THE REAL NUMBERS
Section 2.4 Applications of the Supremum Property
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.
(b) If the suprema or infima of two sets are involved, it is often necessary to establish results in two stages, working with one set at a time. Here is an example.
Suppose thatAandBare nonempty subsets ofR that satisfy the property:
ab for alla2Aand allb2B:
We will prove that
supAinfB:
For, givenb2B, we haveabfor alla2A. This means thatbis an upper bound ofA, so that supAb. Next, since the last inequality holds for allb2B, we see that the number supAis a lower bound for the setB. Therefore, we conclude that supAinfB. &
Functions
The idea of upper bound and lower bound is applied to functions by considering the range of a function. Given a function f :D!R, we say that f is bounded aboveif the set fðDÞ ¼ff xð Þ:x2Dgis bounded above inR; that is, there existsB2Rsuch thatfðxÞ Bfor allx2D. Similarly, the functionfisbounded belowif the setf(D) is bounded below.
We say thatfisboundedif it is bounded above and below; this is equivalent to saying that there existsB2R such thatjf xð Þj Bfor allx2D.
The following example illustrates how to work with suprema and infima of functions.
2.4.2 Examples Suppose thatfandg are real-valued functions with common domain DR. We assume thatfandgare bounded.
(a) IffðxÞ gðxÞfor allx2D, then supf Dð Þ supg Dð Þ, which is sometimes written:
sup
x2DfðxÞ sup
x2DgðxÞ:
We first note thatfðxÞ gðxÞ supgðDÞ, which implies that the number supg(D) is an upper bound forf(D). Therefore, supfðDÞ supgðDÞ.
(b) We note that the hypothesisfðxÞ gðxÞfor allx2Din part (a) does not imply any relation between supf(D) and infg(D).
For example, iffðxÞ:¼x2andgðxÞ:¼xwithD¼fx:0x1g, thenfðxÞ gðxÞ for allx2D. However, we see that supfðDÞ ¼1 and infgðDÞ ¼0. Since supgðDÞ ¼1, the conclusion of (a) holds.
(c) IffðxÞ gðyÞfor allx,y2D, then we may conclude that supfðDÞ infgðDÞ, which we may write as:
sup
x2DfðxÞ inf
y2DgðyÞ:
(Note that the functions in (b) do not satisfy this hypothesis.)
The proof proceeds in two stages as in Example 2.4.l(b). The reader should write out
the details of the argument. &
Further relationships between suprema and infima of functions are given in the exercises.
The Archimedean Property
Because of your familiarity with the setRand the customary picture of the real line, it may seem obvious that the setNof natural numbers isnotbounded inR. How can we prove this 2.4 APPLICATIONS OF THE SUPREMUM PROPERTY 41
‘‘obvious’’ fact? In fact, we cannot do so by using only the Algebraic and Order Properties given in Section 2.1. Indeed, we must use the Completeness Property ofRas well as the Inductive Property ofN (that is, if n2N, thennþ12N).
The absence of upper bounds forNmeans that given any real numberxthere exists a natural numbern(depending on x) such thatx<n.
2.4.3 Archimedean Property If x2R,then there exists nx2Nsuch that xnx. Proof. If the assertion is false, thennxfor alln2N; therefore,xis an upper bound ofN. Therefore, by the Completeness Property, the nonempty setNhas a supremumu2R. Subtracting 1 fromugives a numberu1, which is smaller than the supremumuofN. Thereforeu1 is not an upper bound ofN, so there existsm2Nwithu1<m. Adding 1 givesu<mþ1, and sincemþ12N, this inequality contradicts the fact thatuis an
upper bound ofN. Q.E.D.
2.4.4 Corollary If S:¼ f1=n:n2Ng,then infS¼0.
Proof. SinceS6¼ ;is bounded below by 0, it has an infimum and we letw:¼infS. It is clear thatw0. For anye>0, the Archimedean Property implies that there existsn2N such that 1=e<n, which implies 1=n<e. Therefore we have
0w1=n<e:
But sincee>0 is arbitrary, it follows from Theorem 2.1.9 thatw¼0. Q.E.D.
2.4.5 Corollary If t>0, there exists nt2N such that0<1=nt<t.
Proof. Since inf 1f =n:n2Ng ¼0 and t>0, thent is not a lower bound for the set f1=n:n2Ng. Thus there existsnt2N such that 0<1=nt<t. Q.E.D.
2.4.6 Corollary If y>0,there exists ny2N such that ny1yny.
Proof. The Archimedean Property ensures that the subsetEy:¼fm2N:y<mgof N is not empty. By the Well-Ordering Property 1.2.1, Ey has a least element, which we denote by ny. Then ny1 does not belong to Ey, and hence we have
ny1y<ny. Q.E.D.
Collectively, the Corollaries 2.4.4–2.4.6 are sometimes referred to as the Archimedean Property ofR.
