THE REAL NUMBERS
Section 2.3 The Completeness Property of R
4. Show thatjxaj<eif and only ifae<x<aþe.
5. Ifa<x<banda<y<b, show thatjxyj<ba. Interpret this geometrically.
6. Find allx2R that satisfy the following inequalities:
(a) j4x5j 13; (b) jx21j 3:
7. Find allx2R that satisfy the equationjxþ1j þ jx2j ¼7.
8. Find all values ofxthat satisfy the following equations:
(a) xþ1¼ j2x1j, (b) 2x1¼ jx5j.
9. Find all values ofxthat satisfy the following inequalities. Sketch graphs.
(a) jx2j xþ1, (b) 3jxj 2x:
10. Find allx2R that satisfy the following inequalities.
(a) jx1j>jxþ1j; (b) jxj þ jxþ1j<2:
11. Sketch the graph of the equationy¼ jxj jx1j.
12. Find allx2R that satisfy the inequality 4<jxþ2j þ jx1j<5.
13. Find allx2R that satisfy bothj2x3j<5 andjxþ1j>2 simultaneously.
14. Determine and sketch the set of pairsðx;yÞinRRthat satisfy:
(a) jxj ¼ jyj; (b) jxj þ jyj ¼1;
(c) jxyj ¼2, (d) jxj jyj ¼2: 15. Determine and sketch the set of pairs (x,y) inRR that satisfy:
(a) jxj jyj; (b) jxj þ jyj 1;
(c) jxyj 2; (d) jxj jyj 2:
16. Lete>0 andd>0, anda2R. Show thatVeðaÞ \VdðaÞandVeðaÞ [VdðaÞareg-neighbor- hoods ofafor appropriate values ofg.
17. Show that ifa;b2R, anda6¼b, then there existe-neighborhoodsUofaandVofbsuch that U\V¼ ;.
18. Show that ifa;b2R then
(a) maxfa;bg ¼12ðaþbþ jabjÞ and minfa;bg ¼12ðaþb jabjÞ:
(b) minfa;b;cg ¼minfminfa;bg;cg:
19. Show that if a;b;c2R, then the ‘‘middle number’’ is midfa;b;cg ¼ minfmaxfa;bg;
maxfb;cg;maxfc;agg.
Suprema and Infima
We now introduce the notions of upper bound and lower bound for a set of real numbers.
These ideas will be of utmost importance in later sections.
2.3.1 Definition LetS be a nonempty subset ofR.
(a) The setSis said to bebounded aboveif there exists a numberu2Rsuch thatsu for all s2S. Each such numberu is called anupper boundofS.
(b) The setSis said to be boundedbelowif there exists a numberw2Rsuch thatws for all s2S. Each such numberwis called alower boundof S.
(c) A set is said to beboundedif it is both bounded above and bounded below. A set is said to beunboundedif it is not bounded.
For example, the setS:¼ fx2R:x<2gis bounded above; the number 2 and any number larger than 2 is an upper bound ofS. This set has no lower bounds, so that the set is not bounded below. Thus it is unbounded (even though it is bounded above).
If a set has one upper bound, then it has infinitely many upper bounds, because ifuis an upper bound ofS, then the numbersuþ1;uþ2;. . .are also upper bounds ofS. (A similar observation is valid for lower bounds.)
In the set of upper bounds ofSand the set of lower bounds ofS, we single out their least and greatest elements, respectively, for special attention in the following definition. (See Figure 2.3.1.)
2.3.2 Definition LetS be a nonempty subset ofR.
(a) IfSis bounded above, then a numberuis said to be asupremum(or aleast upper bound) ofS if it satisfies the conditions:
(1) uis an upper bound ofS, and
(2) ifvis any upper bound of S, thenuv.
(b) IfSis bounded below, then a numberwis said to be aninfimum(or agreatest lower bound) ofS if it satisfies the conditions:
(10) wis a lower bound ofS, and
(20) iftis any lower bound ofS, thentw.
It is not difficult to see thatthere can be only one supremum of a given subset S ofR. (Then we can refer tothesupremum of a set instead ofasupremum.) For, suppose thatu1
andu2are both suprema ofS. Ifu1<u2, then the hypothesis thatu2is a supremum implies thatu1 cannot be an upper bound ofS. Similarly, we see thatu2<u1 is not possible.
Therefore, we must have u1¼u2. A similar argument can be given to show that the infimum of a set is uniquely determined.
If the supremum or the infimum of a setS exists, we will denote them by supS and infS:
Figure 2.3.1 infSand supS
2.3 THE COMPLETENESS PROPERTY OFR 37
We also observe that ifu0is an arbitrary upper bound of a nonempty setS, then supSu0. This is because supS is the least of the upper bounds ofS.
First of all, it needs to be emphasized that in order for a nonempty setSinRto have a supremum, it must have an upper bound. Thus, not every subset ofR has a supremum;
similarly, not every subset ofR has an infimum. Indeed, there are four possibilities for a nonempty subset Sof R: it can
(i) have both a supremum and an infimum, (ii) have a supremum but no infimum, (iii) have an infimum but no supremum, (iv) have neither a supremum nor an infimum.
We also wish to stress that in order to show thatu¼supSfor some nonempty subsetS ofR, we need to show thatboth(1) and (2) of Definition 2.3.2(a) hold. It will be instructive to reformulate these statements.
The definition ofu¼supSasserts thatuis an upper bound ofSsuch thatuvfor any upper boundvofS. It is useful to have alternative ways of expressing the idea thatuis the
‘‘least’’ of the upper bounds ofS. One way is to observe that any numbersmallerthanuis notan upper bound ofS. That is, ifz<u, thenzisnotan upper bound ofS. But to say thatz is not an upper bound ofSmeans there exists an elementszinSsuch thatz<sz. Similarly, ife>0, thenueis smaller thanuand thus fails to be an upper bound of S.
