The set is closed under the operation. Open and closed sets Types of sets on the real line

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Open and closed sets

Appendix 1 . Open and closed sets

A bunch of M on a straight line is called open, if each of its points is contained in this set together with some interval. Closed is called a set containing all its limit points (i.e., such that any interval containing this point intersects with the set at least one more point). For example, a segment is a closed set, but is not open, and an interval, on the contrary, is an open set, but is not closed. There are sets that are neither open nor closed (for example, a half-interval). There are two sets that are both closed and open at the same time - this is empty and all Z(prove that there are no others). It is easy to see that if M open, then [` M] (or Z \ M- addition to the set M before Z) is closed. Indeed, if [` M] is not closed, then it does not contain some of its limit points m. But then m O M, and each interval containing m, intersects with the set [` M], i.e., has a point that does not lie in M, which contradicts the fact that M- open. Similarly, also directly from the definition, it is proved that if M closed, then [` M] open (check!).

We now prove the following important theorem.

Theorem. Any open set M can be represented as a union of intervals with rational ends (that is, with ends at rational points).

Proof . Consider the union U all intervals with rational ends that are subsets of our set. Let us prove that this union coincides with the entire set. Indeed, if m- a point from M, then there is an interval ( m 1 , m 2) M M, containing m(this follows from the fact that M- open). It is possible to find a rational point on any interval. Let on ( m 1 , m) - This m 3 , on ( m, m 2) is m 4 . Then the point m covered by union U, namely, the interval ( m 3 , m 4). Thus, we have proved that each point m from M covered by union U. Moreover, as is obvious from the construction U, no point not contained in M, not covered U. Means, U and M match.

An important consequence of this theorem is the fact that any open set is countable combining intervals.

Nowhere dense sets and sets of measure~zero. Cantor set>

Annex 2 . Nowhere dense sets and sets of measure zero. Cantor set

A bunch of A called nowhere tight, if for any different points a and b there is a segment [ c, d] M [ a, b], not intersecting with A. For example, the set of points in the sequence a n = [ 1/(n)] is nowhere dense, but the set of rational numbers is not.

Baer's theorem. A segment cannot be represented as a countable union of nowhere dense sets.

Proof . Suppose there is a sequence A k nowhere dense sets such that i A i = [a, b]. Let's construct the following sequence of segments. Let be I 1 is some segment nested in [ a, b] and not intersecting with A one . By definition, a nowhere dense set on the interval I 1 there is a segment that does not intersect with the set A 2. Let's call it I 2. Next, on the segment I 2 take in a similar way the segment I 3 , not intersecting with A 3 , etc. The sequence I k nested segments have a common point (this is one of the main properties of real numbers). This point, by construction, does not lie in any of the sets A k, so these sets do not cover the entire segment [ a, b].

Let's call the set M having measure zero, if for any positive e there is a sequence I k intervals with a total length less than e, covering M. Obviously, any countable set has measure zero. However, there are also uncountable sets that have measure zero. Let us construct one such, very well-known, called the Cantor one.

Let's take a cut. Let's divide it into three equal parts. Throw out the middle segment (Fig. 11, a). There will be two segments of the total length [ 2/3]. With each of them we will perform exactly the same operation (Fig. 11, b). There will be four segments of the total length [ 4/9] = ([ 2/3]) \ B 2 . Continuing so on (Fig. 11, in–e) to infinity, we obtain a set that has measure less than any given positive measure, i.e. measure zero. One can establish a one-to-one correspondence between the points of this set and infinite sequences of zeros and ones. If during the first "throwing out" our point fell into the right segment, we put 1 at the beginning of the sequence, if in the left - 0 (Fig. 11, a). Further, after the first "throwing out", we get a small copy of the large segment, with which we do the same: if our point after throwing out fell into the right segment, we put 1, if into the left - 0, etc. (check the mutual uniqueness) , rice. eleven, b, in. Since the set of sequences of zeros and ones has the cardinality of the continuum, the Cantor set also has the cardinality of the continuum. Moreover, it is easy to prove that it is nowhere dense. However, it is not true that it has a strict measure zero (see the definition of a strict measure). The idea behind the proof of this fact is as follows: take the sequence a n, tending to zero very quickly. For this, for example, the sequence a n = [ 1/(2 2 n)]. Then we prove that this sequence cannot cover the Cantor set (do it!).

