Solution: {1}, {2}, {1, 2},

Example 2 Find the subset of the set {1, 2}

Solutions: {{ 1}, {2}, {1, 2}, and so on.

Definition: A set without any elements is called an empty set. Representation: using symbols? express

Properties of empty sets:

An empty set is a subset of all sets.

For any set a, an empty set is a subset of a;

A: {}? A

For any set a, the union of empty set and a is a:

A: A ∨ {} = A.

For any set a, the intersection of an empty set and a is an empty set:

Some things don't exist, they are empty sets.

A: A ∩ {} = {}

For any set a, an empty set and the cartesian product of a are empty sets;

A: A × {} = {}

The only subset of an empty set is the empty set itself:

A: A? {} ? A = {}

The number of elements in an empty set (that is, its potential) is zero; In particular, the empty set is limited:

|{}| = 0

In set theory, two sets are equal if they have the same elements; Then only one set can have no elements, that is, the empty set is unique.

Considering that an empty set is a subset of a real number line (or any topological space), it is both an open set and a closed set. The boundary point set of an empty set is an empty set and its subset, so an empty set is a closed set. The interior point set of an empty set is also an empty set and a subset of it, so an empty set is an open set. In addition, an empty set is a compact set because all finite sets are compact.

The closure of an empty set is an empty set.

Noun interpretation

The Concept and Operation of Sets: Lecture 1

Perspective of test sites

1. Understand the concepts of set, subset, complement, intersection and union.

2. Understand the meaning of empty set and complete set.

3. Understand the meaning of ownership, tolerance and equality. Master the related terms and symbols, and use them to represent some simple sets correctly.

4. To solve the set problem, we must first correctly understand the related concepts of the set, especially the three elements of the elements in the set; For the set {x|x∈P} given by the description, we should firmly grasp the representative element x before the vertical line and its property p; We should attach importance to the role of graphic method and solve the problem intuitively through the combination of numbers and shapes.

5. Pay attention to the particularity of empty set. In solving problems, if the set is not an empty set, we should consider the possibility of an empty set, such as A B, there are two possibilities: A= or A≦, and we should discuss them separately at this time.

Case analysis

Question 1. Correctly understand and apply the concept of set.

Understanding the concept of set and correctly applying the properties of set are the key to solve this kind of problem.

Example 1. Given the set m = {y | y = x2+ 1, x ∈ r}, n = {y | y = x+ 1, x ∈ r}, then M∩N= ().

A.(0, 1), (1, 2) B. {(0, 1), (1, 2)} C. {y | y = 1, or y = 2.

Revelation: Sets M and N are represented by descriptive method, and the elements are real numbers Y instead of real number pairs (x, y), so M and N represent the range of functions Y = x2+ 1 (x ∈ r) and Y = x+ 1 (x ∈ r) respectively, and find M.

Solution: m = {y | y = x2+ 1, x ∈ r} = {y | y ≥ 1}, n = {y | y = x+ 1, x ∈ r} = {y |.

∴m∩n = { y | y≥ 1 }∩{ y | y∈r } = { y | y≥ 1 }，∴ D .

Comments: ① Finding M∩N in this question often leads to solving the equation.

So it is wrong to choose B, because in the understanding of the concept of set, we only pay attention to the * * * same attributes of the elements that make up the set, and ignore what the elements of the set are. In fact, the elements of m and n are numbers rather than points, so m and n are number sets rather than point sets. ② A set is composed of elements, so we should pay attention to distinguish {x | y = x2+65438.

Example 2. If P={y|y=x2, x∈R}, q = {y | y = x2+ 1, x ∈ r}, then p ∈ q is equal to ().

I don't know.

Enlightenment: Similar to the above topic, set P is the range set of y=x2(x∈R) and set Q is the range set of y = x2+ 1 (x ∈ r), so the meaning of P∩Q is very clear.

Solution: In fact, the representative elements in P and Q are both Y, which respectively represent the range of functions Y = x2 and Y = x2+ 1. From P={y|y≥0} and Q={y|y≥ 1}, Q P is known, that is, P ∩ Q.