The following statements about an upper bounduof a setS are equivalent:
(1) ifvis any upper bound of S, thenuv, (2) ifz<u, thenzis not an upper bound ofS, (3) ifz<u, then there existssz2S such thatz<sz, (4) ife>0, then there existsse2S such thatue<se. Therefore, we can state two alternate formulations for the supremum.
2.3.3 Lemma A number u is the supremum of a nonempty subset S ofRif and only if u satisfies the conditions:
(1) su for all s2S,
(2) if v<u,then there exists s02S such that v<s0.
For future work with limits, it is useful to have this condition expressed in terms of e>0. This is done in the next lemma.
2.3.4 Lemma An upper bound u of a nonempty set S inRis the supremum of S if and only if for every e>0 there exists an se2S such that ue<se.
Proof. Ifuis an upper bound ofSthat satisfies the stated condition and ifv<u, then we pute:¼uv. Thene>0, so there existsse2Ssuch thatv¼ue<se. Therefore,vis not an upper bound of S, and we conclude thatu¼supS.
Conversely, suppose thatu¼supSand lete>0. Sinceue<u, thenueis not an upper bound of S. Therefore, some element se of S must be greater thanue; that is,
ue<se. (See Figure 2.3.2.) Q.E.D.
It is important to realize that the supremum of a set may or may not be an element of the set. Sometimes it is and sometimes it is not, depending on the particular set. We consider a few examples.
2.3.5 Examples (a) If a nonempty setS1has a finite number of elements, then it can be shown thatS1has a largest elementuand a least elementw. Thenu¼supS1andw¼infS1,
and they are both members ofS1. (This is clear ifS1has only one element, and it can be proved by induction on the number of elements inS1; see Exercises 12 and 13.) (b) The setS2 :¼ fx:0x1gclearly has 1 for an upper bound. We prove that 1 is its supremum as follows. Ifv<1, there exists an elements02 S2such thatv<s0. (Name one such elements0.) Thereforevis not an upper bound ofS2and, sincevis an arbitrary number v<1, we conclude that supS2¼1. It is similarly shown that infS2¼0. Note that both the supremum and the infimum ofS2are contained inS2.
(c) The set S3:¼ fx:0<x<1g clearly has 1 for an upper bound. Using the same argument as given in (b), we see that supS3¼1. In this case, the setS3doesnotcontain its supremum. Similarly, inf S3¼0 is not contained inS3. &
The Completeness Property ofR
It is not possible to prove on the basis of the field and order properties of R that were discussed in Section 2.1 that every nonempty subset of R that is bounded above has a supremum inR. However, it is a deep and fundamental property of the real number system that this is indeed the case. We will make frequent and essential use of this property, especially in our discussion of limiting processes. The following statement concerning the existence of suprema is our final assumption aboutR. Thus, we say thatR is acomplete ordered field.
2.3.6 The Completeness Property ofR Every nonempty set of real numbers that has an upper bound also has a supremum inR.
This property is also called theSupremum PropertyofR. The analogous property for infima can be deduced from the Completeness Property as follows. Suppose that Sis a nonempty subset ofRthat is bounded below. Then the nonempty setS:¼ sf :s2Sgis bounded above, and the Supremum Property implies thatu:¼supSexists inR. The reader should verify in detail that –uis the infimum ofS.
Exercises for Section 2.3
1. Let S1:¼ fx2R:x0g. Show in detail that the setS1has lower bounds, but no upper bounds. Show that infS1¼0.
2. LetS2:¼ fx2R:x>0g. DoesS2have lower bounds? DoesS2have upper bounds? Does infS2exist? Does supS2exist? Prove your statements.
3. Let S3¼ f1=n:n2 NÞ. Show that sup S3 ¼ 1 and inf S3 0. (It will follow from the Archimedean Property in Section 2.4 that infS3¼0.)
4. LetS4:¼f1 1ð Þn=n:n2Ng. Find infS4and supS4. Figure 2.3.2 u¼supS
2.3 THE COMPLETENESS PROPERTY OFR 39
5. Find the infimum and supremum, if they exist, of each of the following sets.
(a) A:¼fx2R: 2xþ5>0g; (b) B:¼x2R: xþ2x2
; (c) C:¼fx2R:x<1=xg; (d) D:¼x2R:x22x5<0
: 6. LetSbe a nonempty subset ofRthat is bounded below. Prove that infS¼ supfs:s2Sg.
7. If a setSRcontains one of its upper bounds, show that this upper bound is the supremum ofS. 8. LetSRbe nonempty. Show thatu2Ris an upper bound ofSif and only if the conditions
t2Randt>uimply thatt2=S.
9. Let SR be nonempty. Show that ifu¼supS, then for every numbern2N the number u1=nis not an upper bound ofS, but the number uþ1=nis an upper bound ofS. (The converse is also true; see Exercise 2.4.3.)
10. Show that if A and B are bounded subsets of R, then A[B is a bounded set. Show that supðA[BÞ ¼sup supf A;supBg.
11. Let S be a bounded set in R and let S0 be a nonempty subset of S. Show that infSinfS0supS0supS.
12. Let SR and suppose that s:¼supS belongs to S. If u2=S, show that supðS[ fugÞ ¼supfs;ug.
13. Show that a nonempty finite set SR contains its supremum. [Hint: Use Mathematical Induction and the preceding exercise.]
14. LetSbe a set that is bounded below. Prove that a lower boundwofSis the infimum ofSif and only if for anye>0 there existst2Ssuch thatt<wþe.