16 MATH 3333–INTERMEDIATE ANALYSIS–BLECHER NOTESdblecher/3333note2.pdf16 MATH...

16 MATH 3333–INTERMEDIATE ANALYSIS–BLECHER NOTES

2 is a prime factor of n. This contradicts the fact that m and n have no common

prime factors. �

Remark A similar proof shows that√p is irrational for any prime number p.

Homework 1.

See course webpage.

3.3. The completeness axiom. The next bit you also saw in Calculus I and it is

in the official syllabus for the “Transitions” course:

Definitions Let S be a nonempty subset of F. A number M ∈ F is called anupper bound for S if x ≤ M, ∀x ∈ S. If there exists an upper bound for S thenwe say that S is bounded above, otherwise S is unbounded above.

Similarly, a number m ∈ F is a lower bound for S, and the set S is boundedbelow, if m ≤ x, ∀x ∈ S. Otherwise S is unbounded below.

The set S is bounded if it is bounded below and above.

An upper bound β of the subset S, which has the property that β ≤ M forevery upper bound M for S, is called the least upper bound or supremum of S. It

is denoted by lub(S) or sup(S).

Similarly, a lower bound α of S, which has the property that α ≥ m for everylower bound m for S, is called the greatest lower bound or infimum of S. It is

denoted by glb(S) or inf(S).

If an upper bound M is an element of S, then it is called the maximum of S

(largest element), denoted by max(S). Clearly this is the least upper bound of

S, since any other upper bound of S is ≥ max(S) (because max(S) ∈ S). Somax(S) = sup(S) in this case. If no upper bounds of S are in S we say that S has

no maximum, or that no maximum exists.

Similarly, if a lower bound m is an element of S, then it is called the minimum

(least element), denoted by min(S). Clearly this is the greatest lower bound of S,

so min(S) = inf(S) in this case.

Remarks. 1) It is not obvious that least upper bounds or greatest lower bounds

exist. In fact they do not in the field F = Q.

2) Any nonempty subset T of a bounded set S is bounded. Proof: If M is an

upper bound for S, it is also an upper bound for T ⊂ S. The same holds for lowerbounds.

We say that an ordered field F has the completeness property (CP) if every

nonempty subset S of F which is bounded above, has a least upper bound in F.

MATH 3333–INTERMEDIATE ANALYSIS–BLECHER NOTES 17

Theorem 3.5. There exists one and only one ordered field that satisfies the com-

pleteness property (CP).

We define the real numbers R to be the one and only one complete ordered field

in the last theorem. A real number is a member, or element, of this one and only

one complete ordered field. We will not prove this theorem (due to Dedekind and

Cantor), it is too lengthy. In practice this means that we take the completeness

property (CP) as an axiom, we are taking it on faith. It also means that all the

properties of the real numbers that we learned in high school, are deducible, or can

be proved, from the 16 properties above (A1–A5,M1–M5, DL, O1-O4, CP). We did

some of this in Theorem 3.2 above.

• It is easy to prove from (CP) that every nonempty subset of R which isbounded below has a greatest lower bound or inf. Indeed in Homework 2

you will prove that if S is a nonempty set, then S is bounded below iff

T = {−x : x ∈ S} is bounded above, and in this case inf(S) = − sup(T ).• A moments thought shows that a nonempty set in R which is boundedabove (resp. below) has a maximum (resp. minimum) iff sup(S) ∈ S (resp.inf(S) ∈ S).

Examples:

(a) The finite set S = {2, 4, 6, 8} is bounded. The upper bounds of this set, areall numbers x ≥ 8. Since the upper bound 8 is in S, we have max(S) =sup(S) = 8. Similarly, the lower bounds of this set, are all numbers x ≤ 2,and min(S) = inf(S) = 2.

By the same argument, any finite set is bounded, and has a maximum

and a minimum, which equal the sup and inf respectively.

(b) The interval S = [1,∞) has no upper bounds, but is bounded below. Sinceone of its lower bounds, namely 1, is in S, we have min(S) = inf(S) = 1.

(c) The interval S = (0, 4] is bounded. The upper bound 4 is in S, so max(S) =

sup(S) = 4. The lower bound 0 is not in the set S. Indeed the set has

no minimum. To see that 0 = inf(S), BWOC (by way of contradiction)

suppose that there existed a lower bound m of S with m > 0. Since m

is a lower bound, m ≤ 4, and so 0 < m2

< m ≤ 4. Hence m2

∈ S. Butthis contradicts the fact that m is a lower bound of S (since m > m

2∈ S).

