P STRUKTUR

Chapter 1

The Real Numbers

1.1 The Completeness Property of R

Example 1.1.1 (Bartle 2.3.5 (a) Page 39). A nonempty set S1 with a finite number of elements willhave a least element, say u, and a largest element, say w. Then, u = inf S1 and w = supS1. Both theinfinimum and supremum, that is u and v, are members of S1.

Example 1.1.2 (Bartle 2.3.5 (b) Page 39). Given S2 = {x : 0 x 1}. Since 0 x 1 for all x S2,we have a lower bound 0, and an upper bound 1.If v < 1 then there exists some s S2 such that v < s, for example s = 1. Thus, v is not an upperbound of S2, and since v < 1 is arbitrary, we conclude that supS2 = 1.If w > 0 then there exists some s S2 such that s < w, for example s = w/2 < w. Thus, w is not alower bound of S2, and since w > 0 is arbitrary, we conclude that inf S2 = 0.Note that the infimum and supremum are members of S2.

Example 1.1.3 (Bartle 2.3.5 (c) Page 39). Given S3 = {x : 0 < x < 1}. Since 0 < x < 1 for all x S3,we have a lower bound 0, and an upper bound 1.If v < 1 there exists some s S3 such that v < s. Thus, v is not an upper bound of S3, and since v < 1is arbitrary, we conclude that supS3 = 1.If w > 0 then there exists some s S3 such that s < w, for example s = w/2 < w. Thus, w is not alower bound of S3, and since w > 0 is arbitrary, we conclude that inf S3 = 0.Note that the infimum and supremum are not members of S3.

Problem 1 (Bartle Exercise 1 for Section 2.3). Let S1 := {x R : x 0}. Show in detail that the set haslower bounds, but no upper bounds. Show that inf S1 = 0.

Since x 0 for all x S1, u = 0 is a lower bound of S1, while S1 has no upper bounds.If v > 0 then there exists some s S1 such that 0 < s < v, for example s = v/2 < v. Thus, v is not alower bound of S1, and since v is arbitrary inf S1 = 0. Also, inf S1 S1.Since S1 has no upper bounds supS1 does not exist.

Problem 2 (Bartle Exercise 2 for Section 2.3). Let S2 := {x R : x > 0}. Does S2 have lower bounds?Does S2 have upper bounds? Does inf S2 exist? Does supS2 exist? Prove your statements.

Since S2 6= and x > 0, the set S2 is bounded below by 0. Then, from the Completeness Property S2also has an infimum. For every > 0, there exists some s S2 such that 0 < s < 0 + ; for example

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s = /2 < . Thus, 0 + is not a lower bound, and since is arbitrary, inf S2 = 0.Since x > 0, the set S2 has no upper bounds, and by definition supS2 does not exist.

Problem 3 (Bartle Exercise 3 for Section 2.3). Let S3 := {1/n : n N}. Show that supS3 = 1 andinf S3 0.

Since 0 < 1/n 1 for all n N, the set S3 has a lower bound 0 and an upper bound 1, and by theCompleteness Property has both an infimum and a supremum.For every > 0 there exists some s S3 such that 1 < s 1; for example s = 1. Thus, 1 is notan upper bound of S3, and since is arbitrary, supS3 = 1.Also, for every > 0 there exists some s S3 such that 0 s 0+ , say s = 0 or s = /2. Thus, 0 + is not a lower bound of S3, and since > 0 is arbitrary, inf S3 = 0

1.

Problem 4 (Bartle Exercise 4 for Section 2.3). Let S4 := {1 (1)n/n : nN}. Find inf S4 and supS4.

Since 1/2 1 (1)n/n 2, the set S4 has a lower bound 1/2 and an upper bound 2, and by theCompleteness Property has an infimum and supremum.For every > 0, there exists some s S4 such that 1/2 s < 1/2 + ; for example s = 1/2 + /2. Thus,1/2 + is not a lower bound, and since > 0 is arbitrary inf S4 = 1/2.For every > 0 there exists s S4 such that 2 < s

2, say s

= 2. Thus, 2 is not an upper boundof S4, and since > 0 is arbitrary, supS4 = 2.

Problem 5 (Bartle Exercise 5(a) for Section 2.3). Let A := {x R : 2x + 5 > 0}. Find the infimum andsupremum, if they exist.

