Definition 1. A point x of an infinite line is called a limit point of the sequence (x n) if in any e-neighborhood of this point there are infinitely many elements of the sequence (x n).

Lemma 1. If x is a limit point of the sequence (x k ), then from this sequence we can select a subsequence (x n k ), converging to the number x.

Comment. The opposite statement is also true. If from the sequence (x k) it is possible to select a subsequence converging to the number x, then the number x is the limit point of the sequence (x k). Indeed, in any e-neighborhood of the point x there are infinitely many elements of the subsequence, and therefore of the sequence itself (x k ).

From Lemma 1 it follows that we can give another definition of the limit point of a sequence, equivalent to Definition 1.

Definition 2. A point x of an infinite line is called a limit point of a sequence (x k), if from this sequence it is possible to select a subsequence converging to x.

Lemma 2. Every convergent sequence has only one limit point, which coincides with the limit of that sequence.

Comment. If the sequence converges, then by Lemma 2 it has only one limit point. However, if (xn) is not convergent, then it can have several limit points (and, in general, infinitely many limit points). Let us show, for example, that (1+(-1) n ) has two limit points.

Indeed, (1+(-1) n )=0,2,0,2,0,2,... has two limit points 0 and 2, because subsequences (0)=0,0,0,... and (2)=2,2,2,... of this sequence have limits of numbers 0 and 2, respectively. This sequence has no other limit points. Indeed, let x be any point on the number axis other than points 0 and 2. Let us take e >0 so

small so that e - neighborhoods of points 0, x and 2 do not intersect. The e-neighborhood of points 0 and 2 contains all the elements of the sequence and therefore the e-neighborhood of point x cannot contain infinitely many elements (1+(-1) n ) and therefore is not a limit point of this sequence.

Theorem. Every bounded sequence has at least one limit point.

Comment. No number x exceeding , is a limiting point of the sequence (x n), i.e. - the largest limit point of the sequence (x n).

Let x be any number greater than . Let us choose e>0 so small that

and x 1 О(x), to the right of x 1 there are a finite number of elements of the sequence (x n) or there are none at all, i.e. x is not a limit point of the sequence (x n ).

Definition. The largest limit point of the sequence (x n) is called the upper limit of the sequence and is denoted by the symbol. It follows from the remark that every bounded sequence has an upper limit.

Similarly, the concept of a lower limit is introduced (as the smallest limit point of the sequence (x n )).

So, we have proven the following statement. Every bounded sequence has upper and lower limits.

Let us formulate the following theorem without proof.

Theorem. In order for the sequence (x n) to be convergent, it is necessary and sufficient that it be bounded and that its upper and lower limits coincide.

The results of this section lead to the following main theorem of Bolzano-Weierstrass.

Bolzano-Weierstrass theorem. From any bounded sequence one can select a convergent subsequence.

Proof. Since the sequence (x n ) is bounded, it has at least one limit point x. Then from this sequence we can select a subsequence converging to the point x (follows from Definition 2 of the limit point).

Comment. From any bounded sequence one can isolate a monotonic convergent sequence.