Appendix 3 . Tasks

Operations on sets

Sets A and B called equal if each element of the set A belongs to the set B, and vice versa. Designation: A = B.

A bunch of A called subset sets B if each element of the set A belongs to the set B. Designation: A M B.

1. For each of the following two sets, indicate whether one is a subset of the other:

2. Prove that the set A if and only then is it a subset of the set B when every element not belonging B, not belong A.

3. Prove that for arbitrary sets A, B and C

a) A M A; b) if A M B and B M C, then A M C;

in) A = B, if and only if A M B and B M A.

The set is called empty if it does not contain any elements. Designation: Zh.

4. How many elements does each of the following sets have:

5. How many subsets does a set of three elements have?

6. Can a set have exactly a) 0; b*) 7; c) 16 subsets?

Association sets A and B x, what x O A or x O B. Designation: A And B.

crossing sets A and B is called a set consisting of such x, what x O A and x O B. Designation: A W B.

difference sets A and B is called a set consisting of such x, what x O A and x P B. Designation: A \ B.

7. Sets are given A = {1,3,7,137}, B = {3,7,23}, C = {0,1,3, 23}, D= (0,7,23,1998). Find sets:

8. Let be A is the set of even numbers, and B is the set of numbers divisible by 3. Find A W B.

9. Prove that for any sets A, B, C

a) A And B = B And A, A W B = B W A;

b) A AND ( B And C) = (A And B)AND C, A Z ( B W C) = (A W B)Z C;

in) A Z ( B And C) = (A W B)AND ( A W C), A AND ( B W C) = (A And B)Z ( A And C);

G) A \ (B And C) = (A \ B)Z ( A \ C), A \ (B W C) = (A \ B)AND ( A \ C).

10. Is it true that for any set A, B, C

Set mappings

If each element x sets X mapped to exactly one element f(x) sets Y, then they say that given display f from many X into the multitude Y. At the same time, if f(x) = y, then the element y called way element x when displayed f, and the element x called prototype element y when displayed f. Designation: f: X ® Y.

11. Draw all possible mappings from the set (7,8,9) to the set (0,1).

Let be f: X ® Y, y O Y, A M X, B M Y. Full preimage of an element y when displayed f is called a set ( x O X | f(x) = y). Designation: f - 1 (y). The image of the set A M X when displayed f is called a set ( f(x) | x O A). Designation: f(A). The prototype of the set B M Y is called a set ( x O X | f(x) O B). Designation: f - 1 (B).

12. To display f: (0,1,3,4) ® (2,5,7,18) given by the picture, find f({0,3}), f({1,3,4}), f - 1 (2), f - 1 ({2,5}), f - 1 ({5,18}).

13. Let be f: X ® Y, A 1 , A 2 M X, B 1 , B 2 M Y. Is it always true that

a) f(X) = Y;

b) f - 1 (Y) = X;

in) f(A 1 and A 2) = f(A 1) And f(A 2);

G) f(A 1 Z A 2) = f(A 1)Z f(A 2);

e) f - 1 (B 1 and B 2) = f - 1 (B 1) And f - 1 (B 2);

e) f - 1 (B 1 Z B 2) = f - 1 (B 1)Z f - 1 (B 2);

g) if f(A 1M f(A 2), then A 1M A 2 ;

h) if f - 1 (B 1M f - 1 (B 2), then B 1M B 2 ?

Composition mappings f: X ® Y and g: Y ® Z is called a mapping that maps to an element x sets X element g(f(x)) sets Z. Designation: g° f.

14. Prove that for arbitrary mappings f: X ® Y, g: Y ® Z and h: Z ® W the following is done: h° ( g° f) = (h° g)° f.

15. Let be f: (1,2,3,5) ® (0,1,2), g: (0,1,2) ® (3,7,37,137), h: (3,7,37,137) ® (1,2,3,5) – mappings shown in the figure:

Draw pictures for the following displays:

a) g° f; b) h° g; in) f° h° g; G) g° h° f.

Display f: X ® Y called bijective if for each y O Y there is exactly one x O X such that f(x) = y.

