Theorem 3.1. The union of any number of open sets is an open set.
Let G k, where k О N are open sets.
3Select any point X O ÎG. By definition of the union of sets, the point X o belongs to one of the sets G k. Since G k is an open set, then there exists e- neighborhood of a point x o, which lies entirely in the set Gk: U(x o , e)Ì G k Þ U(x o ,e)Ì G.
Got that any point x o ÎG– internal, which means that G– open set. 4
Theorem 3.2 . Intersection finite number open non-empty sets – open sets.
Let G k (k = 1,2, …,n) are open sets.
Let us prove that is an open set.
3Select any point X O ÎG. By definition of intersection of sets X o belongs to each of the sets G k. Since every set G k open, then in any set G k exists e k- neighborhood of a point X O :U(x o , e k)Ì G k. Lots of numbers( e 1 , e 2 ,…, e n) is finite, so there is a number e = min{e 1 ,e 2 ,…,e n). Then e- neighborhood of a point X o is in every e k- neighborhood of a point X O :U(x o , e)M U e(x o , e k) Þ U(x o , e)Ì G.
Got that X o – interior point of the set G, which means that G– open set. 4
Remark 3.1. The intersection of an infinite number of open sets may not be an open set.
Example 3.1. Let in space R G k =(2 – 1/k; 4+ 1/k), Where k= 1,2,…,n,…. G 1 =(1;5), G 2(1.5;4.5), Segment Ì G k and is not an open set, points 2 and 4 are not internal.
Theorem 3.3 . The intersection of any collection of closed non-empty sets is a closed set.
Let Fk- closed sets.
Let us prove that the set is closed, i.e. it contains all its limit points.
3Let X F. From the definition of intersection of sets it follows that in any e- neighborhood of a point X o there are infinitely many points of each set Fk, which means that X o – limit point of each set Fk. Due to the closedness of the sets Fk dot
X O О F k "k Þ x O Î F. Since the point X F, and that means a lot F closed. 4
Theorem 3.4. The union of a finite number of closed sets is a closed set.
Let each set Fk closed.
Let us prove that the set is closed, i.e., if X o – limit point of the set F, That X O О F.
3Let X o – any limit point of the set F, then at any e- neighborhood of a point X o there are infinitely many points of the set. Since the number of sets Fk finite, then X o belongs to at least one of the sets Fk, i.e. X o is the limit point for this set.
Due to isolation Fk dot X o belongs Fk, and therefore many. Since the point X o is chosen arbitrarily, then all limit points belong to the set F, which means a lot F closed. 4
Remark 3.2. The union of an infinite number of closed sets can be an open set.
Example 3.2 . In space R: F k =
F 1 =; F 2 = ; …. The interval (2;5) is an open set.
Let us accept without proof Theorems 3.5 and 3.6 related to the complement of the set E to many X: C x E=CE.
Theorem 3.5 . If the set E closed, then its complement SE open set.
Example 3.3 . E=, C R E =(- ¥, 2)È (5,+¥ ).
Theorem 3.6 . If the set E open, then its complement SE closed set.
Example 3.4 . E=(2,5), C R E =(-¥, 2]È[ 5, +¥ ).
One of the main tasks of the theory of point sets is the study of properties various types point sets. Let's get acquainted with this theory using two examples and study the properties of the so-called closed and open sets.
The set is called closed , if it contains all its limit points. If a set does not have a single limit point, then it is also considered 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 for it.
Let's give 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 consider the empty set to be 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 unique limit point x=0, which belongs to the set.
The main task is to find out how an arbitrary closed or open set is structured. To do this, we will 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 above, then it contains its supremum. Similarly, if a closed set is bounded below, then it contains its infimum.
Let E be an arbitrary set of points on a line. Let us 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 a set F is closed, then its complement CF is open and vice versa.
Proposition 4 shows that there is a very close connection between closed and open sets: some are complements of others. Because of this, it is sufficient to study only closed or only open sets. 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.
Let's start studying the properties of closed sets. Let's introduce one definition. Let F be a closed set. An interval (a, b) having the property that none of its points belong to the set F, but the points a and b belong to F, is called an adjacent interval of the set F.