Since 2x+ 5 > 0 we have x > 5/2, and so A has a lower bound 5/2, but no upper bounds. Thus, bythe Completeness Property, A also has an infimum, but the supremum does not exist. For every > 0 thereexists a A such that 5/2 a < 5/2 + , say a = 5/2 + /2. Thus, 5/2 + is not a lower boundof A, and since is arbitrary, inf A = 5/2.

Problem 6 (Bartle Exercise 5(b) for Section 2.3). Todo...

Problem 7 (Bartle Exercise 5(c) for Section 2.3). Todo...

Problem 8 (Bartle Exercise 5(d) for Section 2.3). Todo...

Problem 9 (Bartle Exercise 6 for Section 2.3). Let S be a nonempty subset of R that is bounded below.Prove that inf S = sup{s : s S}.

Let S be a nonempty subset of R that is bounded below. Then, by the Completeness Property of R aninfimum of S exists, say w := inf S. Thus w s for each s S. But this implies that s w for alls S, so that if we define S := {s : s S}, then w is an upper bound for S. We must show that u isthe supremum of S. Let v < w for all v S. Then w < v, and since w = inf S there must exist somet S such that w < t < v. But this implies that there exists some t S such that v < t < w. Thusv is not an upper bound of S, and since v is arbitrary, w = sup S. Hence, inf S = sup{s : s S}.

Problem 10 (Bartle Exercise 7 for Section 2.3). If a set S R contains one of its upper bounds, show thatthis upper bound is the supremum of S.

Let u S be an upper bound of S. If v is any upper bound of S, then s v for all s S. But u S, sou v, and since u is an upper bound then it must be the least upper bound; that is, u := supS.

Problem 11 (Bartle Exercise 8 for Section 2.3). Let S R be nonempty. Show that u R is an upperbound of S if and only if the conditions t R and t > u imply that t / S.

1The question asks to show that inf S3 0. The infimum inf S3 = 0 is for the case of n . If n = 2, then inf S3 = 1/2.Perhaps ask lecturer to just elaborate a little more on this problem at consultation.

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Let S R be nonempty. The proof proceeds as follows:Assume u R is an upper bound of S. Then s u for all s S. Thus, if t > u for all t R, then t > s forall s S. Hence t / S.Conversely, assume that t > u t / S, and suppose, to the contrary, that u is not an upper bound of S.Then, since S is nonempty there must exist some s S such that s > u. But s > u s / S, which is acontradiction. Hence, u is an upper bound of S.

Problem 12 (Bartle Exercise 9 for Section 2.3). Let S R be nonempty. Show that if u = supS, then forevery n N the number u 1/n is not an upper bound of S, but the number u+ 1/n is an upper bound ofS2.

Let S R be a nonempty set, and let u := supS. For all n N we have 1/n > 0, so that u 1/n < ufor all n N. Thus, there must exist some s S such that u 1/n < s < u and so u 1/n is not an upperbound of S. However, u < u+ 1/n for all n N, and since u := supS, we have that s u < u+ 1/n for alls S. Hence, u+ 1/n is an upper bound of S for all n N.

Problem 13 (Bartle Exercise 10 for Section 2.3). Show that if A and B are bounded subsets of R, thenA B is a bounded set. Show that sup(A B) = sup{supA, supB}.

Let A R and B R. Let u := supA and v := supB. Also, let w := sup{u, v}.For all a A and all b B we have a u w and b v w respectively. Thus w is an upperbound for A B. We need to show that w is the supremum (least upper bound) of A B. If z is anyupper bound of A B, then z is also an upper bound of A and B; that is, u z and v z. Hencew z, and since z is any upper bound of A B, w must be the least upper bound, and we conclude thatsup(A B) = sup{supA, supB}.

Problem 14 (Bartle Exercise 11 for Section 2.3). Let S be a bounded set in R and let S0 be a nonemptysubset of S. Show that inf S inf S0 supS.

Since the set S is bounded, the Completeness Property of R ensures that S has an infimum and asupremum. If s0 S0, then s0 S, so that s0 supS for all s0 S0. This implies that supS0 supS.Similarly, inf S s0 for all s0 S0, so that inf S inf S0. Hence, inf S inf S0 supS0 supS.