16. Let be f: X ® Y, g: Y ® Z. Is it true that if f and g are bijective, then g° f bijectively?

17. Let be f: (1,2,3) ® (1,2,3), g: (1,2,3) ® (1,2,3), are the mappings shown in the figure:

18. For each two of the following sets, find out if there is a bijection from the first to the second (assume that zero is a natural number):

a) the set of natural numbers;

b) the set of even natural numbers;

c) the set of natural numbers without the number 3.

metric space called a set X with a given metric r : X× X ® Z

1) " x,y O X r( x,y) i 0, and r ( x,y) = 0 if and only if x = y (non-negativity ); 2) " x,y O X r( x,y) = r ( y,x) (symmetry ); 3) " x,y,z O X r( x,y) + r ( y,z) and r ( x,z) (triangle inequality ). 19 19. X

a) X = Z, r ( x,y) = | x - y| ;

b) X = Z 2 , r 2 (( x 1 ,y 1),(x 2 ,y 2)) = C (( x 1 - x 2) 2 + (y 1 - y 2) 2 };

in) X = C[a,ba,b] functions,

open(respectively, closed) radius ball r in space X centered on a point x is called a set U r (x) = {y O x:r( x,y) < r) (respectively, B r (x) = {y O X:r( x,y) Ј r}).

internal point sets U M X U

open neighborhood this point.

limit point sets F M X F.

closed

20. Prove that

21. Prove that

b) set union A closure A

Display f: X ® Y called continuous

22.

23. Prove that

F (x) = inf y O F r( x,y

F.

24. Let be f: X ® Y– . Is it true that its inverse is continuous?

Continuous one-to-one mapping f: X ® Y homeomorphism. spaces X, Yhomeomorphic.

25.

26. For which couples X, Y f: X ® Y, which does not stick together points (i.e. f(x) № f(y) at x № y investments)?

27*. local homeomorphism(i.e., each point x plane and f(x) of the torus, there are neighborhoods U and V, what f maps homeomorphically U on the V).

Metric spaces and continuous mappings

metric space called a set X with a given metric r : X× X ® Z, satisfying the following axioms:

1) " x,y O X r( x,y) i 0, and r ( x,y) = 0 if and only if x = y (non-negativity ); 2) " x,y O X r( x,y) = r ( y,x) (symmetry ); 3) " x,y,z O X r( x,y) + r ( y,z) and r ( x,z) (triangle inequality ). 28. Prove that the following pairs ( X,r ) are metric spaces:

a) X = Z, r ( x,y) = | x - y| ;

b) X = Z 2 , r 2 (( x 1 ,y 1),(x 2 ,y 2)) = C (( x 1 - x 2) 2 + (y 1 - y 2) 2 };

in) X = C[a,b] is the set of continuous on [ a,b] functions,

open(respectively, closed) radius ball r in space X centered on a point x is called a set U r (x) = {y O x:r( x,y) < r) (respectively, B r (x) = {y O X:r( x,y) Ј r}).

internal point sets U M X is a point that is contained in U together with some ball of non-zero radius.

A set all of whose points are interior is called open. An open set containing a given point is called neighborhood this point.

limit point sets F M X a point is called such, in any neighborhood of which there are infinitely many points of the set F.

A set that contains all of its limit points is called closed(compare this definition with the one given in appendix 1).

29. Prove that

a) a set is open if and only if its complement is closed;

b) a finite union and a countable intersection of closed sets is closed;

c) a countable union and a finite intersection of open sets are open.

30. Prove that

a) the set of limit points of any set is a closed set;

b) set union A and the set of its limit points ( closure A) is a closed set.

Display f: X ® Y called continuous if the preimage of every open set is open.

31. Prove that this definition agrees with the definition of continuity of functions on the line.

32. Prove that

a) distance to the set r F (x) = inf y O F r( x,y) is a continuous function;

b) the set of zeros of the function of point a) coincides with the closure F.

33. Let be f: X ® Y

Continuous one-to-one mapping f: X ® Y, the inverse of which is also continuous, is called homeomorphism. spaces X, Y, for which such a mapping exists, are called homeomorphic.

34. For each pair of the following sets, determine whether they are homeomorphic:

35. For which couples X, Y spaces from the previous problem there is a continuous mapping f: X ® Y, which does not stick together points (i.e. f(x) № f(y) at x № y Such displays are called investments)?

36*. Think of a continuous plane-torus mapping that would be local homeomorphism(i.e., each point x plane and f(x) of the torus, there are neighborhoods U and V, what f maps homeomorphically U on the V).

Completeness. Baer's theorem

Let be X is a metric space. Subsequence x n its elements is called fundamental, if

37. Prove that the convergent sequence is fundamental. Is the converse true?

The metric space is called complete if every fundamental sequence converges in it.