We will also include improper intervals as adjacent intervals, or if point a or 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.
Let us 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 intersection form:
Each of the sets is F and 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 is entirely located on a half-interval, it is bounded below. Let us denote its lower bound by b. According to Proposition 3, which means. Further, since b is the infimum of the set, 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. So, we have constructed a half-interval (x, b) not containing 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.
From the previous it follows that any closed set on a line is obtained by removing a certain number of intervals from the line, namely adjacent intervals of the set F. Since each interval contains at least one rational point, and there is a countable set of all rational points on the line, it is easy make sure that the number of all adjacent intervals is at most countable. From here we get the final conclusion. Every closed set on a line is obtained by removing from the line at most a countable set of disjoint intervals.
By virtue of Proposition 4, it immediately follows that every open set on a line is nothing more than 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.
Open and closed sets
Appendix 1 . Open and closed sets
Many M on a straight line is called open, if each of its points is contained in this set along with a certain interval. Closed a set is called that contains all its limit points (i.e., such that any interval containing this point intersects the set at least at 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 that’s it Z(prove that there are no others). It's easy to see that if M open, then [` M] (or Z \ M- addition to set M to Z) is closed. Indeed, if [` M] is not closed, then it does not contain any limit point of its own m. But then m ABOUT M, and each interval containing m, intersects with the set [` M], i.e. has a point not lying in M, and this contradicts the fact that M– open. Similarly, also directly from the definition, it is proved that if M is closed, then [` M] open (check!).
Now we will 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- some point from M, then there is an interval ( m 1 , m 2) M M containing m(this follows from the fact that M– open). On any interval you can find a rational point. Let on ( m 1 , m) - This m 3, on ( m, m 2) – this is m 4. Then point m covered by union U, namely, the interval ( m 3 , m 4). Thus, we have proven that each point m from M covered by union U. Moreover, as it obviously follows 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>
Appendix 2 . Nowhere dense sets and sets of measure zero. Cantor set
Many A called nowhere dense, 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 a set rational numbers- No.
Baire'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 And i A i = [a, b]. Let's construct the following sequence of segments. Let I 1 – some segment embedded in [ a, b] and does not intersect with A 1. By definition, a nowhere dense set on an interval I 1 there is a segment that does not intersect with the set A 2. Let's call him I 2. Further, on the segment I 2, similarly take the segment I 3, not intersecting with A 3, etc. Sequence I k nested segments have a common point (this is one of the main properties real numbers). By construction, this point does not lie in any of the sets A k, which means that 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's build one, very famous, called Cantor's.
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| Rice. 11 |
Let's take a segment. Let's divide it into three equal parts. Let's throw out the middle segment (Fig. 11, A). There will be two segments of total length [2/3]. We will perform exactly the same operation with each of them (Fig. 11, b). There will be four segments left with total length [ 4/9] = ([ 2/3]) \ B 2 . Continuing like this (Fig. 11, V–e) to infinity, we obtain a set that has a measure less than any predetermined positive measure, i.e., measure zero. It is possible to 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 falls into the right segment, we will put 1 at the beginning of the sequence, if in the left - 0 (Fig. 11, A). Next, after the first “throwing out”, we get a small copy of the large segment, with which we do the same thing: if our point after throwing out falls into the right segment, put 1, if in the left - 0, etc. (check the one-to-one relationship) , rice. 11, b, V. Since the set of sequences of zeros and ones has cardinality continuum, the Cantor set also has cardinality continuum. Moreover, it is easy to prove that it is not dense anywhere. However, it is not true that it has strict measure zero (see the definition of strict measure). The idea of proving this fact is as follows: take the sequence a n, tending to zero very quickly. For example, the sequence a n = [ 1/(2 2 n)]. Then we will prove that this sequence cannot cover the Cantor set (do it!).
Appendix 3 . Tasks
Set Operations
Sets A And B are called equal, if each element of the set A belongs to the set B, and vice versa. Designation: A = B.
Many A called subset sets B, if each element of the set A belongs to the set B. Designation: A M B.
1.
