Some solution mathematical problems involves finding intersection and union number sets. In the article below we will consider these actions in detail, including specific examples. The acquired skill will be applicable to solving inequalities with one variable and systems of inequalities.

The simplest cases

When we talk about the simplest cases in the topic under consideration, we mean finding the intersection and union of numerical sets, which are a set of individual numbers. In such cases, it will be sufficient to use the definition of intersection and union of sets.

Definition 1

Union of two sets is a set in which each element is an element of one of the original sets.

Intersection of many is a set that consists of all common elements original sets.

From these definitions the following rules logically follow:

To form a union of two numerical sets with a finite number of elements, it is necessary to write down all the elements of one set and add to them the missing elements from the second set;

To create the intersection of two numerical sets, it is necessary to check the elements of the first set one by one to see if they belong to the second set. Those of them that turn out to belong to both sets will constitute the intersection.

The set obtained according to the first rule will include all elements belonging to at least one of the original sets, i.e. will become the union of these sets by definition.

The set obtained according to the second rule will include all the common elements of the original sets, i.e. will become the intersection of the original sets.

Let's consider the application of the resulting rules using practical examples.

Example 1

Initial data: numerical sets A = (3, 5, 7, 12) and B = (2, 5, 8, 11, 12, 13). It is necessary to find the union and intersection of the original sets.

Solution

1. Let us define the union of the original sets. Let's write down all the elements, for example, of set A: 3, 5, 7, 12. Let's add to them the missing elements of set B: 2, 8, 11 and 13. Ultimately, we have a numerical set: (3, 5, 7, 12, 2, 8, 11, 13). Let's order the elements of the resulting set and get the desired union: A ∪ B = (2, 3, 5, 7, 8, 11, 12, 13).
2. Let us define the intersection of the original sets. According to the rule, we will go through all the elements of the first set A one by one and check whether they are included in the set B. Let's consider the first element - the number 3: it does not belong to the set B, which means it will not be an element of the desired intersection. Let's check the second element of set A, i.e. number 5: it belongs to the set B, which means it will become the first element of the desired intersection. The third element of set A is the number 7. It is not an element of the set B, and, therefore, is not an element of intersection. Consider the last element of set A: the number 1. It also belongs to the set B, and accordingly will become one of the intersection elements. Thus, the intersection of the original sets is a set consisting of two elements: 5 and 12, i.e. A ∩ B = (5, 12).

Answer: union of the original sets – A ∪ B = (2, 3, 5, 7, 8, 11, 12, 13); intersection of the original sets - A ∩ B = (5, 12).

All of the above applies to working with two sets. As for finding the intersection and union of three or more sets, the solution to this problem can be reduced to sequentially finding the intersection and union of two sets. For example, to determine the intersection of three sets A, B, and C, it is possible to first determine the intersection of A and B, and then find the intersection of the resulting result with the set C. Using an example, it looks like this: let the numerical sets be given: A = (3, 9, 4, 3, 5, 21), B = (2, 7, 9, 21) and C = (7, 9, 1, 3 ) . The intersection of the first two sets will be: A ∩ B = (9, 21), and the intersection of the resulting set with the set A ∩ B = (9, 21). As a result: A ∩ B ∩ C = ( 9 ) .

However, in practice, in order to find the union and intersection of three or more simple numerical sets that consist of a finite number of individual numbers, it is more convenient to apply rules similar to those indicated above.

That is, to find a union of three or more sets of the specified type, it is necessary to add the missing elements of the second set to the elements of the first set, then the third, etc. For clarification, let's take numerical sets: A = (1, 2), B = (2, 3), C = (1, 3, 4, 5). The number 3 from set B will be added to the elements of the first set A, and then the missing numbers 4 and 5 from set C. Thus, the union of the original sets: A ∪ B ∪ C = (1, 2, 3, 4, 5).

As for solving the problem of finding the intersection of three or more numerical sets that consist of a finite number of individual numbers, it is necessary to go through the numbers of the first set one by one and step by step check whether the number in question belongs to each of the remaining sets. For clarification, consider number sets:

A = (3, 1, 7, 12, 5, 2) B = (1, 0, 2, 12) C = (7, 11, 2, 1, 6) D = (1, 7, 15, 8, 2, 6).

Let's find the intersection of the original sets. Obviously, set B has the fewest elements, so these are the ones we will check to determine whether they are included in the remaining sets. Number 1 of set B is an element of other sets, and therefore is the first element of the desired intersection. The second number of set B - number 0 - is not an element of set A, and, therefore, will not become an element of intersection. We continue checking: number 2 of set B is an element of other sets and becomes another part of the intersection. Finally, the last element of set B - the number 12 - is not an element of set D and is not an element of intersection. Thus, we get: A ∩ B ∩ C ∩ D = ( 1 , 2 ) .

The coordinate line and number intervals as a union of their parts

Let's mark an arbitrary point on the coordinate line, for example, with coordinates - 5, 4. Specified point will divide the coordinate line into two numerical intervals - two open rays (-∞, -5,4) and (-5,4, +∞) and the point itself. It is easy to see that, in accordance with the definition of a union of sets, any real number will belong to the union (- ∞, - 5, 4) ∪ (- 5, 4) ∪ (- 5, 4, + ∞). Those. the set of all real numbers R = (- ∞ ; + ∞) can be represented in the form of the union obtained above. Conversely, the resulting union will be the set of all real numbers.