38. Is it true that a space homeomorphic to a complete space is complete?

39. Prove that a closed subspace of a complete space is itself complete; a complete subspace of an arbitrary space is closed in it.

40. Prove that in a complete metric space a sequence of nested closed balls with radii tending to zero has a common element.

41. Is it possible in the previous problem to remove the condition of completeness of space or of the radii of balls tending to zero?

Display f metric space X in itself is called compressive, if

42. Prove that the contraction map is continuous.

43. a) Prove that the contraction mapping of a complete metric space into itself has exactly one fixed point.

b) A map of Russia at a scale of 1:20,000,000 is placed on a map of Russia at a scale of 1:5,000,000. Prove that there is a point whose images on both maps coincide.

44*. Does there exist an incomplete metric space in which the statement of the problem is true, eh?

A subset of a metric space is called dense everywhere if its closure coincides with the entire space; nowhere tight– if its closure has no non-empty open subsets (compare this definition with the one given in Appendix 2).

45. a) Let a, b, a , b О Z and a < a < b < b. Prove that the set of continuous functions on [ a,b], which are monotone on , is nowhere dense in the space of all continuous functions on [ a,b] with uniform metric.

b) Let a, b, c, e O Z and a < b, c> 0, e > 0. Then the set of continuous functions on [ a,b], such that

46. (Generalized Baire's theorem .) Prove that a complete metric space cannot be represented as a union of a countable number of nowhere dense sets.

47. Prove that the set of continuous, non-monotone on any non-empty interval and nowhere differentiable functions defined on the interval is everywhere dense in the space of all continuous functions on with uniform metric.

48*. Let be f is a differentiable function on the segment . Prove that its derivative is continuous on an everywhere dense set of points. This definition Lebesgue measures zero. If the countable number of intervals is replaced by a finite one, then we get the definition Jordanian measures zero.

Let us now prove some special properties of closed and open sets.

Theorem 1. The sum of a finite or countable number of open sets is an open set. The product of a finite number of open sets is an open set,

Consider the sum of a finite or countable number of open sets:

If , then P belongs to at least one of Let Since is an open set, then some -neighborhood of P also belongs to The same -neighborhood of P also belongs to the sum g, whence it follows that g is an open set. Consider now the final product

and let P belong to g. Let us prove, as above, that some -neighborhood P belongs to g. Since P belongs to g, then P belongs to all . Since are open sets, then for any there is some -neighborhood of the point belonging to . If the number is taken equal to the smallest of the number of which is finite, then the -neighborhood of the point P will belong to all and, consequently, to g. Note that it is impossible to assert that the product of a countable number of open sets is an open set.

Theorem 2. The set CF is open and the set CO is closed.

Let us prove the first assertion. Let P belong to CF. It is necessary to prove that some neighborhood P belongs to CF. This follows from the fact that if there were points F in any -neighborhood P, the point P, which does not belong by condition, would be the limit point for F and, due to its closedness, would have to belong, which leads to a contradiction.

Theorem 3. The product of a finite or countable number of closed sets is a closed set. The sum of a finite number of closed sets is a closed set.

Let us prove, for example, that the set

closed. Passing to additional sets, we can write

By the theorem, open sets, and, by Theorem 1, the set is also open, and thus the complementary set g is closed. Note that the sum of a countable number of closed sets may also be a non-closed set.

Theorem 4. A set is an open set and a closed set.

It is easy to check the following equalities:

From them, by virtue of the previous theorems, Theorem 4 follows.

We will say that a set g is covered by a system M of some sets if every point g is included in at least one of the sets of the system M.

Theorem 5 (Borel). If a closed bounded set F is covered by an infinite system a of open sets O, then from this infinite system one can extract a finite number of open sets that also cover F.