The Existence of ffiffiffi p2
The importance of the Supremum Property lies in the fact that it guarantees the existence of real numbers under certain hypotheses. We shall make use of it in this way many times. At the moment, we shall illustrate this use by proving the existence of a positive real numberx such thatx2¼2; that is, the positive square root of 2. It was shown earlier (see Theorem 2.1.4) that such anxcannot be a rational number; thus, we will be deriving the existence of at least one irrational number.
2.4.7 Theorem There exists a positive real number x such that x2¼2.
Proof. Let S:¼s2R :0s;s2<2
. Since 12S, the set is not empty. Also, S is bounded above by 2, because ift>2, thent2>4 so thatt2= S. Therefore the Supremum Property implies that the setShas a supremum inR, and we letx:¼supS. Note thatx>1.
We will prove thatx2¼2 by ruling out the other two possibilities:x2<2 andx2>2.
First assume thatx2<2. We will show that this assumption contradicts the fact that x¼supSby finding ann2Nsuch thatxþ1=n2S, thus implying thatxis not an upper bound forS. To see how to choosen, note that 1=n21=nso that
xþ1 n
2
¼x2þ2x n þ 1
n2x2þ1
nð2xþ1Þ:
Hence if we can choose nso that 1
nð2xþ1Þ<2x2;
then we getðxþ1=nÞ2<x2þ ð2x2Þ ¼2. By assumption we have 2x2>0, so that ð2x2Þ=ð2xþ1Þ>0. Hence the Archimedean Property (Corollary 2.4.5) can be used to obtain n2N such that
1
n<2x2 2xþ1:
These steps can be reversed to show that for this choice ofnwe havexþ1=n2S, which contradicts the fact that xis an upper bound ofS. Therefore we cannot havex2<2.
Now assume thatx2>2. We will show that it is then possible to findm2Nsuch that x1=mis also an upper bound ofS, contradicting the fact thatx¼supS. To do this, note that
x1 m
2
¼x22x mþ 1
m2>x22x m: Hence if we can choose mso that
2x
m <x22;
thenðx1=mÞ2>x2 ðx22Þ ¼2. Now by assumption we havex22>0, so that ðx22Þ=2x>0. Hence, by the Archimedean Property, there existsm2N such that
1
m<x22 2x :
These steps can be reversed to show that for this choice ofmwe have ðx1=mÞ2>2.
Now if s2S, then s2<2<ðx1=mÞ2, whence it follows from 2.1.13(a) that s<x1=m. This implies that x1/m is an upper bound forS, which contradicts the fact thatx¼supS. Therefore we cannot havex2>2.
Since the possibilities x2<2 and x2>2 have been excluded, we must have
x2¼2. Q.E.D.
By slightly modifying the preceding argument, the reader can show that ifa>0, then there is a uniqueb>0 such thatb2¼a. We callbthepositive square rootofaand denote it byb¼ ffiffiffi
pa
orb¼a1=2. A slightly more complicated argument involving the binomial 2.4 APPLICATIONS OF THE SUPREMUM PROPERTY 43
theorem can be formulated to establish the existence of a uniquepositiventh rootofa, denoted by ffiffiffi
a pn
ora1=n, for eachn2N.
Remark If in the proof of Theorem 2.4.7 we replace the set S by the set of rational numbersT :¼ fr2Q :0r; r2 <2g, the argument then gives the conclusion thaty:¼ supT satisfies y2¼2. Since we have seen in Theorem 2.1.4 thatycannot be a rational number, it follows that the setTthat consists of rational numbers does not have a supremum belonging to the setQ. Thus the ordered fieldQ of rational numbers doesnotpossess the Completeness Property.
Density of Rational Numbers inR
We now know that there exists at least one irrational real number, namely ffiffiffi p2
. Actually there are ‘‘more’’ irrational numbers than rational numbers in the sense that the set of rational numbers is countable (as shown in Section 1.3), while the set of irrational numbers is uncountable (see Section 2.5). However, we next show that in spite of this apparent disparity, the set of rational numbers is ‘‘dense’’ inRin the sense that given any two real numbers there is a rational number between them (in fact, there are infinitely many such rational numbers).
2.4.8 The Density Theorem If x and y are any real numbers with x<y, then there exists a rational number r2Q such that x<r<y.
Proof. It is no loss of generality (why?) to assume thatx>0. Sinceyx>0, it follows from Corollary 2.4.5 that there exists n2N such that 1=n<yx. Therefore, we have nxþ1<ny. If we apply Corollary 2.4.6 to nx>0, we obtain m2N with m1nx<m. Therefore,mnxþ1<ny, whencenx<m<ny. Thus, the rational
numberr:¼m=nsatisfies x<r<y. Q.E.D.
To round out the discussion of the interlacing of rational and irrational numbers, we have the same ‘‘betweenness property’’ for the set of irrational numbers.