Note that it is possible to attach a given point to any of the open rays, then it will become simple numerical beam(- ∞ , - 5 , 4 ] or [ - 5 , 4 , + ∞) . In this case, the set R will be described by the following unions: (- ∞ , - 5 , 4 ] ∪ (- 5 , 4 , + ∞) or (- ∞ , - 5 , 4) ∪ [ - 5 , 4 , + ∞). .

Similar reasoning is valid not only with respect to a point on a coordinate line, but also with respect to a point on any numerical interval. That is, if we take any internal point of any arbitrary interval, it can be represented as a union of its parts obtained after division given point, and the point itself. For example, a half-interval (7, 32] and a point 13 belonging to this numerical interval are given. Then the given half-interval can be represented as a union (7, 13) ∪ (13) ∪ (13, 32] and vice versa. We can include the number 13 in any of the intervals and then the given set (7, 32 ] can be represented as (7, 13 ] ∪ (13, 32 ] or (7, 13 ] ∪ (13, 32 ]. We can also take not the internal point of a given half-interval, and its end (the point with coordinate 32), then the given half-interval can be represented as the union of the interval (7, 32) and a set of one element (32) Thus: (7, 32] = (7, . 32) ∪ ( 32 ) .

Another option: when not one, but several points are taken on a coordinate line or a numerical interval. These points will divide the coordinate line or numerical interval into several numerical intervals, and the union of these intervals will form the original sets. For example, points on the coordinate line are given with coordinates - 6, 0, 8, which will divide it into intervals: (- ∞, - 6), (- 6, 0), (0, 8), (8, + ∞) . In this case, the set of all real numbers, the embodiment of which is the coordinate line, can be represented as a combination of the resulting intervals and the indicated numbers:

(- ∞ , - 6) ∪ { - 6 } ∪ (- 6 , 0) ∪ { 0 } ∪ (0 , 8) ∪ { 8 } ∪ (8 , + ∞) .

The topic of finding the intersection and union of sets can be clearly understood if you use images of given sets on a coordinate line (unless we are talking about the simplest cases discussed at the very beginning of the article).

We will look at a general approach that allows us to determine the result of the intersection and union of two number sets. Let us describe the approach in the form of an algorithm. We will consider its steps gradually, each time citing the next stage of solving a specific example.

Example 2

Initial data: given numerical sets A = (7, + ∞) and B = [ - 3, + ∞). It is necessary to find the intersection and union of these sets.

Solution

1. Let us depict the given numerical sets on coordinate lines. They need to be placed one above the other. For convenience, it is generally accepted that the origin points of the given sets coincide, and the location of the points relative to each other remains preserved: any point with a larger coordinate lies to the right of the point with a smaller coordinate. Moreover, if we are interested in the union of sets, then the coordinate lines are combined on the left by the square bracket of the set; if you are interested in intersection, then use the curly brace of the system.

In our example, to write the intersection and union of numerical sets we have: and

Let's draw another coordinate line, placing it under the existing ones. It will be needed to display the desired intersection or union. On this coordinate line, all the boundary points of the original numerical sets are marked: first with dashes, and later, after clarifying the nature of the points with these coordinates, the dashes will be replaced by punctured or non-punctured points. In our example, these are points with coordinates - 3 and 7.

And

The points that are depicted on the lower coordinate line in the previous step of the algorithm make it possible to consider the coordinate line as a set of numerical intervals and points (we talked about this above). In our example, we represent the coordinate line as a set of five numerical sets: (- ∞, - 3), (- 3), (- 3, 7), (7), (7, + ∞).

Now you need to check one by one whether each of the written sets belongs to the desired intersection or union. The resulting conclusions are marked in stages on the lower coordinate line: when the gap is part of an intersection or union, a hatch is drawn above it. When a point enters an intersection or union, the stroke is replaced by a solid point; if the point is not part of the intersection or union, it is punctured. In these actions you must adhere to the following rules:

A gap becomes part of the intersection if it is simultaneously part of set A and set B (or in other words, if there is shading above this gap on both coordinate lines representing sets A and B);

A point becomes part of the intersection if it is simultaneously part of each of the sets A and B (in other words, if the point is a non-punctured or internal point of any interval of both numerical sets A and B);

A gap becomes part of a union if it is part of at least one of the sets A or B (in other words, if there is shading over this gap on at least one of the coordinate lines representing the sets A and B.

A point becomes part of a union if it is part of at least one of the sets A and B (in other words, the point is a non-punctured or interior point of any interval of at least one of the sets A and B).

Briefly summarizing: the intersection of numerical sets A and B is the intersection of all numerical intervals of sets A and B, over which shading is simultaneously present, and all individual points belonging to both set A and set B. The union of numerical sets A and B is the union of all numerical intervals , over which at least one of the sets A or B has shading, as well as all unpunctured individual points.

1. Let's go back to the example and define the intersection of given sets. To do this, let's check the sets one by one: (- ∞ , - 3) , ( - 3 ) , (- 3 , 7) , ( 7 ) , (7 , + ∞) . Let's start with the set (- ∞, - 3), clearly highlighting it in the drawing:

This gap will not be included in the intersection because it is not part of either set A or set B (no shading). And so our drawing retains its original appearance:

Consider the following set (-3). The number - 3 is part of set B (not a punctured point), but is not part of set A, and therefore will not become part of the desired intersection. Accordingly, on the lower coordinate line we make a point with coordinate - 3:

We evaluate the following set (- 3, 7).

