Tag: introduction to set

Assume $ A \subset B$ is true. 

Then, every element of $A$ i.e. $a _1,a _2, ... , a _n$ in A are also in B.

So, number of elements in $B$ will always be greater than no. of elements in $A$

And $P(A)$ will contain less number of subsets than $P(B)$

Hence, $n[P(A)] <  n[P(B)]$

$\displaystyle n\left( A\quad \cup \quad B \right) =n\left( A \right) +n\left( B \right) -n\left( A\quad \cap \quad B \right) $

If, $n(A)=n(B)$

a.  $n(A-B) = n(B-A)$. As, the no. of elements are same, if we subtract A from B or B from A, we will get the same no. of elements

b.  $n(AB)\neq n(A)+n(B)$. It is $n(A\cup B)=n(A)+n(B)$

c.  $n(A-B)= \phi$

d.  $n(AB)\neq n(B) - n(A-B)$. It is $n(AB) = n^2$ where n is the no. of elements of these sets

$$n\left( A \right) =5$ 

If there is a finite number of n elements in $A,$ then the power set $P (A)$ has $2^n$ elements. 

$n(A\cup B)=n(A)+n(B)-n(A\cap B)$
$=200+300-100$
$=400$
$n(A'\cap B')=n(A\cup B)'$
                    $=n(U) - n(A\cup B)$
$300=n(U)-400$
$n(U)=700$

Using De Morgan's law,
$\displaystyle n(A'\cup B')= n(A\cap  B)' $ $\displaystyle

= n(U)-n(A\cap B)=  n(U)-100 $ $\displaystyle = 700-100= 600$