For each two of the following sets, indicate whether one is a subset of the other:
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2. Prove that the set A if and only if is a subset of the set B, when every element not belonging to B, does 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, That A M C;
V) 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: F.
4.
How many elements does each of the following sets have:
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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 ABOUT A or x ABOUT B. Designation: A AND B.
By crossing sets A And B is called a set consisting of such x, What x ABOUT A And x ABOUT B. Designation: A Z B.
By difference sets A And B is called a set consisting of such x, What x ABOUT A And x P B. Designation: A \ B.
7. Given sets A = {1,3,7,137}, B = {3,7,23}, C = {0,1,3, 23}, D= (0,7,23,1998). Find the sets:
| A) A AND B; | b) A Z B; | V) ( A Z B)AND D; |
| G) C Z ( D Z B); | d) ( A AND B)Z ( C AND D); | e) ( A AND ( B Z C))Z D; |
| and) ( C Z A)AND (( A AND ( C Z D))Z B); | h) ( A AND B) \ (C Z D); | And) A \ (B \ (C \ D)); |
| To) (( A \ (B AND D)) \ C)AND B. |
8. Let A is the set of even numbers, and B– set of numbers divisible by 3. Find A Z B.
9. Prove that for any sets A, B, C
A) A AND B = B AND A, A Z B = B Z A;
b) A AND ( B AND C) = (A AND B)AND C, A Z ( B Z C) = (A Z B)Z C;
V) A Z ( B AND C) = (A Z B)AND ( A Z C), A AND ( B Z C) = (A AND B)Z ( A AND C);
G) A \ (B AND C) = (A \ B)Z ( A \ C), A \ (B Z C) = (A \ B)AND ( A \ C).
10. Is it true that for any sets A, B, C
| A) A Z ZH = F, A I F = A; | b) A AND A = A, A Z A = A; | V) A Z B = A Y A M B; |
| G) ( A \ B)AND B = A; 7 d) A \ (A \ B) = A Z B; | e) A \ (B \ C) = (A \ B)AND ( A Z C); | |
| and) ( A \ B)AND ( B \ A) = A AND B? |
Set mappings
If each element x sets X exactly one element is matched f(x) sets Y, then they say that it is 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 f: X ® Y, y ABOUT Y, A M X, B M Y. Full prototype of the element y when displayed f is called a set ( x ABOUT X | f(x) = y). Designation: f - 1 (y). The image of the multitude A M X when displayed f is called a set ( f(x) | x ABOUT A). Designation: f(A). The prototype of the set B M Y is called a set ( x ABOUT X | f(x) ABOUT 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}).
| a) b) c) |
13. Let 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;
V) f(A 1 I A 2) = f(A 1)And f(A 2);
G) f(A 1 W A 2) = f(A 1)Z f(A 2);
d) f - 1 (B 1 I B 2) = f - 1 (B 1)And f - 1 (B 2);
e) f - 1 (B 1 W B 2) = f - 1 (B 1)Z f - 1 (B 2);
g) if f(A 1) M f(A 2), then A 1 M A 2 ;
h) if f - 1 (B 1) M f - 1 (B 2), then B 1 M B 2 ?
Composition mappings f: X ® Y And g: Y ® Z is called a mapping that associates 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 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:
| f: g: h: |
Draw pictures for the following displays:
A) g° f; b) h° g; V) f° h° g; G) g° h° f.
Display f: X ® Y called bijective, if for each y ABOUT Y there is exactly one x ABOUT X such that f(x) = y.
16. Let f: X ® Y, g: Y ® Z. Is it true that if f And g are bijective, then g° f bijectively?
17. Let f: (1,2,3) ® (1,2,3), g: (1,2,3) ® (1,2,3), – mappings shown in the figure:
18. For each two of the following sets, find out whether there is a bijection from the first to the second (assuming that zero is a natural number):
a) many 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 ABOUT X r( x,y) i 0, and r ( x,y) = 0 if and only if x = y (non-negativity ); 2) " x,y ABOUT X r( x,y) = r ( y,x) (symmetry ); 3) " x,y,z ABOUT X r( x,y) + r ( y,z) i 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 };
V) X = C[a,ba,b] functions,
Open(respectively, closed) ball of radius r in space X centered at a point x called a set U r (x) = {y ABOUT x:r ( x,y) < r) (respectively, B r (x) = {y ABOUT X:r ( x,y) Ј r}).