Definition v.7. A point x € R on the number line is called a limit point of a sequence (xn) if for any neighborhood U (x) and any natural number N it is possible to find an element xn belonging to this neighborhood with a number greater than LG, i.e. x 6 R - limit point if. In other words, a point x will be a limit point for (xn) if any of its neighborhoods contains elements of this sequence with arbitrarily large numbers, although perhaps not all elements with numbers n > N. Therefore, the following statement is quite obvious. Statement b.b. If lim(xn) = 6 6 R, then b is the only limit point of the sequence (xn). Indeed, by virtue of Definition 6.3 of the limit of a sequence, all its elements, starting from a certain number, fall into any arbitrarily small neighborhood of point 6, and therefore elements with arbitrarily large numbers cannot fall into the neighborhood of any other point. Consequently, the condition of definition 6.7 is satisfied only for a single point 6. However, not every limit point (sometimes called a thin condensed point) of a sequence is its limit. Thus, the sequence (b.b) has no limit (see example 6.5), but has two limit points x = 1 and x = - 1. The sequence ((-1)pp) has two infinite points +oo and as limit points -with the extended number line, the union of which is denoted by one symbol oo. That is why we can assume that the infinite limit points coincide, and the infinite point oo, according to (6.29), is the limit of this sequence. Limit points of the sequence number line. Proof of the Weierstrass test and the Cauchy criterion. Let the sequence (jn) be given and let the numbers k form an increasing sequence of positive integers. Then the sequence (Vnb where yn = xkn> is called a subsequence of the original sequence. Obviously, if (i„) has the number 6 as a limit, then any of its subsequences has the same limit, since starting from a certain number all elements of both the original sequence and any of its subsequences fall into any chosen neighborhood of point 6. At the same time, any limit point of a subsequence is also a limit point for the sequence. Theorem 9. From any sequence that has a limit point, one can choose a subsequence that has this limit point as its limit. Let b be the limit point of the sequence (xn), then, according to Definition 6. 7 limit point, for each n there is an element belonging to the neighborhood U (6, 1/n) of point b of radius 1/n. The subsequence composed of points ijtj, ...1 ..., where zjfcn€U(6, 1/n) Vn 6 N, has a limit at point 6. Indeed, for an arbitrary e > 0, one can choose N such that. Then all elements of the subsequence, starting with the number km, will fall into the ^-neighborhood U(6, e) of point 6, which corresponds to condition 6.3 of the definition of the limit of the sequence. The converse theorem is also true. Limit points of the sequence number line. Proof of the Weierstrass test and the Cauchy criterion. Theorem 8.10. If some sequence has a subsequence with limit 6, then b is the limit point of this sequence. From Definition 6.3 of the limit of a sequence it follows that, starting from a certain number, all elements of the subsequence with limit b fall into a neighborhood U(b, ​​e) of arbitrary radius e. Since the elements of the subsequence are simultaneously elements of the sequence (xn)> the elements xn fall within this neighborhood with as many arbitrarily large numbers, and this, by virtue of Definition 6.7, means that b is the limit point of the sequence (n). Remark 0.2. Theorems 6.9 and 6.10 are also valid in the case when the limit point is infinite, if, when proving the merto neighborhood of U(6, 1 /n), we consider the neighborhood (or neighborhoods). The condition under which a convergent subsequence can be isolated from a sequence is established by the following theorem. Theorem 6.11 (Bolzano - Weierstrass). Every bounded sequence contains a subsequence converging to a finite limit. Let all the elements of the sequence (an) be between the numbers a and 6, i.e. xn € [a, b] Vn € N. Divide the segment [a , b] in half. Then at least one of its halves will contain an infinite number of elements of the sequence, since otherwise the entire segment [a, b] would contain a finite number of them, which is impossible. Let ] be that of the halves of the segment [a , 6], which contains an infinite set of elements of the sequence (zn) (or if both halves are such, then any of them). Similarly, from the segment containing an infinite set of elements of the sequence, etc. Continuing this process, we will construct a system of nested segments with bn - an = (6- a)/2P. According to the principle of nested segments, there is a point x that belongs to all these segments. This point will be the limit point for the sequence (xn) - In fact, for any e-neighborhood U(x, e) = (xx + e) ​​point x there is a segment C U(x, e) (it is enough just to choose n from the inequality (, containing an infinite number of elements of the sequence (sn). According to definition 6.7, x is the limit point of this sequence. Then, by Theorem 6.9, there is a subsequence converging to the point x. The method of reasoning used in the proof of this theorem (it is sometimes called the Bolzano-Weyer-Strass lemma) and associated with the sequential bisection of the segments under consideration is known as the Bolzano method. This theorem greatly simplifies the proof of many complex theorems. It allows you to prove a number of key theorems in a different (sometimes simpler) way. Appendix 6.2. Proof of the Weierstrass test and the Cauchy criterion First, we prove Statement 6.1 (Weierstrass test for the convergence of a bounded monotonic sequence). Let us assume that the sequence (jn) is non-decreasing. Then the set of its values ​​is bounded above and, by Theorem 2.1, has a supremum which we denote by sup(xn) be R. Due to the properties of the supremum (see 2.7) Limit points of the sequence are the number line. Proof of the Weierstrass test and the Cauchy criterion. According to Definition 6.1 for a non-decreasing sequence we have or Then > Ny and taking into account (6.34) we obtain that corresponds to Definition 6.3 of the limit of the sequence, i.e. 31im(sn) and lim(xn) = 66R. If the sequence (xn) is non-increasing, then the course of the proof is similar. Now let's move on to proving the sufficiency of the Kochia criterion for the convergence of a sequence (see Statement 6.3), since the necessity of the criterion condition follows from Theorem 6.7. Let the sequence (jn) be fundamental. According to Definition 6.4, given an arbitrary € > 0, one can find a number N(s) such that m^N and n^N imply. Then, taking m - N, for Vn > N we obtain € £ Since the sequence under consideration has a finite number of elements with numbers not exceeding N, it follows from (6.35) that the fundamental sequence is bounded (for comparison, see the proof of Theorem 6.2 on the boundedness of a convergent sequence ). For a set of values ​​of a bounded sequence, there are infimum and supremum bounds (see Theorem 2.1). For the set of element values ​​for n > N, we denote these faces an = inf xn and bjy = sup xn, respectively. As N increases, the exact infimum does not decrease, and the exact supremum does not increase, i.e. . Do I get an air conditioning system? segments According to the principle of nested segments, there is a common point that belongs to all segments. Let's denote it by b. Thus, with From comparison (6. 36) and (6.37) as a result we obtain that corresponds to Definition 6.3 of the limit of the sequence, i.e. 31im(x„) and lim(sn) = 6 6 R. Bolzano began to study fundamental sequences. But he did not have a rigorous theory of real numbers, and therefore he was unable to prove the convergence of the fundamental sequence. Cauchy did this, taking for granted the principle of nested segments, which Cantor later substantiated. Not only is the criterion for the convergence of a sequence given the name Cauchy, but the fundamental sequence is often called the Cauchy sequence, and the principle of nested segments is named after Cantor. Questions and tasks 8.1. Prove that: 6.2. Give examples of nonconvergent sequences with elements belonging to the sets Q and R\Q. 0.3. Under what conditions do the terms of arithmetic and geometric progressions form decreasing and increasing sequences? 6.4. Prove the relationships that follow from table. 6.1. 6.5. Construct examples of sequences tending to the infinite points +oo, -oo, oo, and an example of a sequence converging to the point 6 € R. c.v. Can an unbounded sequence not be b.b.? If yes, then give an example. at 7. Construct an example of a divergent sequence consisting of positive elements that has neither a finite nor an infinite limit. 6.8. Prove the convergence of the sequence (jn) given by the recurrent formula sn+i = sin(xn/2) under the condition “1 = 1. 6.9. Prove that lim(xn)=09 if sn+i/xn-»g€ .