Problem 15 (Bartle Exercise 12 for Section 2.3). Let S R and suppose that s := supS belongs to S. Ifu / S, show that sup (S {u}) = sup{s, u}.

Since u / S and s S we have that u 6= s. Thus we consider only the two cases that u < s ands < u.Now, if u < s then s = sup{s, u} since s is an upper bound which is contained in the set {s, u}. Also,for all s S we have that s u < s so that s is an upper bound of the set S {u}, and since s iscontained in this set, s = sup (S {u}).Similarly, if s < u, then u = sup{s, u} and consequently u = sup (S {u}).

Problem 16 (Bartle Exercise 13 for Section 2.3). Todo...


1.2 Applications of the Supremum Property

Example 1.2.1 (Bartle 2.4.1 (a) Page 40). Let S be a nonempty subset of R that is bounded above, andlet a be any number in R. Define the set a+ S := {a+ s : s S}. Show that sup(a+ S) = a+ supS.Solution: Let u := supS. Then for all s S we have a + s a + u, and so a+ u is an upper bound of

2The converse is also true. See Exercise 2.4.3 in Bartle

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the set a+ S. Consequently, sup(a+ S) a+ u.Now, if v is any upper bound of a+ S, then a+ s v for all s S. Thus s v a for all s S, and sov a is an upper bound for S. But since u := supS, we have that u v a, and so a + u v. Sincev is any upper bound of a+ S we have a + u sup(a + S). Now we have that sup(a + S) a + u anda+ u sup(a+ S), which means that sup(a+ S) = a+ supS.

Example 1.2.2 (Bartle 2.4.1 (b) Page 41). Suppose A and B are nonempty subsets of R that satisfy theproperty

a b for all a A, b B

Prove thatsupA supB

Solution: For any given b B we have that a b for all a A. Thus b is an upper bound of A, and sosupA b, where supA is the least upper bound of A. Since supA b holds for all b B, we have thatsupA is a lower bound of the set B. Hence, we conclude that supA supB.

Example 1.2.3 (Bartle 2.4.2 Page 41). Suppose that f and g are real-valued functions with commondomain D R. We assume that f and g are bounded.

1. Show that if f(x) g(x) for all x D, then sup(D) sup g(D). Solution: Firstly, f(x) g(x) sup g(D). This implies that sup g(D) is an upper bound for f(D). Hence, sup f(D) sup g(D).

Example 1.2.4 (Bartle 2.4.2 (b) Page 41). Todo...

Example 1.2.5 (Bartle 2.4.2 (c) Page 41). Todo...

Problem 18 (Bartle Exercise 1 for Section 2.4). Show that sup{1 1/n : n N} = 1.

Let S = {1 1/n : n N}. Since 1 1/n < 1 for all n N we have an upper bound of 1 for S. Thus theCompleteness Property ensures that a supremum of S exists. For every > 0, the Archimedean Propertyimplies that there exists some n N such that 1/ < n. Thus 1/n < so that 1 < 1 1/n < 1.Since 1 1/n S we know that 1 is not an upper bound of S, and since > 0 is arbitrary, supS = 1.

Problem 19 (Bartle Exercise 2 for Section 2.4). If S := {1/n 1/m : n,m N}, find inf S and supS.

Since 1 1/n 1/m 1, the set S has a lower bound 1 and an upper bound 1. By the CompletenesProperty the set S has an infimum and a supremum.Infimum: We have that 1/n 1/m 1/n 1 > 1. Then 1 is indeed a lower bound. For every > 0there exists an n N such that 0 < 1/n < . Thus 1 < 1/n 1 < 1 = 1+ . But 1/n 1 S. Hence,1 + is not a lower bound of S, and since > 0 is arbitrary, inf S = 1.Supremum: The set S has the property S = S. Thus, we can invoke the earlier result inf S = sup{s :s S}. Since inf S = 1 we have 1 = sup{s : s S} = sup(S) = supS. This implies thatsupS = 1.

Problem 20 (Bartle Exercise 3 for Section 2.4). Let S R be nonempty. Prove that if a number u in Rhas the properties: (i) for every n N the number u 1/n is not an upper bound of S, and (ii) for everynumber n N the number u+ 1/n is an upper bound of S, then u = supS.3

3This is the converse of Exercise 2.3.9 in Bartle

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Todo...