Example 3. If P={y|y=x2, x∈R}, Q={(x, y)|y=x2, x∈R}, there must be ().

A.P∩Q= B.P Q C.P=Q D.P Q

Revelation: Some students come to the conclusion that P=Q as soon as they come into contact with this question. This is because they only see that Y = X2 and X ∈ R in the two sets are the same, but they don't notice that the elements that make up the two sets are different. P set is the set of function values, and Q set is the set of points on Y = X2 and X ∈ R, which means that the element is not a thing at all.

Solution: The correct solution should be: p represents the range of function y=x2, and q represents the point set composed of points on parabola y=x2, so p ∩ q =. You should choose a.

Example 4 If, then = ()

A.{3} B.{ 1} C. D.{- 1}

Thinking enlightenment:

Solution: D.

Comments: To solve this kind of problem, we must first determine the known set.

Question 2. Mutual difference of set elements

The dissimilarity of the elements of a set is an important attribute of the set. Teaching practice tells us that the dissimilarity of set elements is often ignored by students in solving problems, which leads to the failure of solving problems. Here, we will give further examples to strengthen our understanding of the differences between set elements.

Example 5. If A = {2 2,4,3-22-+7}, B = {1,+1, 2-2+2, -(2-3-8), 3+2+3+7}, A ∩.

Revelation: ∫ A ∩ B = {2,5}, ∴ 3-22-+7 = 5, from which we can get =2 or = 1 A = {2, 4, 5}, what elements are there in set B, and do they satisfy the difference of elements?

When = 1, 2-2+2 = 1, which is contrary to the anisotropy of elements, should be omitted = 1.

When =- 1, B={ 1, 0,5,2,4}, which is inconsistent with a ∩ b = {2,5}, so =- 1 is omitted.

When =2, a = {2 2,4,5} and b = {1, 3,2,5,25}, then a ∩ b = {2 2,5} satisfies the problem.

So =2 is what you want.

Example 6. The known sets a = {,+b, +2b}, b = {,c, C2}. If A=B, the value of c is _ _ _ _ _.

Enlightenment: The key to solving the problem of C evaluation is to have the mathematical thought of equation. This problem should be based on the certainty, mutual dissimilarity and disorder of the same elements in two sets.

Solution: Discuss in two situations.

(1) If +b = c and ++2b= c2, if b is eliminated, +c2-2c = 0,

When =0, all three elements in set B are zero, which contradicts the difference of elements, so ≠ 0.

∴ C2-2C+ 1 = 0, that is, c= 1, but when c= 1, the three elements in b are the same, and there is no solution at this time.

(2) If +b = C2 and ++2b= c, if B is eliminated, it is: 2c2-c-= 0.

∫≠0, ∴ 2c2-c- 1 = 0, that is, (c- 1) (2c+ 1) = 0, and c≠ 1, so c =-.

Comments: Solving the equal set problem is easy to produce contradictory additional solutions, which need to be tested and corrected after solving the problem.

Example 7. Given the set a = {x | x2-3x+2 = 0}, b = {x | x2-x+- 1 = 0}, A∪B=A, the value is _ _ _ _.

Revelation: There are four possibilities to deduce B from A∪B=A, and then it is the value.

Solution: ∫A∪B = A,

∫a = {1, 2}, ∴ B= or B={ 1} or B={ 1, 2}.

If B=, then △

If B={ 1}, let △=0 get =2, then 1 is the root of the equation;

If B={2}, let △=0 get =2, where 2 is not the root of the equation.

If b = {1, 2}, then △ > 0 is ∈R and ≠2. Substitute x= 1 into ∈R of the equation and x=2 into = 3 of the equation.

The total value is 2 or 3.

Comments: In this question, you can't write B={ 1,-1} directly, because-1 may be equal to 1, which contradicts the set elements. In addition, consider that set B may be an empty set or a single element set.

Question 3. Pay attention to the method of proving and judging the relationship between the two groups.