It is part of set B (there is shading above the interval), but is not included in set A (there is no shading above the interval): it will not be included in the desired intersection, which means that no new marks appear on the lower coordinate line:

The next set to check is (7). It is part of the set B (the point with coordinate 7 is an internal point of the interval [ - 3, + ∞)), but is not part of the set A (punctured point), thus, the interval in question will not become part of the desired intersection. Let us mark the point with the coordinate 7 as punched out:

And finally, we check the remaining gap (7, + ∞).

The gap is included in both sets A and B (hatching is present above the gap), therefore, it becomes part of the intersection. We shade the place above the considered gap:

Ultimately, an image of the desired intersection of the given sets was formed on the lower coordinate line. Obviously it is the set of all real numbers more number 7, i.e.: A ∩ B = (7, + ∞).

1. Next step Let's define the union of the given sets A and B. We will sequentially check the sets (- ∞ , - 3), ( - 3), (- 3, 7), ( 7), (7, + ∞), establishing the fact of their inclusion or non-inclusion in the desired union.

The first set (- ∞, - 3) is not part of any of the original sets A and B (there are no shadings above the intervals), therefore, the set (- ∞, - 3) will not be included in the desired union:

The set ( - 3) is included in the set B, which means it will be included in the desired union of the sets A and B:

The set (- 3 , 7) is integral part set B (hatching is present above the interval) and becomes an element of the union of sets A and B:

The set 7 is included in the numerical set B, therefore it will also be included in the desired union:

The set (7, + ∞), being an element of both sets A and B at the same time, becomes another part of the desired union:

Based on the final image of the union of the original sets A and B, we obtain: A ∩ B = [ - 3 , + ∞) .

Having some practical experience in applying the rules for finding intersections and unions of sets, the described checks are easily carried out orally, which allows you to quickly write down the final result. Let us demonstrate with a practical example what its solution looks like without detailed explanations.

Example 3

Initial data: sets A = (- ∞ , - 15) ∪ ( - 5 ) ∪ [ 0 , 7) ∪ ( 12 ) and B = (- 20 , - 10) ∪ ( - 5 ) ∪ (2 , 3) ​​∪ (17). It is necessary to determine the intersection and union of the given sets.

Solution

Let us mark the given numerical sets on the coordinate lines in order to be able to obtain an illustration of the required intersection and union:

Answer: A ∩ B = (- 20, - 15) ∪ (- 5) ∪ (2, 3); A ∪ B = (- ∞ , - 10) ∪ ( - 5 ) ∪ [ 0 , 7 ] ∪ ( 12 , 17 ) .

It is also clear that with sufficient understanding of the process, the specified algorithm can be optimized. For example, in the process of finding the intersection, you don’t have to waste time checking all the intervals and sets that represent individual numbers, limiting yourself to considering only those intervals and numbers that make up the set A or B. Other intervals will not be included in the intersection in any case, i.e. To. are not part of the original sets. Let's illustrate what has been said using a practical example.

Example 4

Initial data: sets A = ( - 2 ) ∪ [ 1 , 5 ] and B = [ - 4 , 3 ] .

It is necessary to determine the intersection of the original sets.

Solution

Let us represent the numerical sets A and B geometrically:

The boundary points of the original sets will divide the number line into several sets:

(- ∞ , - 4) , { - 4 } , (- 4 , - 2) , { - 2 } , (- 2 , - 1) , { 1 } , (1 , 3) , { 3 } , (3 , 5) , { 5 } , (5 , + ∞) .

It is easy to see that the numerical set A can be written by combining some of the listed sets, namely: ( - 2), (1, 3), (3) and (3, 5). It will be enough to check these sets for their inclusion also in set B in order to find the desired intersection. Those that will be included in set B and become elements of intersection. Let's check.

It is absolutely clear that ( - 2) is part of the set B, because the point with coordinate - 2 is an internal point of the segment [ - 4, 3). The interval (1, 3) and the set (3) are also included in set B (there is a shading above the interval, and the point with coordinate 3 is boundary and not punctured for set B). The set (3, 5) will not be an intersection element, because is not included in set B (there is no shading above it). Let's note all of the above in the drawing:

As a result, the desired intersection of two given sets will be the union of sets, which we will write as follows: ( - 2 ) ∪ (1 , 3 ] .

Answer: A ∩ B = ( - 2 ) ∪ (1 , 3 ] .

At the end of the article, we will also discuss how to solve the problem of finding the intersection and union of several sets (more than 2). Let us reduce it, as recommended earlier, to the need to determine the intersection and union of the first two sets, then the resulting result with the third set, and so on. Or you can use the algorithm described above with the only difference that checking the occurrence of intervals and sets that represent individual numbers must be carried out not by two, but by all given sets. Let's look at an example.

Example 5

Initial data: sets A = (- ∞, 12], B = (- 3, 25], D = (- ∞, 25) ꓴ (40). It is necessary to determine the intersection and union of the given sets.

Solution

We display the given numerical sets on coordinate lines and place a curly bracket on the left side of them, denoting intersection, as well as a square bracket, denoting union. Below we display coordinate lines with boundary points of numerical sets marked with strokes:

Thus, the coordinate line is represented by the following sets: (- ∞, - 3), (- 3), (- 3, 12), (12), (12, 25), (25), (25, 40), ( 40 ) , (40 , + ∞) .

We begin to look for intersections, alternately checking the written sets to see if they belong to each of the original ones. All three given sets include the interval (- 3, 12) and the set (- 12): they will become the elements of the desired intersection. Thus, we get: A ∩ B ∩ D = (- 3 , 12 ] .