Internal point sets U M X U
open surroundings this point.
Limit point sets F M X F.
closed
20. Prove that
21. Prove that
b) union of a set A short circuit A
Display f: X ® Y called continuous
22.
23. Prove that
F (x) = inf y ABOUT F r( x,y
F.
24. Let 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. at each point x plane and f(x) torus there are such neighborhoods U And V, What f homeomorphically maps U on 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 ABOUT X r( x,y) i 0, and r ( x,y) = 0 if and only if x = y (non-negativity ); 2) " x,y ABOUT X r( x,y) = r ( y,x) (symmetry ); 3) " x,y,z ABOUT X r( x,y) + r ( y,z) i 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 };
V) X = C[a,b] – set of continuous on [ a,b] functions,
Open(respectively, closed) ball of radius r in space X centered at a point x called a set U r (x) = {y ABOUT x:r ( x,y) < r) (respectively, B r (x) = {y ABOUT 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 this point, called surroundings this point.
Limit point sets F M X is a point such that any neighborhood of which contains infinitely many points of the set F.
A set that contains all 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) the finite union and countable intersection of closed sets is closed;
c) the countable union and finite intersection of open sets are open.
30. Prove that
a) the set of limit points of any set is a closed set;
b) union of a set A and the set of its limit points ( short circuit A) is a closed set.
Display f: X ® Y called continuous, if the inverse image of every open set is open.
31. Prove that this definition is consistent with the definition of continuity of functions on a line.
32. Prove that
a) distance to set r F (x) = inf y ABOUT F r( x,y) is a continuous function;
b) the set of zeros of the function in item a) coincides with the closure F.
33. Let 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 mappings are called investments)?
36*. Come up with a continuous mapping from a plane to a torus that would be local homeomorphism(i.e. at each point x plane and f(x) torus there are such neighborhoods U And V, What f homeomorphically maps U on V).
Completeness. Baire's theorem
Let X– metric space. Subsequence x n its elements are called fundamental, If
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37. Prove that the convergent sequence is fundamental. Is the opposite statement 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 one is complete?
39. Prove that a closed subspace of a complete space is itself complete; the 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 the tendency of the radii of the balls to zero?
Display f metric space X called into oneself compressive, If
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42. Prove that the contraction map is continuous.
43. a) Prove that a contraction mapping of a complete metric space into itself has exactly one fixed point.
b) Place a map of Russia at a scale of 1:20,000,000 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*. Is there an incomplete metric space in which the statement of problem eh is true?
A subset of a metric space is called dense everywhere, if its closure coincides with the entire space; nowhere dense– if its closure does not have non-empty open subsets (compare this definition with the one given in Appendix 2).
45. a) Let a, b, a , b O Z And a < a < b < b. Prove that the set continuous functions on [ a,b], monotone on , 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
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46. (Generalized Baire's theorem .) Prove that a complete metric space cannot be represented as the union of a countable number of nowhere dense sets.
47. Prove that the set of continuous, non-monotonic 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 a uniform metric.
48*. Let f– differentiable function on the interval. Prove that its derivative is continuous on an everywhere dense set of points. This is the definition Lebesgue measures zero. If the countable number of intervals is replaced by a finite one, we get the definition Jordanova 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. The same -neighborhood of P also belongs to the sum g, from which it follows that g is an open set. Let us now consider the final product
and let P belong to g. Let us prove, as above, that some -neighborhood of P also belongs to g. Since P belongs to g, then P belongs to everyone. Since - are open sets, then for any there is some -neighborhood of the point belonging to . If the number is taken to be equal to the smallest of which the number is finite, then the -neighborhood of the point P will belong to everyone and, consequently, to g. Note that we cannot claim 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's prove the first statement. 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 of P, the point P, which does not belong by condition, would be a limit point for F and, due to its closedness, should 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. Moving on to additional sets, we can write
By theorem, sets are open, and, by Theorem 1, the set is also open, and thus the additional set g is closed. Note that the sum of a countable number of closed sets may also turn out to be an open set.