We prove this theorem from the converse. Let us assume that no finite number of open sets from the system a covers and reduce this to a contradiction. Since F is a bounded set, then all points of F belong to some finite two-dimensional interval . Let us divide this closed interval into four equal parts, dividing the intervals in half. Each of the obtained four intervals will be taken closed. Those points of F that fall on one of these four closed intervals will, by virtue of Theorem 2, represent a closed set, and at least one of these closed sets cannot be covered by a finite number of open sets from the system a. We take one of the above four closed intervals where this circumstance takes place. We again divide this interval into four equal parts and argue in the same way as above. Thus, we obtain a system of nested intervals, of which each next one is the fourth part of the previous one, and the following circumstance takes place: the set of points F belonging to any k cannot be covered by a finite number of open sets from the system a. With an infinite increase in k, the gaps will shrink indefinitely to some point P, which belongs to all the gaps. Since, for any k, they contain an uncountable set of points, the point P is a limit point for and therefore belongs to F, since F is a closed set. Thus the point P is covered by some open set belonging to the system a. Some -neighborhood of the point P will also belong to the open set O. For sufficiently large values ​​of k, the intervals D will fall inside the above -neighborhood of the point P. Thus, these will be completely covered by only one open set O of the system a, and this contradicts the fact that the points belonging to for any k cannot be covered by a finite number of open sets belonging to a. Thus the theorem is proved.

Theorem 6. An open set can be represented as the sum of a countable number of half-open gaps in pairs without common points.

Recall that a half-open gap in the plane is a finite gap defined by inequalities of the form .

Let's put on the plane a grid of squares with sides parallel to the axes, and with a side length equal to one. The set of these squares is a countable set. We choose from these squares those squares all of whose points belong to a given open set O. The number of such squares may be finite or countable, or there may be no such squares at all. We divide each of the remaining squares of the grid into four identical squares and from the newly obtained squares we again select those whose all points belong to O. We again divide each of the remaining squares into four equal parts and select those squares whose all points belong to O, etc. Let us show that any point P of the set O will fall into one of the chosen squares, all of whose points belong to O. Indeed, let d be a positive distance from P to the boundary of O. When we reach squares whose diagonal is less than , then we can obviously assert that the point P has already fallen into a square, all of whose volumes belong to O. If the chosen squares are considered half-open, then they will not have pairwise common points, and the theorem is proved. The number of selected squares will necessarily be countable, since the finite sum of half-open gaps is obviously not an open set. Denoting by DL those half-open squares that we obtained as a result of the above construction, we can write

Let two sets X and Y be given, coinciding or not.

Definition. The set of ordered pairs of elements, of which the first belongs to X and the second to Y, is called Cartesian product of sets and is denoted.

Example. Let be
,
, then

.

If a
,
, then
.

Example. Let be
, where R is the set of all real numbers. Then
is the set of all Cartesian coordinates of points in the plane.

Example. Let be
is a family of sets, then the Cartesian product of these sets is the set of all ordered strings of length n:

If , then. Items from
are row vectors of length n.

Algebraic structures with one binary operation

1 Binary algebraic operations

Let be
is an arbitrary finite or infinite set.

Definition. binary algebraic operation ( internal law of composition) on the
is called an arbitrary but fixed mapping of a Cartesian square
in
, i.e.

(1)

(2)

Thus, any ordered pair

. The fact that
, is written symbolically as
.

As a rule, binary operations are denoted by symbols
etc. As before, the operation
means "addition", and the operation "" means "multiplication". They differ in the form of notation and, perhaps, in axioms, which will be clear from the context. Expression
will be called a product, and
- sum of elements and .

Definition. A bunch of
is called closed under the operation if for any .

Example. Consider the set of non-negative integers
. As binary operations on
we will consider the usual operations of addition
and multiplications. Then the sets
,
will be closed under these operations.

Comment. As follows from the definition, the assignment of the algebraic operation * on
, is equivalent to the closedness of the set
regarding this operation. If it turns out that the set
is not closed under the given operation *, then in this case we say that the operation * is not algebraic. For example, the operation of subtraction on the set of natural numbers is not algebraic.

Let be
and
two sets.

Definition. Outer law compositions on the set is called mapping

, (3)

those. the law by which any element
and any element
element is assigned
. The fact that
, denoted by the symbol
or
.

Example. Matrix multiplication
per number
is an external law of composition on the set
. Multiplication of numbers in
can be considered both as an internal law of composition and as an external one.

distributive regarding the internal law of composition * in
, if

The external law of composition is called distributive with respect to the internal law of composition * in Y, if

Example. Matrix multiplication
per number
is distributive both with respect to matrix addition and with respect to number addition, because,.

1. Properties of Binary Operations

Binary algebraic operation  on a set
called:

Comment. The properties of commutativity and associativity are independent.

Example. Consider the set of integers . Operation on define according to the rule
. Let's choose numbers
and perform the operation on these numbers:

those. The operation  is commutative, but not associative.