The union of the given sets will make up the following sets: (- ∞ , - 3) - element of set A; ( - 3 ) – element of set A; (- 3, 12) – element of set A; ( 12 ) – element of set A; (12, 25) – element of set B; (25) is an element of set B and (40) is an element of set D. Thus, we get: A ∪ B ∪ D = (- ∞ , 25 ] ∪ ( 40 ) .

Answer: A ∩ B ∩ D = (- 3, 12 ]; A ∪ B ∪ D = (- ∞, 25 ] ∪ ( 40 ) .

Note also that the desired intersection of numerical sets is often the empty set. This happens in cases where the given sets do not include elements that simultaneously belong to all of them.

Example 6

Initial data: A = [ - 7 , 7 ] ; B = ( - 15 ) ∪ [ - 12 , 0) ∪ ( 5 ) ; D = [ - 15 , - 10 ] ∪ [ 10 , + ∞) ; E = (0, 27) . Determine the intersection of given sets.

Solution

Let us display the original sets on coordinate lines and the boundary points of these sets on the additional line with strokes.

The marked points will divide the number line into sets: (- ∞ , - 15) , ( - 15 ) , (- 15 , - 12) , ( - 12 ) , (- 12 , - 10) , ( - 10 ) , (- 10 , - 7) , ( - 7 ) , ( - 7 , 0) , ( 0 ) , (0 , 5) , ( 5 ) , (5 , 7) , ( 7 ) , (7 , 10) , ( 10 ) , (10, 27) , (27) , (27, + ∞) .

None of them is simultaneously an element of all the original sets; therefore, the intersection of the given sets is the empty set.

Answer: A ∩ B ∩ D ∩ E = Ø.

It is convenient to represent sets in the form of circles, which are called Euler circles.

In the figure, the intersection set of the sets X and Y is colored orange.

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An operation on sets is a rule, as a result of which a new set is uniquely obtained from given sets.

Let us denote an arbitrary operation by *. Set obtained from given sets A and B, written in the form A*B. The resulting set and the operation itself are usually called one term.

Comment. For basic numerical operations, two terms are used: one denotes the operation itself as an action, the other denotes the number obtained after performing the action. For example, the operation denoted by + is called addition, and the number resulting from addition is called a sum of numbers. Similarly, the sign of the multiplication operation, and the result a b - product of numbers a and b. However, less often this difference is not taken into account and they say “Consider the sum of numbers,” meaning not a specific result, but the operation itself.

Intersection operation.The intersection of sets A and B AglV, consisting of all objects, each of which belongs to both sets A And IN simultaneously.

In other words, AsV - is the set of all.g such that heA And heV:

Merge operation.Union of sets A and B is called a set denoted by A" and B, consisting of all objects, each of which belongs to at least one set A or IN.

The union operation is sometimes denoted by a + sign and is called set addition.

Difference operations.The difference between sets A and B is called a set denoted by AB, consisting of all objects, each of which lies in A, but doesn't lie IN.

Expression ApV read "A in intersection with IN», AkjB- “And in association with B", AB - "A without IN".

Example 7.1.1. Let A = {1, 3,4, 5, 8,9}, IN = {2,4, 6, 8}.

Then AkjB= (1,2, 3,4, 5, 6, 8, 9), AcB=( 4,8}, AB= (1.3, 5, 9), YAL = (2.6).”

Based on these operations, two more important operations can be identified.

Addition operation. Let AQS. Then the difference S.A. called addition of set A to S and is designated A s.

Let any set under consideration be a subset of some set U. Addition to such a fixed (in the context of solving a particular problem) set U simply mean A. The notation is also used SA, With A, A."

Example 7.1.2. The complement of the set (1, 3,4, 5, 8, 9) to the set of all decimal digits is (0, 2, 6, 7).

Complementing the set Q to the set R there is a set of 1.

The complement of a set of squares to a set of rectangles is the set of all rectangles having unequal adjacent sides.

We see that the operations of union, intersection and complement of sets correspond to the logical operations of disjunction, conjunction and negation.

Symmetric difference operation.The symmetric difference of sets A and B is called a set denoted by A®B, consisting of all objects, each of which belongs to exactly one of the sets A and B:

It is easy to see that the symmetric difference is the union of two sets AB And VA. The same set can be obtained if we first combine the sets A And IN, and then remove common elements from the set.

Example 7.1.3. Let real numbers be given a Then for the corresponding numerical intervals we have:

Note that since the segment [A; b] contains a number c> and the interval (c;d) point With does not contain th number With lies in difference [A; b] without [with; cf. But the difference, for example, (2;5), does not contain the number 3, since it lies in the segment. We have (2;5)=(2;3).

Let disjoint sets be given A And IN. Since n is the sign of the intersection operation, then the entry A(bb incorrect. It is also incorrect to say that sets have no intersection. There is always an intersection; it is defined for any sets. The fact that the sets do not intersect means that their intersection is empty (that is, by performing the indicated operation, we obtain an empty set). If the sets intersect, then their intersection is not empty. We conclude:

Let us generalize the operations of intersection union to the case when there are more than two sets.

Let the system be given TO sets. The intersection of sets of a given system is the set of all elements, each of which lies in all sets of their TO.

The union of sets of a given system is the set of all elements, each of which lies in at least one set of them TO.

Let the sets of the system TO numbered by elements of some index family /. Then any set of TO can be designated A,-, Where iel. If the set is finite, then the set of first natural numbers (1,2,...,u) is used as /. In general, / can be infinite.

Then in the general case the union of sets A for all iel denote (J A( , and the intersection - f]A i .

