8-Complements of a set, Cumulative, associative properties of sets.

Last Updated on September 10, 2026 by Maged kamel

Complements Of A Set, Cumulative, Associative, Distributive Of Sets.

The complement of a set.

This is a new example: if we have a universal set U={1,2,3,4,5,6,7,8,9,10}, and the set A, which is another set, where A={1,2,3,4,5}, what is the new set, which is called A’?A’ represents all the elements that exist in U but are not present in set A if we draw U as a rectangle and set A inside it.

If we draw a circle around A, then A’ is called the complement of A. The complement of a set is the set of elements not present in set A that do exist in U.

The complement of a set.

The union and intersection of sets.

The concepts of union and intersection are well known. The intersection of two sets is the set of all elements that belong to both sets.

For a set called P = {1,2,3,4,5,6,7,8,9,10}.

For another set, called Q, {2,4,6,8,9,10,12,14,16,18,20}.

The intersection of the two sets is the set of common elements: {2, 4, 6, 8, 10}.

P ∩ Q: objects that belong to set P and set Q; the symbol is close to the letter P. The union symbol resembles the letter U.

The definition of a union is the set of all elements that belong to either or both sets. We will list all elements without repetition. P ∪ Q will include 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 16, 18, 20.

Union and intersection of two sets.

How to use the Venn diagram?

Using the Venn diagram, draw each set in a circle and examine the common elements for the intersection. This will help us investigate the relationships in a diagram.

Another example:: if we have P = {3,6,9,12,15,18} and Q = {2,4,6,8,10,12}. If we use the highlighter to identify common elements between the two sets, the values are (6, 12). This can be expressed as P ∩ Q, or the intersection of sets P and Q. If we draw set P as a circle, it contains the elements 3, 6, 9, 12, 15, 18.

We draw the common elements 6, 122)ont the right edge. We draw the Q elements in another circle. What are the remaining elements of P, not included in the intersection?

The elements are 3,9, 15, and 18, while the remaining elements from Q are (2,4,8,10). In other words, 6 is part of the Area enclosed by the intersection of P and Q. The enclosed Area shows 6 and 12.

So 6 belongs to the intersection of P and Q, but 8 does not. It is written in this symbol 8 b∉P ∩ Q, or the intersection of P and Q. While the union relation for the universal set U is drawn as a rectangle.

Inside the rectangle, we can draw two circles for P and Q. All of the elements of both P and Q are thus included. The U will consist of U ={3,6,9,12,15,18,2,4,8,10}.

Using Venn diagram for intersection and union

Solved example#10.

A new example. Let P= {3,9, 27}, Q={2,3,10,18,27,28}. R={2,10,28}. The intersection between P and Q is required, or P∩ R & Q∩ R and P∩ R. What are the common elements between P and Q? These elements are (3,27). For Q∩ R, these elements are (2, 28,10).

The total is three elements. While between P and R, there are none. Then we can express that the intersection between P and R is Ø. PR = Ø. This is called a disjoint set, meaning the sets are not joined.

Solved problem-10 for intersection of given sets.

Cumulative, associative, and distributive properties of a set.

A new subject is the set’s commutative, associative, and distributive properties. The commutative relation can be expressed as A∩ B= B∩A.

Also, A ∪ B = B ∪ A. You can reverse the order of A and B, but remember that the operation is the same on both sides (intersection on the left, union on the right). For the Associative relation, if we have three sets A, B, and C, it is required to make a union between these sets. This can be written as A ∪ (B ∪ C) = (A ∪ B)∪ C. A ∩ (B ∩ C)= (A ∩ B) ∩C.

The distributive property is the mix. The mix between union and intersection. It is called the distributive property: A ∪ (B ∩ C) = (A ∪ B) ∩ C. Another relation is A ∩B ∪ C) = (A ∩ B )∪C. We discussed the commutative law, then the associative law, and finally the distributive law.

What is cumulative and associative?

Set identity law.

There is a new item: the Identity Law for sets. The first relation is A ∪ A =A. The second relation is A ∩A=A. The third relation is A∪Ø =A. This is a true statement. While A ∩Ø=Ø., the subject of the power set is: if we have set A = {1, 2, 3, 4}. What is the subset?

Set identity laws.

There are plenty of combinations of subsets as follows:
The first four numbers are selected, from 1 to 4., And then select {1,2}& {1,3} {1,4} and also select {1},{2},{3},{4}, then & select the same set order {1,2,3,4}, {1,2}& {1,3}&, (1,4}, other collection {1,2,3} {1,3,4}, How many selections so far, we have a total of 10 selections.

Starting with number2: :{2,3},{2,4}. Starting with the number 3:{3,4}, then select Ø. The total number of selections = 5 + 5 + 6 = 16. In the end, if we have a set.
The total number of subsets that can be created = the number of elements raised to the power of 2, or 4×4=16.Make as power 2 is why the Expression of power set is used, since 4^2=16. For instance, if one set has three elements, then the number of subsets is 3^2=9.

Power set of a matrix.

Ordered pairs.

The new item is the ordered pair. For the Cartesian coordinates x and y, we first draw the x-axis, then the y-axis, and then plot the point A at (5, 2).