2.4.9 Corollary If x and y are real numbers with x<y,then there exists an irrational number z such that x<z<y.
Proof. If we apply the Density Theorem 2.4.8 to the real numbersx= ffiffiffi p2
andy= ffiffiffi p2
, we obtain a rational numberr6¼0 (why?) such that
xffiffiffi
p2<r< yffiffiffi p2: Thenz:¼r ffiffiffi
p2
is irrational (why?) and satisfiesx<z<y. Q.E.D.
Exercises for Section 2.4
1. Show that supf11=n:n2Ng ¼1.
2. IfS:¼ f1=n1=m:n;m2Ng, find infSand supS.
3. LetSRbe nonempty. Prove that if a numberuinRhas the properties: (i) for everyn2Nthe numberu1=nis not an upper bound ofS, and (ii) for every numbern2Nthe numberuþ1=n is an upper bound ofS, thenu¼supS. (This is the converse of Exercise 2.3.9.)
4. LetSbe a nonempty bounded set inR.
(a) Leta>0, and letaS:¼ fas:s2Sg. Prove that
infðaSÞ ¼ainfS; supðaSÞ ¼asupS:
(b) Letb<0 and letbS¼ fbs:s2Sg. Prove that
infðbSÞ ¼bsupS; supðbSÞ ¼binfS:
5. LetSbe a set of nonnegative real numbers that is bounded above and letT:¼ fx2:x2Sg.
Prove that ifu¼supS, thenu2¼supT. Give an example that shows the conclusion may be false if the restriction against negative numbers is removed.
6. LetX be a nonempty set and letf :X!R have bounded range in R. Ifa2R, show that Example 2.4.l(a) implies that
supfaþfðxÞ:x2Xg ¼aþsupffðxÞ:x2Xg:
Show that we also have
inffaþfðxÞ:x2Xg ¼aþinfffðxÞ:x2Xg:
7. LetAandBbe bounded nonempty subsets ofR, and letAþB:¼ faþb:a2A;b2Bg. Prove that supðAþBÞ ¼supAþsupBand infðAþBÞ ¼infAþinfB.
8. LetXbe a nonempty set, and letfandgbe defined onXand have bounded ranges inR. Show that supffðxÞ þgðxÞ:x2Xg supffðxÞ:x2Xg þsupfgðxÞ:x2Xg
and that
infffðxÞ:x2Xg þinffgðxÞ:x2Xg infffðxÞ þgðxÞ:x2Xg:
Give examples to show that each of these inequalities can be either equalities or strict inequalities.
9. LetX¼Y :¼ fx2R:0<x<1g. Defineh:XY !Rbyhðx;yÞ:¼2xþy.
(a) For eachx2X, findfðxÞ:¼supfhðx;yÞ:y2Yg; then find infffðxÞ:x2Xg. (b) For eachy2Y, findgðyÞ:¼inffhðx;yÞ:x2Xg; then find supfgðyÞ:y2Yg. Compare
with the result found in part (a).
10. Perform the computations in (a) and (b) of the preceding exercise for the functionh:XY!R defined by
hðx;yÞ:¼ 0 ifx<y;
1 ifxy:
11. LetXandYbe nonempty sets and leth:XY!Rhave bounded range inR. Letf:X!R andg:Y!R be defined by
fðxÞ:¼supfhðx;yÞ:y2Yg; gðyÞ:¼inffhðx;yÞ:x2Xg:
Prove that
supfgðyÞ:y2Yg infffðxÞ:x2Xg:
We sometimes express this by writing supy inf
x hðx;yÞ inf
x sup
y hðx;yÞ:
Note that Exercises 9 and 10 show that the inequality may be either an equality or a strict inequality.
2.4 APPLICATIONS OF THE SUPREMUM PROPERTY 45
12. LetXandYbe nonempty sets and leth:XY!Rhave bounded range inR. LetF:X!R andG:Y!R be defined by
FðxÞ:¼supfhðx;yÞ:y2Yg; GðyÞ:¼supfhðx;yÞ:x2Xg:
Establish thePrinciple of the Iterated Suprema:
supfhðx;yÞ:x2X;y2Yg ¼supfFðxÞ:x2Xg ¼supfGðyÞ:y2Yg We sometimes express this in symbols by
supx;yhðx;yÞ ¼sup
x sup
y hðx;yÞ ¼sup
y sup
x hðx;yÞ:
13. Given anyx2R, show that there exists aunique n2Zsuch thatn1x<n. 14. Ify>0, show that there existsn2Nsuch that 1=2n<y.
15. Modify the argument in Theorem 2.4.7 to show that there exists a positive real numberysuch thaty2¼3.
16. Modify the argument in Theorem 2.4.7 to show that ifa>0, then there exists a positive real numberzsuch thatz2¼a.
17. Modify the argument in Theorem 2.4.7 to show that there exists a positive real numberusuch thatu3¼2.
18. Complete the proof of the Density Theorem 2.4.8 by removing the assumption thatx>0.
19. Ifu>0 is any real number andx<y, show that there exists a rational numberrsuch that x<ru<y. (Hence the setfru:r2Qgis dense inR.)