The relationship between sets is a problem that we often encounter and must solve in the process of solving mathematical problems, so we should pay attention to it. The relationship between elements and sets defines a series of concepts that reflect the relationship between sets. So when proving (judging) the relationship between two sets, we should return to the relationship between elements and sets.

Example 8. Set A = {| = 3n+2, n ∈ z} and set B = {b | b = 3k- 1, k ∈ z}, then the relationship between sets a and b is _ _ _ _ _ _ _ _ _.

Solution: Let ∈A be = 3n+2 = 3 (n+1)-1(n ∈ z),

∴ n ∈ z, ∴ n+ 1 ∈ z, ∴∈ b, therefore. ①

Let b∈B, then b = 3k-1= 3 (k-1)+2 (k ∈ z),

∵ K ∈ Z, ∴ K- 1 ∈ Z. ∴ B ∈ A, so (2)

From ① and ②, we know that A = B.

Comments: It is explained here that in the process of ∈B or ∈ A, the key is to change (or make up for) the form before reasoning.

Example 9 If A, B and C are three sets, then there must be ().

Answer. The fourth century BC.

【 Objective 】 This topic mainly investigates the operation of the relationship between sets.

Solution: I know, so I choose A.

Example 10. Set a set, then the number of set b is ().

Answer. 1

【 Objective 】 This topic examines the union operation of sets and the number of subsets, and also examines the idea of equivalent transformation.

Solution: then set b must contain element 3, that is, this problem can be transformed into the problem of finding the number of subsets of the set, so there is a set B*** that satisfies the problem condition. So I chose C.

Example 1 1. Remember that the solution set of inequality is, and the solution set of inequality is.

(i) If so, ask;

(II) If yes, find the range of positive numbers.

Revelation: first solve the inequality and find the sum of sets.

Solution: (1) from, from.

㈡。

You, you, you, so,

That is, the range of values is.

Question 4. Pay attention to the particularity and special function of empty set.

An empty set is a special and important set, which contains no elements, is a subset of any set, and is the proper subset of any non-empty set. Obviously, the intersection of an empty set and any set is an empty set, and the union with any set is still equal to this set. When there is a set relationship in which empty sets participate, its particularity is easily overlooked, which leads to mistakes in solving problems.

Example 12. If a = {x | x2-3x+2 = 0}, b = {x | x-2 = 0}, and A∪B=A, then the set c composed of real numbers is _ _ _ _ _.

Solution: x= 1 or x2-3x+2 = 0 2. When x= 1, =2, when x=2, = 1.

This result is incomplete. The above solution only pays attention to the fact that b is a non-empty set. In fact, when B=, it still satisfies A∪B=A, and when =0, B=, it conforms to the topic and should be supplemented, so the correct answer is c = {0, 1, 2}.

Example 13. Known set,. If, the range of real numbers is.

Revelation: first determine the known sets a and B.

Solution:

So the range of real numbers is.

Example 14. Given the set a = {x | x2+(m+2) x+ 1 = 0, x ∈ r}, if A∩ =, the range of real number m is _ _ _ _ _.

Revelation: From the point of view of equation, set A is the solution set of real coefficient quadratic equation x2+(m+2) x+ 1 = 0 about X, and x=0 is not the solution of the equation, so we can know from A∩ = that the equation has only two negative roots or no real roots, so we can transform the discriminant into inequality about m and find the range of m respectively.

Solution: Equation X2+(m+2) x+ 1 = 0 has no zero roots, so the equation has only two negative roots or no real roots.

Or δ=(m+2)2-4.

Comments: The error that easily occurs in this question is that the equation has only two negative roots from A∩ = (because the product of these two roots is 1, because the equation has no zero roots), and the item A= is omitted, which requires a comprehensive and accurate understanding and understanding of assembly language.

Example 15. It is known that the set A = {x | x2-3x- 10 ≤ 0} and the set B = {x | p+65438 ≤ x ≤ 2p- 1}. If B A, the range of real number p is _ _ _ _.

Solution: -2 ≤ x ≤ 5 from x2-3x- 10 ≤ 0.