The ordered pair is the sequence of the axes arrangement. We go right five spaces, then up two spaces. Here, the sequence is essential.

This is called an ordered pair. Pair means two. If we select (5,2), it will give another point, not point A. Set A, with elements {a,b}, and set B, with elements {c, d}, will be considered equal if a=c and b=d.
 

Illustration of Ordered pairs.

You can use ordered pairs to solve equations, as in the example. For a given (x-3,y-2), given=set {4,5}. We need to find the x- and y-values. Since both sides are equal,x- 3 = 44 and y-2=5.

A solved problem for ordered pairs.

Cartesian Product of two sets.

The Cartesian Product of two sets. If we multiply two sets, we write it as A × B.

A new pair is constructed with x,y coordinates, such that x is a part of A and y is a part of B. As an example, if set A is the ordered pair{7,8}, while set B is another, the coordinates are {2,4,6}. We need to find A × B. Following the order of operations, we can write,(7,2),(7,4),(7,6)& (8,2),(8,4),(8,6). This is the result of × A × B. In the end, we have 6 ordered pairs.

Set A is composed of two elements. Set B is composed of three elements. Then AXB = 2 × 3 = 6.

The ordered pair is the sequence of the axes’ arrangement: we go 5 spaces to the right, then 2 spaces up. Here, the sequence is important.

This is called an ordered pair. Pair means two. If we select (5,2), it will give another point, not point A. Set A, with elements {a,b}, and set B, with elements {c, d}, will be considered equal if a=c and b=d.

What is the Cartesian products of two sets?

We can use a table to compute the cross Product A × B by placing the elements of A in the first column and the elements of B in the second, third, and fourth columns; in our example, set B has three elements.

Starting with 2, the first subset is 2 with the second-column value b, then 2 with the third-column value b, and finally 2 with the last-column value b.

Please move to the second Row, where we have 3, then combine it with 2, again with 4, and finally with 2 and 6. The final answer is shown in the next slide image.

A Solved problem for cartesian product.

Another example of the Cartesian Product of Two Sets.

In another example, part of the ordered pair is given, and you are required to find A × B in terms of the given ordered pair and three subsets. Suppose A and B are two sets. The AxB consists of 6 elements, meaning that one set has 3 elements and the other has 2, since 2×3=6.

Given only three elements of the Product A×B, determine the full sets of elements of AxB and A×A. From the given elements, we call these elements (a1,b1),(a2,b2), and (a3,b3).

The first three terms are A’sterms, so A = {2, 3, 4}. Then B should have only two elements. Since A × B = 6, these elements can be written as B={5,7}. B cannot have three elements, as 7 is repeated. AxB ={(2,5)(2,7), (3,5)(3,7), (4,3),(4,7) }, each ordered pair is different from the other pairs.

For the AxA={(2,2)(2,3), 2,4), (3,2)(3,3),( 3,4),(4,2)(4,3),( 4,4) }.

A solved problem for AxB and AxA for given three elements.

Again, a table facilitates estimating the cross Product AxB.

Set of all ordered n-tubles

Use of a table to facilitate the estimation of the cross Product of AxA.

Using table for AxA-Given solved problem.

What are the N-Tuples?

N-tuples, when we have three coordinates x,y, and z. We want to show a point in space, so we need three coordinates.

For the point in space with coordinates (a1, a2, a3), the order is: start with a1 on the x-axis, then a2 in the y-direction, and finally a3 in the z-direction.3 tuples. A vector can represent this from the origin, pointing to the point.

R3 is a linear algebra subject. Rn is in the n-dimensional space. R3 is the set of all ordered triples of real numbers. R1 is a vector in the x-direction that can also be defined as the set of all real numbers in the 1-space.
R2 is a vector in the 2-space (x and y), written as a square of R.
The definition of R2 is the set of all ordered pairs; it differs from the first definition of R1, which is the set of all real numbers.

R2 can be positive or negative, and many choices can represent it. The vector will be in the (x,y) direction. The next slide shows R3. A new term, R4, is presented.

Illustration of N-tuples.

The next image shows a definition of vectors in Rn quoted from Prof. Ron Larson’s handbook.

Source from linear algebra textbook for vectors in Rn.

Definition of a vector space.

R4 is the set of all ordered quadruples of real numbers. Numbers, the point in R4 can be written as (a1, a2, a3, a4).

For Rn, or a vector in space, the point can be represented by (a1,a2,a3,…, an), called an n-tuple. For the quadruples R4, the number of operations that can be performed is 1, as we are going to see from the book of elementary linear algebra.

The first operation is closure under addition; the second is the commutative property u+v=v+u; and the third is u(v+w)=u+(v+w), U+Ø=U((additive identity)). U+(-U)=0, additive inverse.

Cu is in V, where c is a constant. C(u+v)=Cu+Cv, by the distributive property. (C+d)U=cu+du. Distributive property, C(du)=Cdu.1*(u)=u. These are the operations.

Complements of a set, Cumulative, associative of sets.

Solved problem in R4.

Our example for R4.The points are given, and we will compute U + V + W, which yields (0, 4, 6, 2).

Example for quadruple.

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The previous post illustrates the Universal set and subsets of sets.

For an external link, see the story of mathematics for sets and set theory.