cap B$ are clearly disjoint and hence
$L.H.S.=n(A\cup B)-n(A-B)+n(B-A)+n(A\cap B)$
or $n(A\cup B)=n\left(A\right)+n\left(B\right)-n(A\cap B)$ by $\left(4\right)$
Cor. $n\left(A\Delta B\right)=n\left\{\right(A-B)\cup (B-A\left)\right\}$ by def.
$=n(A-B)+n(B-A)-n\left\{(A-B)\cap (B-A)\right\}$
$=n\left(A\right)-n(A\cap B)+n\left(B\right)-n(B\cap A)-0$
$=n\left(A\right)+n\left(B\right)-2n(A\cap B)$
Order of Finite Sets :
$n\left(A\right)$ means the total number of elements in $A$. They may belong to $B$ also. It does not mean the number of elements which exclusively belong to $A$.
We give below certain results which are obvious by the help of Venn diagram. If $A,B,C$ be different sets and $U$ be the universal set, then
$\left(1\right)n\left(A^{\prime }\right)=n\left(U\right)-n\left(A\right)$
$\left(2\right)n(A\cap B^{\prime })=n(A-B)=n\left(A\right)-n(A\cap B)$
$\left(3\right)n(A\cup B)=n\left(A\right)+n\left(B\right)-n(A\cap B)$
or $n(A\cap B)=n\left(A\right)+n\left(B\right)-n(A\cup B)$
$\left(4\right)$ $n(A\cup B\cup C)=n\left(A\right)+n\left(B\right)+n\left(C\right)-\{n(A\cap B)-n(B\cap C)-n(C\cap A)\}+n(A\cap B\cap C)$
i.e. $S_1-S_2+S_3$
Proof. $n(A\cup B\cup C)=n(A\cup P)$
where $P=B\cup C$
$=n\left(A\right)+n\left(P\right)-n(A\cap P)$. Put for $P$
$=n\left(A\right)+n(B\cup C)-n\{A\cap (B\cup C)\}$
$=n\left(A\right)+n(B\cup C)-n\{(A\cap B)\cup (A\cap C)\}$ Dist. Law
$=n\left(A\right)+n(B\cup C)-n\{L\cup M\}$
$=n\left(A\right)+n\{n\left(B\right)+n\left(C\right)-n(B\cap C)\}-\{n\left(L\right)+n\left(M\right)-n(L\cap M)\}$
Now put for $L$ and $M$ and
$L\cap M=(A\cap B)\cap (A\cap C)=A\cap (B\cap C)$
$\therefore n(A\cup B\cup C)=n\left(A\right)+n\left(B\right)+n\left(C\right)-n(B\cap C)-n(A\cap B)-n(A\cap C)+n(A\cap B\cap C)$
$=S_1-S_2+S_3$
Note. If $A,B,C$ are all disjoint, then
$n(A\cup B\cup C)=n\left(A\right)+n\left(B\right)+n\left(C\right)$ only.
This can be extended to any number of disjoint sets.
$\left(5\right)n(A^{\prime }\cap B^{\prime }\cap C^{\prime })=n(A\cup B\cup C{)}^{\prime }$
$=n\left(U\right)-n(A\cup B\cup C)$
$=n\left(U\right)-\left\{S_1-S_2+S_3\right\}$
Above stands for the number of elements which do not belong to any of the sets $A,B$ and $C$.
$\left(6\right)n(A\cap B^{\prime }\cap C^{\prime })$. This stands for the number of elements which belong to $A$ only i.e. they do not belong to both $B$ and $C$.
$n(A\cap B^{\prime }\cap C^{\prime })=n(A\cap (B\cup C{)}^{\prime })$
$=n\left(A\right)-n\{A\cap (B\cup C)\}$
$=n\left(A\right)-n\{(A\cap B)\cup (A\cap C)\}$
$=n\left(A\right)-n(P\cup Q)$
$=n\left(A\right)-\{n\left(P\right)+n\left(Q\right)-n(P\cap Q)\}$
$=n\left(A\right)-n(A\cap B)-n(A\cap C)+n\{(A\cap B)\cap (A\cap C)\}$
$=n\left(A\right)-n(A\cap B)-n(A\cap C)+n(A\cap B\cap C)$
$\therefore n(A\cap B^{\prime }\cap C^{\prime })=n\left(A\right)-n(A\cap B)-n(A\cap C)+n\{A\cap B\cap C\}$
$\left(7\right)n(A\cap B\cap C^{\prime })$. This stands for number of elements which belong to both $A$ and $B$ but do not belong to $C$
$n(A\cap B\cap C^{\prime })=n(P\cap C^{\prime })$
$=n\left(P\right)-n(p\cap C)$ by $\left(2\right)$
$=n(A\cap B)-n(A\cap B\cap C)$