Divide the segment [ a 0 ,b 0 ] in half into two equal segments. At least one of the resulting segments contains an infinite number of terms of the sequence. Let's denote it [ a 1 ,b 1 ] .

In the next step, we will repeat the procedure with the segment [ a 1 ,b 1 ]: divide it into two equal segments and choose from them the one on which an infinite number of terms of the sequence lie. Let's denote it [ a 2 ,b 2 ] .

Continuing the process we obtain a sequence of nested segments

in which each subsequent one is half of the previous one, and contains an infinite number of terms of the sequence ( x k } .

The lengths of the segments tend to zero:

By virtue of the Cauchy-Cantor principle of nested segments, there is a single point ξ that belongs to all segments:

By construction on each segment [a m ,b m ] there is an infinite number of terms of the sequence. Let's choose sequentially

while observing the condition of increasing numbers:

Then the subsequence converges to the point ξ. This follows from the fact that the distance from to ξ does not exceed the length of the segment containing them [a m ,b m ] , where

Extension to the case of a space of arbitrary dimension

The Bolzano-Weierstrass theorem is easily generalized to the case of a space of arbitrary dimension.

Let a sequence of points in space be given:

(the lower index is the sequence member number, the upper index is the coordinate number). If the sequence of points in space is limited, then each of the numerical sequences of coordinates:

also limited ( - coordinate number).

By virtue of the one-dimensional version of the Bolzano-Weirstrass theorem from the sequence ( x k) we can select a subsequence of points whose first coordinates form a convergent sequence. From the resulting subsequence, we once again select a subsequence that converges along the second coordinate. In this case, convergence along the first coordinate will be preserved due to the fact that every subsequence of a convergent sequence also converges. And so on.

After n we get a certain sequence of steps

which is a subsequence of , and converges along each of the coordinates. It follows that this subsequence converges.

Story

Bolzano-Weierstrass theorem (for the case n= 1) was first proven by the Czech mathematician Bolzano in 1817. In Bolzano's work, it acted as a lemma in the proof of the theorem on intermediate values ​​of a continuous function, now known as the Bolzano-Cauchy theorem. However, these and other results, proven by Bolzano long before Cauchy and Weierstrass, went unnoticed.