Hence 0 is the greatest lower bound: that is, 0 = inf(S).

(d) The interval S = [1, 2). As in (b) we have 1 = min(S) = inf(S). Clearly

2 is an upper bound of S. BWOC, suppose that there existed an upper

bound M of S with M < 2. Since M is an upper bound of S we have

M ≥ 1. If one draws a picture of the number line one sees the problem:if M < 2 then numbers between M and 2 like the average M+2

2, are in S


so are supposed to be ≤ M , but they are bigger than M . More precisely,Hence

1 ≤ M = M +M2

<M + 2

2<

2 + 2

2= 2.

So M+22

∈ S. This contradicts the fact thatM is an upper bound of S (sinceM < M+2

2∈ S). So 2 is the least upper bound of S: that is, 2 = sup(S).

In the examples above we have used the fact that supS is an upper bound β for

S with the property that for any real number t < β, t is not an upper bound for

S. There is another test that is very useful:

Principle: β = sup(S) iff the following two conditions hold: (a) β is an upper

bound for S, and (b) If t ∈ R with t < β, then ∃x ∈ S s.t. t < x ≤ β.

Principle in terms of ǫ: β = sup(S) iff the following two conditions hold: (a)

β is an upper bound for S, and (b) For every ǫ > 0, ∃x ∈ S s.t. β − ǫ < x ≤ β.

Proof of ‘Principle’: This is immediate from the fact mentioned above the ‘Prin-

ciple’, and the fact that saying that t ‘is NOT an upper bound for S’ is the same

as saying ∃x ∈ S s.t. t < x ≤ β. �

Proof of ‘Principle in terms of ǫ’: In the ‘Principle’, simply set t = β − ǫ, orequivalently, ǫ = β− t. Thus for example, for the (⇐) direction of the ‘Principle interms of ǫ’, if (a) and (b) there hold, and if t < β, set ǫ = β− t > 0. By hypothesis,∃x ∈ S s.t. β − ǫ < x ≤ β. Since β − ǫ = t, (b) of the ‘Principle’ holds. So by the‘Principle’, β = sup(S).

To illustrate this ‘Principle’, lets redo the proof that 2 = sup(S) in Example (d)

above, where S = [1, 2). Note that (a), 2 is an upper bound for [1, 2). If t < 2 then

either t < 1 in which case t < 1 ≤ 2, or 1 ≤ t < 2 in which case

1 ≤ t = t+ t2

<t+ 2

2<

2 + 2

2= 2.

Thus we have found an x ∈ S s.t. t < x ≤ 2: in the first case x = 1 and in thesecond case x = t+2

2. Thus (b) in the ‘Principle’ holds. So by the ‘Principle’,

2 = sup(S).

Of course there is a similar ‘Principle’ for infimums, namely:

Principle: α = inf(S) iff the following two conditions hold: (a) α is a lower

bound for S, and (b) If t ∈ R with t > α, then ∃x ∈ S s.t. t > x ≥ α.

Principle in terms of ǫ: α = inf(S) iff the following two conditions hold: (a)

α is a lower bound for S, and (b) For every ǫ > 0, ∃x ∈ S s.t. α+ ǫ > x ≥ α.


To illustrate this ‘Principle’, lets redo the proof that 0 = inf(S) in Example (c)

above, where S = (0, 4]. Note that (a), 0 is a lower bound for S. If t ∈ R witht > 0 then either t > 4 in which case t > 4 > 0, or t ≤ 4 in which case

0 <t

2< t ≤ 4.

Thus we have found an x ∈ S s.t. 0 ≤ x < t: in the first case x = 4 and in the secondcase x = t

2. Thus (b) in the ‘Principle’ holds. So by the ‘Principle’, 0 = inf(S).

Theorem 3.6. Let x, y ∈ R s.t. x ≤ y + ǫ, ∀ǫ > 0. Then x ≤ y.

Proof. BWOC, suppose that y < x. Let ǫ = x−y2

> 0. Then

y + ǫ = y +x− y2

=2y + x− y

2=

x+ y

2<

x+ x

2= x,

which contradicts the hypothesis that x ≤ y + ǫ, ∀ǫ > 0. Hence, x ≤ y. �

Remark. With almost exactly the same proof, it follows that: If x, y ∈ R s.t.x < y + ǫ, ∀ǫ > 0, then x ≤ y.