Let the totality TO final, then K= In this case

write AyjA 2 v...KjA„ And AG4 2 (^---G4p-

Example 7.1.4. Let us consider the intervals of the number line А| = [-oo;2], L 2 =H°; 3], L 3 = u

Both spaces are surrounded by square brackets, which means their boundaries belong to them.

For clarity, we list all the integers belonging to the intervals [−2; 3] and :

−2, −1, 0, 1, 2, 3 ∈ [−2; 3]

4, 5, 6, 7 ∈

It can be seen that the numerical intervals [−2; 3] and do not have total numbers. Therefore, their intersection will be the empty set:

[−2; 3] ∩ = Ø

If we depict numerical intervals [−2; 3] and on the coordinate line, you can see that they do not intersect anywhere:

Example 7. Given a set of one element (2). Find its intersection with the interval (−3; 4)

A set consisting of one element (2) is depicted on the coordinate line as a filled circle, and the numerical interval (−3; 4) is an interval whose boundaries do not belong to it. This means that the boundaries −3 and 4 will be depicted as empty circles:

The intersection of the set (2) and the numerical interval (−3; 4) will be a set consisting of one element (2), since element 2 belongs to both the set (2) and the numerical interval (−3; 4)

{ 2 } ∩ (−3; 4) = { 2 }

In fact, we have already dealt with the intersection of numerical intervals when solving systems linear inequalities. Remember how we solved them. First, we found many solutions to the first inequality, then many solutions to the second. Then we found many solutions that satisfy both inequalities.

Essentially, the set of solutions satisfying both inequalities is the intersection of the sets of solutions to the first and second inequalities. The role of these sets is taken by numerical intervals.

For example, to solve a system of inequalities, we must first find the sets of solutions to each inequality, then find the intersection of these sets.

IN in this example solving the first inequality x≥ 3 is the set of all numbers that are greater than 3 (including the number 3 itself). In other words, the solution to the inequality is the numerical interval

A general decision system will be the intersection of the sets of solutions to the first and second inequalities, that is, the intersection of numerical intervals

If we depict the set of solutions of the system on the coordinate line, we will see that these solutions belong to the interval, which in turn is the intersection of the intervals

Therefore, as an answer we indicated that the values ​​of the variable x belong to the numerical interval, that is, the intersection of the sets of solutions to the first and second inequalities

x ∈

Example 2. Solve inequality

All inequalities included in the system have already been resolved. It is only necessary to indicate those solutions that are common to all inequalities.

The solution to the first inequality is the numerical interval (−∞; −1) .

The solution to the second inequality is the numerical interval (−∞; −5) .

The solution to the third inequality is the numerical interval (−∞; 4) .

The solution to the system will be the intersection of numerical intervals (−∞; −1), (−∞; −5) and (−∞; 4). In this case, this intersection is the interval (−∞; −5) .

(−∞; −1) ∩ (−∞; −5) ∩ (−∞; 4) = (−∞; −5)

The figure shows numerical intervals and the inequalities by which these numerical intervals are defined. It can be seen that numbers belonging to the interval (−∞; −5) simultaneously belong to all original intervals.

Let's write the answer to the system using a numerical interval:

x ∈ (−∞; −5)

Example 3. Solve inequality

Solving the first inequality y> 7 is the numerical interval (7; +∞) .

Solving the second inequality y< 4 является числовой промежуток (−∞; 4) .

The solution to the system will be the intersection of the numerical intervals (7; +∞) and (−∞; 4).

In this case, the intersection of the numerical intervals (7; +∞) and (−∞; 4) is an empty set, since these numerical intervals do not have common elements:

(7; +∞) ∩ (−∞; 4) = ∅

If you depict the numerical intervals (7; +∞) and (−∞; 4) on the coordinate line, you can see that they do not intersect anywhere:

Union of sets

The union of two (or several) original sets is a set that consists of elements belonging to at least one of the original sets.

In practice, the union of sets consists of all elements belonging to the original sets. That is why they say that the elements of such a set belong to at least one of the original sets.

Consider the set A with elements 1, 2, 3 and set B with elements 4, 5, 6.

A = { 1, 2, 3 }

B = { 4, 5, 6 }

Let's define a new set C A and all elements of the set B

C = { 1, 2, 3, 4, 5, 6 }

In this case unification sets A And B is a set C and is denoted as follows:

A∪B=C

The symbol ∪ means union and replaces the conjunction OR. Then the expression A∪B=C can be read like this:

Elements belonging to set A OR set B, there are elements belonging to set C.

The definition of a union states that the elements of such a set belong to at least one of the original sets. This phrase can be taken literally.

Let's return to the set we created C, which includes all elements of the sets A And B. Let’s take element 5 from this set as an example. What can you say about it?

If 5 is an element of the set C, and the set WITH is a union of sets A And B, then we can confidently say that element 5 belongs to at least one of the sets A And B. The way it is:

A = { 1, 2, 3 }

B = { 4, 5 , 6 }

C = { 1, 2, 3, 4, 5 , 6 }

Let's take another element from the set WITH, for example, element 2. What can you say about it?

If 2 is an element of the set C, and the set WITH is a union of sets A And B, then we can confidently say that element 2 belongs to at least one of the sets A And B. The way it is:

A = {1, 2 , 3}

B = {4, 5, 6}

C = { 1, 2 , 3, 4, 5, 6 }

If we want to combine two or more sets and suddenly discover that one or more elements belong to each of these sets, then the repeated elements will be included in the union only once.