Only half a century later, Weierstrass, independently of Bolzano, rediscovered and proved this theorem. Originally called Weierstrass's theorem, before Bolzano's work became known and accepted.

Today this theorem bears the names of Bolzano and Weierstrass. This theorem is often called Bolzano-Weierstrass lemma, and sometimes limit point lemma.

The Bolzano-Weierstrass theorem and the concept of compactness

The Bolzano-Weierstrass theorem establishes the following interesting property of a bounded set: every sequence of points M contains a convergent subsequence.

When proving various propositions in analysis, they often resort to the following technique: they determine a sequence of points that has some desired property, and then select a subsequence from it that also has it, but is already convergent. For example, this is how Weierstrass’s theorem is proved that a function continuous on an interval is bounded and takes its greatest and least values.

The effectiveness of such a technique in general, as well as the desire to extend Weierstrass’s theorem to arbitrary metric spaces, prompted the French mathematician Maurice Fréchet to introduce the concept in 1906 compactness. The property of bounded sets in , established by the Bolzano-Weierstrass theorem, is, figuratively speaking, that the points of the set are located quite “closely” or “compactly”: having made an infinite number of steps along this set, we will certainly come as close as we like to some -some point in space.

Frechet introduces the following definition: set M called compact, or compact, if every sequence of its points contains a subsequence converging to some point of this set. It is assumed that on the set M the metric is defined, that is, it is

A proof of the Bolzano-Weierstrass theorem is given. To do this, the lemma on nested segments is used.

Content

See also: Lemma on nested segments

From any bounded sequence of real numbers it is possible to select a subsequence that converges to a finite number. And from any unbounded sequence - an infinitely large subsequence converging to or to .

The Bolzano-Weierstrass theorem can be formulated this way.

From any sequence of real numbers it is possible to select a subsequence that converges either to a finite number, or to or to .

Proof of the first part of the theorem

To prove the first part of the theorem, we will apply the nested segment lemma.

Let the sequence be bounded. This means that there is a positive number M, so that for all n,
.
That is, all members of the sequence belong to the segment, which we denote as . Here . Length of the first segment. Let's take any element of the sequence as the first element of the subsequence. Let's denote it as .

Divide the segment in half. If its right half contains an infinite number of elements of the sequence, then take the right half as the next segment. Otherwise, let's take the left half. As a result, we get a second segment containing an infinite number of elements of the sequence. The length of this segment. Here, if we took the right half; and - if left. As the second element of the subsequence, we take any element of the sequence belonging to the second segment with a number greater than n 1 . Let's denote it as ().

In this way we repeat the process of dividing the segments. Divide the segment in half. If its right half contains an infinite number of elements of the sequence, then take the right half as the next segment. Otherwise, let's take the left half. As a result, we get a segment containing an infinite number of elements of the sequence. The length of this segment. As an element of the subsequence, we take any element of the sequence belonging to a segment with a number greater than n k.

As a result, we obtain a subsequence and a system of nested segments
.
Moreover, each element of the subsequence belongs to the corresponding segment:
.

Since the lengths of the segments tend to zero as , then according to the lemma on nested segments, there is a unique point c that belongs to all segments.

Let us show that this point is the limit of the subsequence:
.
Indeed, since the points and c belong to a segment of length , then
.
Since, then according to the intermediate sequence theorem,
. From here
.

The first part of the theorem has been proven.

Proof of the second part of the theorem

Let the sequence be unlimited. This means that for any number M, there is an n such that
.

First, consider the case when the sequence is unbounded on the right. That is, for any M > 0 , there exists n such that
.

As the first element of the subsequence, take any element of the sequence greater than one:
.
As the second element of the subsequence, we take any element of the sequence greater than two:
,
and to .
And so on. As the kth element of the subsequence we take any element
,
and .
As a result, we obtain a subsequence, each element of which satisfies the inequality:
.

We enter the numbers M and N M, connecting them with the following relations:
.
It follows that for any number M one can choose a natural number, so that for all natural numbers k >
It means that
.

Now consider the case when the sequence is bounded on the right. Since it is unbounded, it must be left unbounded. In this case, we repeat the reasoning with minor amendments.

We choose a subsequence so that its elements satisfy the inequalities:
.
Then we enter the numbers M and N M, connecting them with the following relations:
.
Then for any number M one can choose a natural number, so that for all natural numbers k > N M the inequality holds.
It means that
.

The theorem has been proven.

See also: