1610912322-b551b095a53deaf3d3fbd1ed05ae9b84 (824701), страница 21
Текст из файла (страница 21)
If the interval I1 could not be covered by a finite set of intervals of thesystem S, then, dividing I1 into two halves, we would find that at least one of thetwo halves, which we denote by I2 , does not admit a finite covering. We now repeatthis procedure with the interval I2 , and so on.In this way a nested sequence I1 ⊃ I2 ⊃ · · · ⊃ In ⊃ · · · of closed intervals arises,none of which admit a covering by a finite subsystem of S. Since the length of theinterval In is |In | = |I1 | · 2−n , the sequence {In } contains intervals of arbitrarilysmall length (see the lemma in Paragraph c of Sect.
2.2.4). But the nested intervaltheorem implies that there exists a point c belonging to all of the intervals In , n ∈ N.Since c ∈ I1 = [a, b] there exists an open interval ]α, β[ = U ∈ S containing c, thatis, α < c < β. Let ε = min{c − α, β − c}. In the sequence just constructed, wefind an interval In such that |In | < ε. Since c ∈ In and |In | < ε, we conclude thatIn ⊂ U = ]α, β[. But this contradicts the fact that the interval In cannot be coveredby a finite set of intervals from the system.2.3.3 The Limit Point Lemma (Bolzano–Weierstrass Principle)We recall that we have defined a neighborhood of a point x ∈ R to be an openinterval containing the point and the δ-neighborhood about x to be the open interval]x − δ, x + δ[.Definition 4 A point p ∈ R is a limit point of the set X ⊂ R if every neighborhoodof the point contains an infinite subset of X.This condition is obviously equivalent to the assertion that every neighborhoodof p contains at least one point of X different from p itself.
(Verify this!)We now give some examples.If X = { n1 ∈ R | n ∈ N}, the only limit point of X is the point 0 ∈ R.For an open interval ]a, b[ every point of the closed interval [a, b] is a limit point,and there are no others.For the set Q of rational numbers every point of R is a limit point; for, as weknow, every open interval of the real numbers contains rational numbers.11 É. Borel (1871–1956) and H. Lebesgue (1875–1941) – well-known French mathematicians whoworked in the theory of functions.2.3 Basic Lemmas on Completeness73Lemma (Bolzano–Weierstrass12 ) Every bounded infinite set of real numbers has atleast one limit point.Proof Let X be the given subset of R.
It follows from the definition of boundednessthat X is contained in some closed interval I ⊂ R. We shall show that at least onepoint of I is a limit point of X.If such were not the case, then each point x ∈ I would have a neighborhoodU (x) containing either no points of X or at most a finite number. The totality ofsuch neighborhoods {U (x)} constructed for the points x ∈ I forms a covering ofI by open intervals U (x).
By the finite covering lemma we can extract a systemU (x1 ), . . . , U (xn ) of open intervals that cover I . But, since X ⊂ I , this same systemalso covers X. However, there are only finitely many points of X in U (xi ), andhence only finitely many in their union. That is, X is a finite set.
This contradictioncompletes the proof.2.3.4 Problems and Exercises1. Show thata) if I is any system of nested closed intervals, thensup a ∈ R | [a, b] ∈ I = α ≤ β = inf b ∈ R | [a, b] ∈ Iand[α, β] =[a, b];[a,b]∈Ib) if I is a system of nested open intervals ]a, b[ the intersectionmay happen to be empty.Hint :]an , bn [=]0, n1 [.]a,b[∈I ]a, b[2. Show thata) from a system of closed intervals covering a closed interval it is not alwayspossible to choose a finite subsystem covering the interval;b) from a system of open intervals covering an open interval it is not alwayspossible to choose a finite subsystem covering the interval;c) from a system of closed intervals covering an open interval it is not alwayspossible to choose a finite subsystem covering the interval.12 B. Bolzano (1781–1848) – Czech mathematician and philosopher.
K. Weierstrass (1815–1897)– German mathematician who devoted a great deal of attention to the logical foundations of mathematical analysis.742The Real Numbers3. Show that if we take only the set Q of rational numbers instead of the completeset R of real numbers, taking a closed interval, open interval, and neighborhood ofa point r ∈ Q to mean respectively the corresponding subsets of Q, then none of thethree lemmas proved above remains true.4. Show that we obtain an axiom system equivalent to the one already given if wetake as the axiom of completenessa) the Bolzano–Weierstrass principleorb) the Borel–Lebesgue principle (Heine–Borel theorem).Hint: The principle of Archimedes and the axiom of completeness in the earlierform both follow from a).c) Replacing the axiom of completeness by the Cauchy–Cantor principle leadsto a system of axioms that becomes equivalent to the original system if we alsopostulate the principle of Archimedes.
(See Problem 21 in Sect. 2.2.2.)2.4 Countable and Uncountable SetsWe now make a small addition to the information about sets that was provided inChap. 1. This addition will be useful below.2.4.1 Countable SetsDefinition 1 A set X is countable if it is equipollent with the set N of natural numbers, that is, card X = card N.Propositiona) An infinite subset of a countable set is countable.b) The union of the sets of a finite or countable system of countable sets is a countable set.Proof a) It suffices to verify that every infinite subset E of N is equipollent with N.We construct the needed bijective mapping f : N → E as follows. There is aminimal element of E1 := E, which we assign to the number 1 ∈ N and denotee1 ∈ E.
The set E is infinite, and therefore E2 := E1 \e1 is nonempty. We assignthe minimal element of E2 to the number 2 and call it e2 ∈ E2 . We then considerE3 := E\{e1 , e2 }, and so forth. Since E is an infinite set, this construction cannotterminate at any finite step with index n ∈ N. As follows from the principle of induction, we assign in this way a certain number en ∈ E to each n ∈ N. The mappingf : N → E is obviously injective.2.4 Countable and Uncountable Sets75It remains to verify that it is surjective, that is, f (N) = E. Let e ∈ E.
The set{n ∈ N | n ≤ e} is finite, and hence the subset of it {n ∈ E | n ≤ e} is also finite. Letk be the number of elements in the latter set. Then by construction e = ek .1 ,...,b) If X1 , . . . , Xn , . . . is a countable system of sets and each setXm = {xmnxm , . . .} is itself countable, then since the cardinality of the set X = n∈N Xn , whichn where m, n ∈ N, is not less than the cardinality of eachconsists of the elements xmn ∈ X can beof the sets Xm , it follows that X is an infinite set. The element xmmidentified with the pair (m, n) of natural numbers that defines it.
Then the cardinalityof X cannot be greater than the cardinality of the set of all such ordered pairs. But themapping f : N × N → N given by the formula (m, n) → (m+n−2)(m+n−1)+ m, as2one can easily verify, is bijective. (It has a visualizable meaning: we are enumeratingthe points of the plane with coordinates (m, n) by successively passing from pointsof one diagonal on which m + n is constant to the points of the next such diagonal,where the sum is one larger.)Thus the set of ordered pairs (m, n) of natural numbers is countable. But thencard X ≤ card N, and since X is an infinite set we conclude on the basis of a) thatcard X = card N.It follows from the proposition just proved that any subset of a countable set iseither finite or countable.
If it is known that a set is either finite or countable, we sayit is at most countable. (An equivalent expression is card X ≤ card N.)We can now assert, in particular, that the union of an at most countable family ofat most countable sets is at most countable.Corollaries1) card Z = card N.2) card N2 = card N.(This result means that the direct product of countable sets is countable.)3) card Q = card N, that is, the set of rational numbers is countable.Proof A rational number mn is defined by an ordered pair (m, n) of integers.
Twopairs (m, n) and (m , n ) define the same rational number if and only if they areproportional. Thus, choosing as the unique pair representing each rational number the pair (m, n) with the smallest possible positive integer denominator n ∈ N,we find that the set Q is equipollent to some infinite subset of the set Z × Z. Butcard Z2 = card N and hence card Q = card N.4) The set of algebraic numbers is countable.Proof We remark first of all that the equality Q × Q = card N implies, by induction,that card Qk = card N for every k ∈ N.An element r ∈ Qk is an ordered set (r1 , . .
. , rk ) of k rational numbers.An algebraic equation of degree k with rational coefficients can be written in thereduced form x k + r1 x k−1 + · · · + rk = 0, where the leading coefficient is 1. Thus762The Real Numbersthere are as many different algebraic equations of degree k as there are differentordered sets (r1 , . . . , rk ) of rational numbers, that is, a countable set.The algebraic equations with rational coefficients (of arbitrary degree) also forma countable set, being a countable union (over degrees) of countable sets. Each suchequation has only a finite number of roots. Hence the set of algebraic numbers is atmost countable. But it is infinite, and hence countable.2.4.2 The Cardinality of the ContinuumDefinition 2 The set R of real numbers is also called the number continuum,13 andits cardinality the cardinality of the continuum.Theorem (Cantor) card N < card R.This theorem asserts that the infinite set R has cardinality greater than that of theinfinite set N.Proof We shall show that even the closed interval [0, 1] is an uncountable set.Assume that it is countable, that is, can be written as a sequence x1 , x2 , .
![](https://s.studizba.com/z.php?f=/templates/si/i/noavatar.png&w=100&h=100&t=1"){kind=avatar}