For example, consider the set A with elements 1, 2, 3, 4 and set B with elements 2, 4, 5, 6.

A = {1, 2 , 3, 4 }

B = {2 , 4 , 5, 6}

We see that elements 2 and 4 simultaneously belong to the set A, and many B. If we want to combine sets A And B, then the new set C will contain elements 2 and 4 only once. It will look like this:

C = { 1, 2, 3, 4, 5, 6 }

In order to avoid mistakes when merging, they usually do this: first, add all the elements of the first set to the new set, then add elements of the second set that do not belong to the first set. Let's try to make such a union with sets A And B .

So, we have the following initial sets:

A = { 1, 2, 3, 4 }

B = { 2, 4, 5, 6 }

Let's define a new set WITH and add all the elements of the set to it A

C = { 1, 2, 3, 4,

Now let's add elements from the set B, which do not belong to the set A. To many A elements 5 and 6 do not belong. Let's add them to the list C

C = { 1, 2, 3, 4, 5, 6 }

Example 2. John's friends are Tom, Fred, Max and George. And Michael's friends are Leo, Tom, Fred and Evan. Find the union of sets of friends of John and Michael.

First, let's define two sets: the set of John's friends and the set of Michael's friends.

Let's define a new set with the name "All of John and Michael's Friends" and add all of John and Michael's friends to it.

Note that Tom and Fred are both friends of John and Michael, so we will add them to the new set only once, since there cannot be two Toms and two Freds at once.

In this case, the set of all friends of John and Michael is the union of the sets of friends of John and Michael.

John's Friends ∪ Michael's Friends = All John and Michael's Friends

Example 3. Given two numerical intervals: [−7; 0] and [−3; 5] . Find their union.

Both spaces are surrounded by square brackets, which means their boundaries belong to them.

For clarity, we list all the integers belonging to these intervals:

−7, −6, −5, −4, −3,−2, −1 , 0 ∈ [−7; 0]

−3,−2, −1 , 0, 1, 2, 3, 4, 5 ∈ [−3; 5]

By combining numerical intervals [−7; 0] and [−3; 5] there will be a numerical interval [−7; 5] , which contains all the numbers in the interval [−7; 0] and [−3; 5] without repeating some of the numbers

−7, −6, −5, −4, −3,−2, −1, 0, 1, 2, 3, 4, 5 ∈ [−7; 5]

Please note that the numbers −3, −2, −1 belonged to both the first interval and the second. But since such elements can only be included in a union once, we included them once.

This means that by combining the numerical intervals [−7; 0] and [−3; 5] there will be a numerical interval [−7; 5]

[−7; 0] ∪ [−3; 5] = [−7; 5]

Let us represent the intervals [−7; 0] and [−3; 5] . On the upper area we mark the numerical interval [−7; 0], on the bottom - the interval [−3; 5]

Previously, we found out that the interval [−7; 5] is the union of the intervals [−7; 0] and [−3; 5] . Here it is useful to recall the definition of a union of sets, which was given at the very beginning. A union is interpreted as a set consisting of all elements belonging to at least one of the original sets.

Indeed, if we take any number from the interval [−7; 5], then it turns out that it belongs to at least one of the intervals: either the interval [−7; 0] or the interval [−3; 5] .

Let's take from the interval [−7; 5] any number, for example the number 2. Since the interval [−7; 5] is the union of the intervals [−7; 0] and [−3; 5], then the number 2 will belong to at least one of these intervals. In this case, the number 2 belongs to the interval [−3; 5]

Let's take another number. For example, the number −4. This number will belong to at least one of the intervals: [−7; 0] or [−3; 5] . In this case it belongs to the interval [−7; 0]

Let's take another number. For example, the number −2. It belongs to both the interval [−7; 0] and the interval [−3; 5] . But on the coordinate line it is indicated only once, since there are no two numbers −2 at the same point.

Not every union of number intervals is a number interval. For example, let's try to find the union of numerical intervals [−2; −1] and .

The idea remains the same - the union of the numerical intervals [−2;−1] and there will be a set consisting of elements belonging to at least one of the intervals: [−2; −1] or . But this set will not be a numerical interval. For clarity, we list all the integers belonging to this union:

[−2; −1] ∪ = { −2, −1, 4, 5, 6, 7 }

We got the set ( −2, −1, 4, 5, 6, 7 ) . This set is not a numerical interval due to the fact that the numbers located between −1 and 4 are not included in the resulting set

The numeric span must contain all numbers from the left border to the right. If one of the numbers is missing, then the numerical interval becomes meaningless. Let's say there is a ruler 15 cm long

This line is a number range because it contains all numbers between 0 and 15, inclusive. Now imagine that on the ruler, after the number 9, the number 12 immediately follows.

This ruler is not a 15cm ruler and is not advisable to use for measuring. Also, it cannot be called a number interval, since it does not contain all the numbers that it should contain.

Solving inequalities containing the sign ≠

Some inequalities contain the sign ≠ (not equal). For example, 2 x≠ 8. To solve this inequality, you need to find the set of values ​​of the variable x, for which the left side not equal right side.

Let's solve inequality 2 x≠ 8. Divide both sides of this inequality by 2, then we get:

We got an equivalent inequality x≠ 4. The solution to this inequality is the set of all numbers, unequal 4. That is, if we substitute in the inequality x≠ 4 is any number that is not equal to 4, then we get the correct inequality.

