operations on sets (set theory) and 4 solved problems

let X = { 1 , 2 , 3 ,4 } and Y = { 3 , 4 , 5 , 6 , 7 , 8 } then find  X ∪ Y , X ∩ Y , X - Y , Y - X 

1 - (X ∪ Y ) union

∪ this symbol is called union means all elements in X and all elements in Y and  no need to repeat the elements

then  X ∪ Y  = { all elements in X and all elements in Y  don't repeat the elements  } ={1, 2, 3 ,4 ,5,6, 7 ,8 }

2- (X ∩ Y) intersection

 ∩ this symbol is called intersection means the common elements between X & Y 

then   X ∩ Y = {  the common elements between X & Y}= { 3 , 4 }
If the intersection of two sets A and B is empty, that is they have no elements in common, then they are said to be disjoint, denoted: A ∩ B = ∅. For example the sets {1, 2} and {3, 4} are disjoint, written
{1, 2} ∩ {3, 4} = ∅.

3 -  (X - Y )difference

 - means elements in X but  not in Y (notes that X is first )

then X - Y = {elements in X but  not in Y }= {1 , 2}

4-( Y - X ) difference

- elements in Y but  not in X (note that Y is first)

then  Y - X  = {elements in Y but  not in X}={5, 6 ,7 ,8}

-----------------------------------------------------------------

let A & B are to sets then A is a subset of  B (A ⊂ B) if all elements of A are found in B

ex : A = { 5 , 8 , 9 } &  B = { 4 , 6 ,8 , 9 , 10 , 11 }

then A⊂B because all elements of A (5, 8 ,9 ) are found in B
if not all elements of A not found in B then  A is not a subset of B (A B)

-----------------------------------------------------------------

Other basic concepts 

• universal set (U) : is a set which contains all objects such as if A = { 5 , 8 , 9 } then U = N (natural number set)
• Complement of set A(A^c)   : it is the set of all number in (U) not in (A)
• Symmetric difference of sets A and B is set of elements in A not in B and elements in B not in A For instance, for the sets {1,2,3} and {2,3,4} , the symmetric difference set is {1,4}
• Power set of a set A is the set whose members are all possible subsets of A. For example, the power set of {1, 2} is { {}, {1}, {2}, {1,2} } .

some solved problems :

1) zero ∈ { X - 3  ,   4 } therefore X = ....................................

because zero ∈ { X - 3  ,   4 } then X should equal 4 or X - 3 but zero ≠ 4 then zero should 

equal X - 3 

then X - 3 = zero then X = 3 

2) { 3 , X + 1} ⊂ {4 , 3 } therefore X = ...................................

BECAUSE 3 = 3 THEN  X + 1 = 4  then X = 3

3) { 2,1}∪ { 3 , X - 1 } = {1,2,3,4}THEN X = ...................................

1 = 1 , 2=2 and 3 =3 then X  -1 = 4 then X = 5 

4)  {2X , 8 }∩ {5,12}= {12}then X = ....................................

the common element is 12 then 12 should be in the two sets then12 = 2X because 8 ≠ 12 then 

X = 6