To make b a, the range of ∴ p is -3 ≤ p ≤ 3.

The above solution ignores the conclusion that an empty set is a subset of any set, that is, when B=, the problem is satisfied.

Should have: ① when B≦, that is, P+ 1 ≤ 2p- 1p ≥ 2.

From B A:-2 ≤ P+ 1 and 2p- 1 ≤ 5. From-3 ≤ P ≤ 3。 ∴ 2 ≤ P ≤ 3。

② when B=, that is, P+ 1 >; 2p- 1 p 0}, find A∪B and a ∩ b.

Solution: ∫ A = {x | x2-5x-6 ≤ 0} = {x |-6 ≤ x ≤1},

b = { x | x2+3x & gt； 0} = {x | x <-3 or x & gt0}. As shown in the figure,

∴a∪b = { x |-6≤x≤ 1 }∩{ x | x & lt； -3 or x>0} = r.

a∩B = { x |-6≤x≤ 1 }∩{ x | x & lt； -3 or x>0} = {x |-6 ≤ X.

Comments: This question is represented by the number axis. According to the range represented by the number axis, the result of the question can be written intuitively and accurately.

Example 18. Let a = {x |-2.

Revelation: You can draw a graph on the number axis for analysis and solution.

Solution: As shown in the figure, assume that the range represented by set B moves on the number axis.

Obviously, if and only if b covers the set {x |- 1

According to the relationship between quadratic inequality and quadratic equation, we can know that-1 and 3 are the two roots of equation x2+x+b = 0.

∴ =-(- 1+3)=-2，b=(- 1)×3=-3。

Comments: Similar to the multiple set problem in this question, if the interval graphic representation on the number axis is used, and the combination of numbers and shapes is adopted, the intuitive and clear problem-solving effect will be obtained.

Special training

First, multiple-choice questions:

1. Let M={x|x2+x+2=0} and =lg(lg 10), then the relationship between {} and m is ().

a 、{ }=M B、M { } C 、{ } M D、M { }

2. Given the complete works =R, a = {x | x-| < 2}, B={x|x- 1|≥3}, and A∩B=, the value range of is ().

a 、[0，2] B 、(-2，2) C 、( 0，2)D 、( 0，2)

3. Given the set M={x|x= 2-3 +2, ∈R}, N={x|x=b2-b, b∈R}, the relationship between m and n is ().

A, M N B, M N C, M=N D, uncertain

4. Let the set A={x|x∈Z and-/kloc-0 ≤ x ≤-1}, and B={x|x∈Z and |x|≤5}, then the number of elements in A∪B is ().

a、 1 1 B、 10 C、 16 D、 15

5. The subset of the set m = {1, 2,3,4,5} is ().

a、 15 B、 16 C、3 1 D、32

6 let M={x|x=, k∈Z}, N={x|x=, k∈Z}, then ()

A M=N B M N C M N D M∩N=

7. Given the set a = {x | x2-4mx+2m+6 = 0, x ∈ r}, if a∩r-≦, find the range of the number m. 。

8. Proposition A: Equation X2+MX+ 1 = 0 has two different negative roots; Proposition B: Equation 4x2+4 (m-2) x+ 1 = 0 has no real roots. Only one of these two propositions holds. Find the range of m.

9 known set a = {x |-2 ≤ x ≤ 7}, b = {x | m+ 1

a-3≤m≤4 B-3 & lt； m & lt4 C2 & lt； m & lt4d 2 & lt； m≤4

10. Let M=, and. Then the range of the real number a is ().

A.a- 1 b . a 1 c . a 1d . a 1

1 1. The number of all sets m that satisfy {,b} m = {,b, c, d} is ().

A.7 B. 6 C. 5 D. 4

12. If the proposition P: x ab, then p is ().

A.xbb.xba or xbc.xba and xbd.xba.