Let's substitute, for example, the number 5

5 ≠ 4 is a true inequality because 5 is not equal to 4

Let's substitute 7

7 ≠ 4 is a true inequality because 7 is not equal to 4

And since inequality x≠ 4 is equivalent to the original inequality 2 x≠ 8, then solutions to the inequality x≠ 4 will also apply to inequality 2 x≠ 8. Let's substitute the same test values ​​5 and 7 into inequality 2 x≠ 8 .

2 × 5 ≠ 8

2 × 7 ≠ 8

x≠ 4 on the coordinate line. To do this, we will cut out point 4 on the coordinate line, and highlight the entire remaining area on both sides with strokes:

Now let's write the answer in the form of a numerical interval. To do this, we will use the union of sets. Any number that is a solution to inequality 2 x≠ 8 will belong either to the interval (−∞; 4) or to the interval (4; +∞). x So we write that the values ​​of the variable belong to (−∞; 4) or (4; +∞) . Let us recall that for the word"or"

x ∈ (−∞; 4) ∪ (4; +∞)

x the symbol ∪ is used , belong to the interval (−∞; 4) or

interval (4; +∞). ≠ Inequalities containing a sign ≠ , can also be solved like ordinary equations. For this sign = replaced by a sign

. Then you get the usual equation. At the end of the solution, the found value of the variable x must be excluded from the set of solutions. x Let's solve the previous inequality 2 ≠ 8 as usual equation. Replace the ≠ sign with the equal sign = , we get equation 2 x = x= 4 .

8 . Divide both sides of this equation by 2, we get x We see that when

Example 2. Solve inequality 3x− 5 ≠ 1 − 2x

, equal to 4, the equation turns into a true numerical equality. For other values, equality will not be observed. These other meanings are what interests us. And to do this, it is enough to exclude the found four from the set of solutions. x Let's move −2

from the right side to the left side, changing the sign, and move −5 from the left side to the right side, again changing the sign:

Let us present similar terms in both parts:

Divide both sides of the resulting inequality by 5 x Solving the inequality unequal 1,2 .

≠ 1.2 is the set of all numbers, x Let us depict the set of solutions to the inequality

x ∈ (−∞; 1,2) ∪ (1,2; +∞)

≠ 1.2 on the coordinate line and write the answer in the form of a numerical interval: x This expression states that the values ​​assumed by the variable , belong to the interval (−∞; 4) belong to the interval (−∞; 1,2)

interval (1,2; +∞)

Solving sets of inequalities Let's consider another type of inequalities called set of inequalities

. You may rarely solve this type of inequalities, but for overall development it is useful to study them.

A set of inequalities is very similar to a system of inequalities. The difference is that in a system of inequalities it is necessary to find many solutions that satisfy each inequality that forms this system. And in the case of a set of inequalities, you need to find many solutions that satisfy at least one

the inequalities that make up this aggregate.

A set of inequalities is indicated by a square bracket. For example, the following notation of two inequalities is a set:

Solving the first inequality x Let's solve this set. First you need to solve each inequality separately.

≥ 3 is a numeric interval. x Multiple meanings , for which it is true at least one

x ∈

This expression says that the variable x, included in
the collection takes all values ​​belonging to the interval . And this is what we need. After all, solving a set means finding a set of solutions that satisfy And in the case of a set of inequalities, you need to find many solutions that satisfy the inequalities that make up this aggregate. And any number in the interval will satisfy at least one inequality.

For example, the number 9 from the interval satisfies the second inequality x≤ 6.

Look closely at the expression x∈ , namely on its right side. After all, the expression is a union of numerical intervals. More precisely, the union of the sets of solutions to the first and second inequalities.

That is, the solution to the set of inequalities is the union of sets solutions of the first and second inequalities.

In other words, the solution to the population will be the union of numerical intervals

The union of numerical intervals is the interval (−∞; +∞) . More precisely, the union of numerical intervals is the entire coordinate line. And the entire coordinate line is all the numbers that can be

= (−∞; +∞)

x∈

x∈ (−∞; +∞)

Let's take any number from the resulting combination and check whether it satisfies at least one inequality.

Let's take the number 8 as an example. It satisfies the first inequality x≥ 3.

8 ≥ 3

Let's take another number, for example, the number 1. It satisfies the second inequality x≤ 6

Let's take another number, for example, the number 5. It also satisfies the first inequality x≥ 3 and second x≤ 6

Example 2

To solve this set, you need to find a set of solutions that satisfy at least one inequality that forms this set.

First, let's find many solutions to the first inequality x< −0,25 . Этим множеством является числовой промежуток (−∞; −0,25) .

x≥ −7 is the numerical interval [−7; +∞).

x∈ (−∞; −0,25) ∪ [−7; +∞)

In other words, the solution to the population will be the union of numerical intervals (−∞; −0.25) and [−7; +∞)

By combining the numerical intervals (−∞; −0.25) and [−7; +∞) is the entire coordinate line. And the entire coordinate line is all the numbers that can be

(−∞; −0,25) ∪ [−7; +∞) = (−∞; +∞)

The answer can be left as we wrote it earlier:

x∈ (−∞; −0,25) ∪ [−7; +∞)

or replace it with a shorter one:

x∈ (−∞; +∞)

Example 3. Solve a set of inequalities

Let's solve each inequality separately:

The set of solutions to the first inequality x < −3 является числовой промежуток (−∞; −3) .

The set of solutions to the second inequality x≤ 0 is the numerical interval (−∞; 0] .