Example. Consider the set
are square matrices of dimension
with real coefficients. As a binary operation * on
Let us consider the operations of matrix multiplication. Let be
, then
, but
, i.e. the operation of multiplication on a set of square matrices is associative but not commutative.

Definition. Element
called single or neutral regarding the operation in question  on
, if

Lemma. If a is the identity element of the set
closed under the operation *, then it is unique.

Proof . Let be is the identity element of the set
, closed under the operation *. Let's assume that in
there is one more element
, then
, as is a single element, and
, as is a single element. Hence,
is the only identity element of the set
.

Definition. Element
called reverse or symmetrical to element
, if

Example. Consider the set of integers with addition operation
. Element
, then the symmetrical element
there will be an element
. Really,.

One of the main tasks of the theory of point sets is the study of the properties of various types of point sets. Let's get acquainted with this theory on two examples and study the properties of the so-called closed and open sets.

The set is called closed if it contains all of its limit points. If a set has no limit points, then it is also considered to be closed. In addition to its limit points, a closed set can also contain isolated points. The set is called open if each of its points is internal to it.

Let's bring examples of closed and open sets .

Every segment is a closed set, and every interval (a, b) is an open set. Improper half-intervals and closed, and improper intervals and open. The entire line is both a closed and an open set. It is convenient to think of the empty set as both closed and open at the same time. Any finite set of points on a line is closed, since it has no limit points.

A set consisting of points:

closed; this set has a single limit point x=0, which belongs to the set.

The main task is to find out how an arbitrary closed or open set works. To do this, we need a number of auxiliary facts, which we will accept without proof.

• 1. The intersection of any number of closed sets is closed.
• 2. The sum of any number of open sets is an open set.
• 3. If a closed set is bounded from above, then it contains its upper bound. Similarly, if a closed set is bounded below, then it contains its lower bound.

Let E be an arbitrary set of points on the line. We call the complement of the set E and denote by CE the set of all points on the line that do not belong to the set E. It is clear that if x is an external point for E, then it is an internal point for the set CE and vice versa.

4. If the set F is closed, then its complement CF is open and vice versa.

Proposition 4 shows that there is a very close relationship between closed and open sets: one is the complement of the other. Because of this, it suffices to study only one closed or one open set. Knowing the properties of sets of one type allows you to immediately find out the properties of sets of another type. For example, any open set is obtained by removing some closed set from a line.

We proceed to the study of the properties of closed sets. We introduce one definition. Let F be a closed set. An interval (a, b) with the property that none of its points belongs to the set F, while the points a and b belong to F, is called an adjacent interval of the set F.

Among adjacent intervals, we will also include improper intervals or, if the point a or the point b belongs to the set F, and the intervals themselves do not intersect with F. Let us show that if a point x does not belong to a closed set F, then it belongs to one of its adjacent intervals.

Denote by the part of the set F located to the right of the point x. Since the point x itself does not belong to the set F, it can be represented in the form of an intersection:

Each of the sets F and is closed. Therefore, by Proposition 1, the set is closed. If the set is empty, then the entire half-interval does not belong to the set F. Let us now assume that the set is not empty. Since this set lies entirely on the half-interval, it is bounded from below. Denote by b its lower bound. According to Proposition 3, which means Further, since b is the infimum of the set, then the half-interval (x, b) lying to the left of the point b does not contain points of the set and, therefore, does not contain points of the set F. Thus, we have constructed a half-interval (x, b) that does not contain points of the set F, and either, or the point b belongs to the set F. Similarly, a half-interval (a, x) is constructed that does not contain points of the set F, and either, or. Now it is clear that the interval (a, b) contains the point x and is an adjacent interval of the set F. It is easy to see that if and are two adjacent intervals of the set F, then these intervals either coincide or do not intersect.

It follows from the above that any closed set on the line is obtained by removing from the line a certain number of intervals, namely, adjacent intervals of the set F. Since each interval contains at least one rational point, and all rational points on the line are a countable set, it is easy make sure that the number of all adjacent intervals is at most countable. From here we get the final conclusion. Any closed set on a line is obtained by removing from the line at most a countable set of disjoint intervals.

By Proposition 4, this immediately implies that any open set on the line is at most a countable sum of disjoint intervals. By virtue of Propositions 1 and 2, it is also clear that any set arranged as indicated above is indeed closed (open).

As can be seen from the following example, closed sets can have a very complex structure.