13. The known set m = {,}. p = {-，2- 1 }； If card (M P)=3, then M P= ()

A.{- 1 } b . { 1 } c . { 0 } d . { 3 }

14. Let the set p = {3 3,4,5}. Q = {4，5，6，7}。 Let P*Q=, then the number of elements in P*Q is ().

A.3 B. 7 C. 10 D. 12

2. Fill in the blanks:

15. If M={} and N={x|, then m ∩ n = _ _ _ _ _ _ _

16. the nonempty set p satisfies the following two conditions: (1)p {1, 2,3,4,5}, (2) if the element ∈p is 6- ∈p, then the number of sets p is _ _ _ _ _.

17. Let A = {1, 2} and B = {x | x A} If expressed by enumeration, the set b is.

18. A set containing three real numbers can be expressed as, then.

Three. Answer the question:

19. Let the set A={(x, y)|y= x+ 1} and B={(x, y)|y=|x|}. If A∩B is a single element set, the range of evaluation.

20. let A = {x | x2+px+q = 0 }≦, m = {1, 3,5,7,9}, n = {1, 4,7, 10}, if a \.

2 1. Given the set M={y|y=x2+ 1, x∈R}, N={y|y=x+ 1, x∈R}, find m ∩ n.

22. the known set A={x|x2-3x+2=0}, B={x|x2-mx+2=0}, A∩B=B, and the realistic number m range.

23. The complete set =R is known and found.

24. Known sets,

And,, find the value of B.

Reference answer

1.C 2。 A 3。 C 4 explosive C 5。 D

6.c analytic pair M divides K into two categories: k=2n or k = 2n+ 1 (n ∈ z), m = {x | x = nπ+, n ∈ z} ∨{ x | x = nπ+, n ∈ z.

For n, k is divided into four categories, k=4n or k=4n+ 1, k=4n+2 and k=4n+3(n∈Z).

N={x|x=nπ+，n∈Z} ∨{ x | x = nπ+，n∈Z } ∨{ x | x = nπ+π，n∈Z } ∨{ x | x = nπ+，n∈Z }

7. Solution: Let the complete set = {m |△ = (-4m) 2-4 (2m+6) ≥ 0} = {m | m ≤-1or m ≥}

If the two roots of the equation x2-4mx+2m+6 = 0 are x 1 and x2 is not negative,

So {m | m ≤- 1} about the complement set {m|m≥} is the demand.

8. Solution: Proposition A is established under the following conditions:

∴ Set a = {m | m & gt2}.

The condition of proposition B is: △ 2 = 16 (m-2) 2- 16.

If there is one and only one proposition A and B, then there is:

( 1)m∈A∩CRB，(2)m∈CRA∩B

If it is (1), it is: a ∩ CRB = {m | m > 2}∩{m|m≤ 1 or m ≥ 3} = {m | m ≥ 3};

If it is (2), there is: b ∩ CRA = {m | 1

Combining (1) and (2), we can know that the range of m is {m | 1 < M≤2, or m ≥ 3}.

9.d When ∵A∪B=A, ∴B A and b≦.

∴, that is, 2 < m ≤ 4

10.C 1 1。 D 12。 B 13。 D 14。 B

2. Fill in the blanks:

15.； 16.7 ; 17.； 18.- 1.

Three. Answer the question:

19.≥ 1 or ≤- 1, prompt: draw.

20.or

2 1. solution: before the set operation, we must first identify the set, that is, identify the characteristics of the elements in the set. M and n are both sets, which can't be mistaken for point sets, thus solving the equation. Secondly, we should simplify the set or clarify its characteristics. m = {y | y = x2+ 1，x ∈R}={y|y≥ 1}，N={y|y=x+ 1，x∈。

22. solution: A = {1, 2}, A ∩ B = BBA of simplified conditions.

Depending on the number of elements in the set, B=, B={ 1} or {2}, B={ 1, 2}.

When B=, δ= m2-8.

When B={ 1} or {2}, m has no solution.

When b = {1, 2}, ∴ m = 3.

To sum up, m=3 or.

24. Answer: Elements in. ∴ It must be the element of B.

The elements in ∵ and ∴ belong to B,

Therefore.

And ... ∴- 1,4 is the two roots of the equation, ∴a=-3,b=-4.