The solution to the set of inequalities will be the union of the sets of solutions to the first and second inequalities.

x∈ (−∞; −3) ∪ (−∞; 0]

In other words, the solution to the population will be the union of the numerical intervals (−∞; −3) and (−∞; 0]

The union of the numerical intervals (−∞; −3) and (−∞; 0] is the numerical interval (−∞; 0]

(−∞; −3) ∪ (−∞; 0] = (−∞; 0]

The answer can be left as we wrote it earlier:

x∈ (−∞; −3) ∪ (−∞; 0]

or replace it with a shorter one:

x∈ (−∞; 0]

Tasks for independent solution

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1 QUESTION:Many is a collection of some elements united by some common feature. Elements of a set can be numbers, figures, objects, concepts, etc.

Sets are denoted by uppercase letters, and elements of the set are denoted by lowercase letters. Elements of sets are enclosed in curly braces.

If element x belongs to the set X, then write x ∈ X (∈ - belongs). If set A is part of set B, then write A⊂ IN (⊂ - contained).

Definition 1 (definition of equality of sets). Sets A and B are equal if they consist of the same elements, that is, if x  A implies x  B and vice versa, x  B implies x  A.

Formally, the equality of two sets is written as follows:

(A=B):= x((xA)  (xB)),

this means that for any object x the relations x A and x B are equivalent.

Here  is the universal quantifier ( x reads "for everyone" x").

Subset

Definition: The set X is subset Y, if any element of the set X belongs to the set Y. This is also called non-strict inclusion.Some properties of the subset:

1. ХХ - reflectivity

2. X  Y & YZ  X  Z - transitivity

3.   X i.e. the empty set is a subset of any set. Universal set Definition: Universal set- this is a set that consists of all elements, as well as subsets of the set of objects in the area under study, i.e.

1. If M  I , That M  I

2. If M  I , That Ώ(M)  I, where under Ώ(M) - all possible subsets of M, or Boolean M, are understood.

The universal set is usually denoted I .

The universal set can be selected independently, depending on the set under consideration and the tasks being solved.

Methods for specifying sets:

1. by listing its elements. Usually finite sets are defined by enumeration.

2. by describing properties common to all elements of this set, and only this set. This property is called characteristic property, and this way of specifying the set description. Thus, you can specify both finite and infinite sets. If we define a set with some property, then it may turn out that only one object has this property or that there is no such object at all. This fact may not be at all obvious.

Topic 2.3 Operations on sets.

Now let's define operations on sets.

1. Intersection of sets.

Definition: The intersection of the sets X and Y is a set consisting of all those, and only those elements, that belong to both the set X and the set Y.

For example: X=(1,2,3,4) Y=(2,4,6) intersection (2,4)

Definition: Sets are called disjoint if they do not have common elements, i.e. their intersection is equal to the empty set.

For example : disjoint sets are the sets of excellent students and unsuccessful ones.

This operation can be extended to more than two sets. In this case, it will be a set of elements that simultaneously belong to all sets.

Intersection properties:

1. X∩Y = Y∩X - commutativity

2. (X∩Y) ∩Z =X∩ (Y∩Z)=X∩Y∩Z - associativity

3. X∩ = 

4. X∩ I = X

2. Union of sets

Definition: The union of two sets is a set consisting of all and only those elements that belong to at least one of the sets X or Y.

For example: X=(1,2,3,4) Y=(2,4,6) by combining (1,2,3,4,6)

This operation can be extended to more than two sets. In this case, it will be the set of elements belonging to at least one of these sets.

Join properties:

1. XUY= YUY - commutativity

2. (X UY)UZ =XU (YUZ)=XUYUZ - associativity

4.XU I = I

From the properties of the operations of intersection and union it is clear that the empty set is similar to zero in the algebra of numbers.

3. Set difference

Definition: This operation, unlike the operations of intersection and union, is defined only for two sets. The difference between the sets X and Y is a set consisting of all those and only those elements that belong to X and do not belong to Y.

For example: X=(1,2,3,4) Y=(2,4,6) difference (1,3)

As we have already seen, the role of zero in set algebra is played by the empty set. Let us define a set that will play the role of unit in the algebra of sets

4. Set completion

The complement of a set X is the difference between I and X.

Add-on properties:

1. The set X and its complement do not have common elements

2. Any element I belongs either to the set X or its complement.

QUESTION 2 Sets of numbers

Integers− numbers used when counting (listing) items: N=(1,2,3,…)

Natural numbers with zero included− numbers used to indicate the number of items: N0=(0,1,2,3,…)

Whole numbers− include integers, numbers opposite to natural ones (i.e. with a negative sign) and zero. Positive integers: Z+=N=(1,2,3,…) Negative integers: Z−=(…,−3,−2,−1) Z=Z−∪(0)∪Z+=(…,−3,−2,−1,0,1,2,3,…)

Rational numbers− numbers represented as a common fraction a/b, where a and b are integers and b≠0. Q=(x∣x=a/b,a∈Z,b∈Z,b≠0) When converted to decimal a rational number is represented by a finite or infinite periodic fraction.

Irrational numbers− numbers that are represented as an infinite non-periodic decimal fraction.

Real numbers− union of rational and irrational numbers: R

Complex numbers C=(x+iy∣x∈R иy∈R), where i is the imaginary unit.

Real number modulus and properties

Modulus of a real number- This absolute value this number.

Simply put, when taking the modulus, you need to remove its sign from the number.

The absolute value of a number a denoted by |a|. Please note: the modulus of a number is always non-negative: |a|≥ 0.

|6| = 6, |-3| = 3, |-10,45| = 10,45