Problem 21 (Bartle Exercise 4 (a) for Section 2.4). Let S be a nonempty bounded set in R. Let a > 0, andlet aS := {as : s S}. Prove that

inf(aS) = a inf S, sup(aS) = a supS

Supremum: Let u := supS. Then s u for all s S, so that as au; that is, au is an upper bound forthe set aS. Consequently, sup(aS) au. If v is any upper bound of aS, then as v for all s S, and sos v/a for all s S; that is, v/a is an upper bound for S. Thus, u v/a, so that au v, and since v is anyupper bound of aS, we have au sup(aS). Combining the inequalities sup(aS) au and au sup(aS), wehave

sup(aS) = a supS

Infimum: Todo...

Problem 22 (Bartle Exercise 4 (b) for Section 2.4). Let S be a nonempty bounded set in R. Let b < 0, andlet bS := {bs : s S}. Prove that

inf(bS) = b supS, sup(bS) = b inf S

Since b < 0, we consider b > 0 and apply the results of the previous exercise (Exercise 2.4.4 (a) Bartle).Define S := {s : s S}. Then inf(bS) = inf(b S). But inf(b S) = b inf S, so that inf(bS) = b inf S.Also from an earlier result (Exercise 2.3.6 Bartle) inf S = supS. Thus, inf(bS) = b( supS) and wehave our final result inf(bS) = b supS.Todo sup(bS) = b inf S...

Problem 23 (Bartle Exercise 5 for Section 2.4). Let S be a set of nonnegative real numbers that is boundedabove and let T := {x2 : x S}. Prove that if u = supS, then u2 = supT . Give an example that shows thatthe conclusion may be false if the restriction against negative numbers is removed.

Todo...

Problem 24 (Bartle Exercise 6 for Section 2.4). Let X be a nonempty set and let f : X R have boundedrange in R. If a R, show that Example 2.4.1(a) (in Bartle) implies that

sup{a+ f(x) : x X} = a+ sup{f(x) : x X}

Show that we also haveinf{a+ f(x) : x X} = a+ inf{f(x) : x X}

Supremum: Let u := supX . Then f(x) u for all x X , so that a + f(x) a + u for all x X ; thatis, a+ u is an upper bound of a+ f(x) for all x X . Thus sup{a+ f(x) : x X} a+ u.Todo: Finish...

Problem 25 (Bartle Exercise 7 for Section 2.4). Let A and B be bounded nonempty subsets of R, and letA+B := {a+ b : a A, b B}. Prove that

sup(A+B) = supA+ supB, inf(A+B) = inf A+ inf B

Supremum: Let u := supA and v := supB. The a u for all a A and b v for all b B. Thus,a + b u + v for all a A, b B so that u + v is an upper bound for the set A + B. Hence, from theCompleteness Property, a supremum for A + B exists, and sup(A + B) u + v = supA + supB. It needsto be shown that u + v is the supremum (least upper bound) of A+ B. We can do so by showing that forevery > 0 we can find a s A + B such that (u + v) < s u + v. If we choose a A such thatu/2 < a u and b B such that v/2 < b v then we have (u/2)+(v/2) < a+ b u+v,or simplified

(u+ v) < a + b u+ v

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But a + b A+B, and since > 0 is arbitrary we know that (u+ v) is not an upper bound of A+B,so that u+ v is the supremum and

sup(A+B) = supA+ supB

Problem 26 (Bartle Exercise 8 for Section 2.4). Let X be a nonempty set, and let f and g be defined on Xand have bounded ranges in R. Show that

sup{f(x) + g(x) : x X} sup{f(x) : x X}+ sup{g(X) : x X}

and thatinf{f(x) : x X}+ inf{g(X) : x X} inf{f(x) + g(x) : x X}

Give examples to show that each of these inequalities can be either equalities or strict inequalitles.

Supremum: Let u := sup f(X) and v := sup g(X). Then f(x) u and g(x) v for all x X . Thusf(x) + g(x) u + v for all x X , so that u + v is an upper bound of the set {f(x) + g(x) : x X}. Thisimplies that {f(x) + g(x) : g X} sup{f(x) + g(x) : x X} u+ v. Hence,

sup{f(x) + g(x) : x X} sup{f(x) : x X}+ sup{g(X) : x X}

Infimum: Let u := inf f(X) and v := inf g(X). Then u f(x) and v g(x) for all x X , and consequentlyu + v f(x) + g(x) for all x X . Then u + v is a lower bound of the set {f(x) + g(x) : x X}, whichimplies that u+ v inf{f(x) + g(x) : x X} {f(x) + g(x) : g X}. Hence,

inf{f(x) : x X}+ inf{g(X) : x X} inf{f(x) + g(x) : x X}





Problem 31 (Bartle Exercise 13 for Section 2.4). Given any x R, show that there exists a unique Z suchthat n 1 x < n.