Theorem 4. A set is an open set and a closed set.
It is easy to check the following equalities:
From these, by virtue of the previous theorems, Theorem 4 follows.
We will say that a set g is covered by a system M of certain 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 it is possible to extract a finite number of open sets that also cover F.
We prove this theorem by inverse. Let us assume that no finite number of open sets from the system a covers and we bring 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. We will take each of the resulting four intervals to be 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 four closed intervals indicated above where this circumstance occurs. We again divide this interval into four equal parts and reason in the same way as above. Thus, we obtain a system of nested intervals of which each next represents a fourth part of the previous one, and the following circumstance holds: 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 intervals will infinitely shrink to a certain point P, which belongs to all intervals. Since for any k they contain an infinite number of points, the point P is a limiting 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 entirely 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 proven.
Theorem 6. An open set can be represented as the sum of a countable number of half-open intervals in pairs without common points.
Recall that we call a half-open interval in a plane a finite interval defined by inequalities of the form .
Let us draw 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. From these squares, let us choose those squares all of whose points belong to a given open set O. The number of such squares may be finite or countable, or perhaps there will 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 points all 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 every point P of the set O will fall into one of the selected squares, all points of which belong to O. Indeed, let d be the positive distance from P to the boundary of O. When we get to squares whose diagonal is less than , then we can, obviously, assert that point P has already fallen into a square, all the volumes of which belong to O. If the selected squares are considered half-open, then they will not have common points in pairs, and the theorem is proven. The number of selected squares will necessarily be countable, since the finite sum of half-open intervals 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
Proof.
1) Indeed, if the point A belongs to the union of open sets, then it belongs to at least one of these sets, which, by the conditions of the theorem, is open. This means that it belongs to a certain neighborhood O(a) of the point A, but then this neighborhood also belongs to the union of all open sets. Therefore, the point A is the internal union point. Because A is an arbitrary union point, then it consists only of internal points, and therefore, by definition, is an open set.
2) Let now X– intersection of a finite number of open sets. If A is a set point X, then it belongs to each of the open sets, and, therefore, is an interior point of each of the open sets. In other words, there are intervals that are entirely contained in the sets. Let us denote by the smallest of the numbers. Then the interval will be contained simultaneously in all intervals, i.e. will be entirely contained in , and in ,..., and in , i.e. . From here and we conclude that any point is an interior point of the set X, i.e. many X is open.
From this theorem it follows that the intersection of a finite number of neighborhoods of a point a is again a neighborhood of this point. Note that the intersection of an infinite number of open sets is not always an open set. For example, the intersection of intervals ,... is a set consisting of one point a, which is not an open set (why?).
A point a is called a limit point of a set X if in any punctured neighborhood of this point there is at least one point of the set X.
So, the point is the limit point of the segment , since in any punctured interval of a point there is a point belonging to this segment. For example, a point that satisfies the inequality . And there are obviously many such points.
It is easy to prove that each point of the segment [ 0, 1] is ultimate point of this segment. In other words, the segment consists entirely of its limit points. A similar statement is true for any segment. Note here that all limit points of the set belong to this segment. It is also obvious that all points of the segment will be limit points for the interval (0, 1 ) (prove it!). However, there are already two limiting points 0 and 1 do not belong to the interval (0, 1). In these examples we see that
the limit points of a set may or may not belong to it. It can be proven that in any punctured neighborhood of a limit point a of a set X there are infinitely many points of the set X.
A set X is called a closed set if it contains all its limit points.
So, every segment is a closed set. Interval (0, 1) is not a closed set, since its two limit points do not belong to it 0 and 1. The set of all rational numbers Q is not closed, since it does not contain some of its limit points. In particular, the number is the limit point of the set Q(prove it!), but Q.
Since each point of the set R is the limit point of this set and belongs to it, then R – closed set.
Every finite set is closed, since the set of its limit points is the empty set Æ , which belongs to the set itself.
Closed sets can be bounded, for example, the segment, and unbounded, for example, the set of real numbers R. True
