difference sets

There are two sets whose elements are overlapped partially. For example,
$A=\left\{ 1,2,3,4,5,6  \right\}$
$B=\left\{ 3,4,5,6,7,8,9  \right\}$
In preceding post as we have studied,
$A\cup B=\left\{ 1,2,3,4,5,6,7,8,9  \right\}$ .
In this case overlapped elements $\left\{ 3,4,5,6  \right\}$  become one element.

If we want to clean the overlapped elements, we use the difference set.
$A-B=A\backslash B=\left\{ 1,2  \right\}$ 
In this case the elements of $B$  $\left\{ 7,8,9 \right\}$  have no contributions to the calculation.
Note that $A-B\ne \left\{ 1,2,-7,-8,-9 \right\}$ 
Of course, $B-A=\left\{ 7,8,9 \right\}$ . We will ignore the elements $\left\{ 1,2 \right\}$ .
Hence $A\cup B=(A-B)\cup B=A\cup (B-A)$ .

These are the differences of the calculations of sets with those of real numbers.

equal sets

A set is decided by the own elements or members. Therefore, if every elements of two sets is according, two sets is equivalent. We shall some examples. The sets are as follows.

$A=\left\{ 1,2,3 \right\}$
$B=\left\{ 4,5,6 \right\}$
$C=\left\{ 1,2,3,4,5,6 \right\}$
$D=\left\{ 1,3,5 \right\}$
$E=\left\{ 2,4,6 \right\}$
$F=\left\{ 1,3 \right\}$
$G=\left\{ 2,4 \right\}$
$H=\left\{ 3,5 \right\}$
$I=\left\{ 6 \right\}$

Then, all below is true.

$A\cup B=C$ , $D\cup E=C$ , $F\cup G\cup H\cup I=C$ , 

$A\cup C=C$ , $B\cup C=C$ , $D\cup C=C$ , $E\cup C=C$ ,
$F\cup C=C$ , $G\cup C=C$ , $H\cup C=C$ , $I\cup C=C$ ,

$A\cap D=F$ , $B\cap E=G$ , $A\cap E=A\cap G$ , $B\cap D=B\cap H$ ,
$A\cup G=F\cup G$ , $E\cap H=G\cup H\cup I$ ,

Please note that, in elementary, overlapped elements should be deleted. i.e,
$A\cup F=\left\{ 1,1,2,3,3 \right\}=\left\{ 1,2,3 \right\}$

notations of sets

A set is a collection of objects of any kind. The most important thing is that elements (i.e. objects themselves ) of a set have been defined mathematically with no doubt.

(a) a set $A$ of integer numbers, whose elements are very great numbers.
(b) a set $A$ of integer numbers, whose elements are greater than $10$ .

You must have understood (a) is a insufficient definition of  a set. Because with (a) we are not able to decide whether the number $1,000,000$  should be in $A$ . (please recall Archimedean properties)

A set $A$  with the definition (b) is expressed by two ways, as are well known.

(1) $\left\{ 11,12,13,\cdots  \right\}$
(2) $\left\{ x | x>10, x\in\mathbb{Z}  \right\}$

(1) is said to be a extensional definition and (2) is connotative. For a finite set, if we adopt a extensional definition, we have to write all elements in principle. However, for a set whose elements are much more or infinite, as it is impossible to do, we can use  "$\cdots$ ". In that case, we have to leave no ambiguities for "$ \cdots $ " .

In a connotative definition $\left\{x | C(x) \right\}$ , $x$  is a variable number satisfied the conditions $C(x)$ . Of course, $\left\{x | C(y) \right\}$ is unacceptable.