In the trivial case where x = 0 then n = 1 is the unique element of Z. If x > 0, then by an earlier result(Corollary 2.4.5 in Bartle), there exists an nx N such that nx 1 < x < nx. If x < 0, then we can applythe same argument to x > 0. Thus, we know that given an arbitrary x R, there exists some nx Z suchthat nx n < nx.We must show that such an nx Z is unique. Suppose, without loss of generality, that mx < nx alsosatisfies mx 1 x < mx. Also, we know that mx nx 1 since nx,mx Z. We now have the inequalitymx 1 x < mx nx 1 x < nx, which is a contradiction as it implies x < x. Hence, the assumptionthat there also exists mx < nx satisfying mx 1 x < mx is false, and so we have uniqueness.

Problem 32 (Bartle Exercise 14 for Section 2.4). If y > 0, show that there exists n N such that 1/2n < y.

For each n N we have that n < 2n. Thus we also have that 1/2n < 1/n. But from an earlier result(Corollary 2.4.5 Bartle), we know that given y > 0 there exists some ny N such that 0 < 1/ny < y. Hence,for each y > 0 we have that 0 < 1/2n < 1/n < y; that is, 1/2n < y.





Problem 37 (Bartle Exercise 19 for Section 2.4). Since x < y and u > 0, we know that x/u < y/u. Then,by the Density Theorem, there exists some r Q such that x/u < r < y/u. But this implies that x < ru < y.Hence the set {ru : r Q} is dense in R.

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1.3 Intervals

Problem 38 (Bartle Exercise 1 for Section 2.5). If I := [a, b] and I := [a, b] are closed intervals in R,show that I I if and only if a a and b b.

Let I I . Then a, b I so that a a, b b, or a a and b b.Conversely, suppose alea and b b. Then for any x I we have a x b, so that a x b. Thusx I for all x I, and hence, I I .

Problem 39 (Bartle Exercise 2 for Section 2.5). If S R is nonempty, show that S is bounded if and onlyif there exists a closed bounded interval I such that S I.

Let S be bounded by a lower bound a and an upper bound b. Then for any s S we have a s b.Hence s [a, b], and so S I where I := [a, b].Conversely, let S I where I is the closed bound interval I := [a, b]. But then a s b for all s S.Hence, S is bounded.

Problem 40 (Bartle Exercise 3 for Section 2.5). If S R is a nonempty bounded set, and IS := [inf S, supS],show that S Is. Moreover, if J is any closed bounded interval containing S, show that IS J .

The set S is nonempty and bounded, so the Completeness Property ensures that the infimum andsupremum of S exist. Since inf S s supS for all s S, we know that s IS = [inf S, supS] for alls S. Hence, S IS .Since J is any closed bounded interval, let J := [a, b]. If S J then a s b for all s S. Thus a is a lowerbound of S and b is an upper bound of S, and by the Completeness Property, the infimum and supremumof S exist. Thus a inf S and supS b. But IS = [inf S, supS], so by Exercise 2.5.1 Bartle, IS J .

Problem 41 (Bartle Exercise 4 for Section 2.5). Leave.

Problem 42 (Bartle Exercise 5 for Section 2.5). Leave.


Problem 44 (Bartle Exercise 7 for Section 2.5). Let In := [0, 1/n] for n N. Prove that n=1In = {0}.

Since 0 [0, 1/n] for all n N, we know that {0} n=1In. Thus we know that

n=1In is nonemptyand contains {0}. Let x n=1 (x is arbitrary). But this is only true if x [0, 1/n], so that 0 x 1/nfor all n N. We must show that x = 0. From the Archimedean Property implies that for every > 0 thereexists some n N such that 0 < 1/n < . Thus we now have that 0 x 1/n < . But since > 0 isarbitrary, we can make it as small as we like, and hence, x = 0.











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