Theorem 3.7. (The Archimedian property) The set N of natural numbers is

unbounded above.

Proof. BWOC, suppose that N is bounded above. By the completeness property

(CP) (see above Theorem 3.5), there exists a least upper bound β for N. By the

‘Principle’ on page 18, ∃n ∈ N s.t. β − 1 < n ≤ β. Then β < n + 1 ∈ N.This contradicts the fact that β is an upper bound for N. Thus, N is unbounded

above. �

Theorem 3.8. (The Archimedian property)

(a) ∀z ∈ R, there exists n ∈ N s.t. n > z.(b) ∀x, y ∈ R, with x > 0, there exists n ∈ N s.t. nx > y.(c) ∀x ∈ R, with x > 0, there exists n ∈ N s.t. 0 < 1/n < x.

Proof. (a) If z ∈ R, then z is NOT an upper bound for N by Theorem 3.7, so thereexists n ∈ N s.t. n > z.

(b) Let z = yx. By (a), ∃n ∈ N s.t. n > z = y

x, so nx > y.

(c) Given x > 0, apply (a) with z = 1/x. So ∃n ∈ N s.t. n > 1/x. Hencex > 1/n > 0, the last ‘>’ because n > 0. �

Example. Let B = {2, 32, 43, 54, · · · }. Show that inf(B) = 1.

Solution. We use the ‘Principle in terms of ǫ’ above (see page 18). Note B =

{1+ 1n: n ∈ N}. Since 1

n> 0 we have 1+ 1

n> 1, so 1 is a lower bound for B. Given

ǫ > 0 choose by Theorem 3.8 (iii), an n ∈ N with 1n< ǫ. Hence 1 < 1 + 1

n< 1 + ǫ.

Since 1 + 1n∈ B, we have verified the conditions in the ‘Principle’ above, so that

inf(B) = 1. �


• Sup’s and inf’s have certain properties. Some are in the homework. Hereis another:

Theorem 3.9. sup(A+ B) = sup(A) + sup(B), if A and B are nonempty sets in

R which are bounded above, and A+B is defined to be {x+ y : x ∈ A, y ∈ B}.

Proof. Let C = {x + y : x ∈ A, y ∈ B}. If x ∈ A and y ∈ B then x ≤ sup(A) andy ≤ sup(B), so x + y ≤ sup(A) + sup(B). So sup(A) + sup(B) is an upper boundfor C. Hence C is bounded above, and sup(C) ≤ sup(A) + sup(B).

Conversely, if x ∈ A and y ∈ B then x + y ∈ C so x + y ≤ sup(C). Thusy ≤ sup(C) − x. Since this is true for all y ∈ B, we see that sup(C) − x is anupper bound for B. Hence sup(B) ≤ sup(C) − x. Thus x ≤ sup(C) − sup(B),for every x ∈ A. We see that sup(C) − sup(B) is an upper bound for A, and sosup(A) ≤ sup(C)− sup(B). Therefore sup(A) + sup(B) ≤ sup(C).

Thus sup(C) = sup(A) + sup(B). �

Remark. Read the proof of this in the textbook, which uses epsilons.

Lemma 3.10. If a and b are real numbers with b − a > 1 then there exists aninteger m with a < m < b.

Proof. Case 1: a > 0. Let S = {n ∈ N : n > a}. This is not empty, by theArchimedian property. By the well-ordering axiom (see the start of Section 3.1

above), S has a minimum, m say. Clearly m > a, and m ∈ N. BWOC, supposethat m ≥ b. Then m > a + 1 > 1, so m ≥ 2. Also, m − 1 ≥ b − 1 > a. Som − 1 ∈ S, which is absurd since m = min(S). This contradiction shows thatm < b so a < m < b.

Case 2: general case. By the Archimedian property 3.8 (a), ∃k ∈ N s.t. k > −a.So a+k > 0. Also (b+k)−(a+k) = b−a > 1. So by Case 1, there exists an integerm with a+ k < m < b+ k. Hence a < m− k < b, and m− k is an integer. �

Theorem 3.11. (Density of the rationals in R) If a and b are real numbers with

a < b, then there exists a rational number r with a < r < b.

Proof. Since b−a > 0, by the Archimedian property 3.8 (b), ∃n ∈ N s.t. n(b−a) >1. So nb − na > 1. By the lemma, with a replaced by na and b replaced by nb,there exists an integer m with na < m < nb. Divide through by n. So a < r < b

where r = mn∈ Q. �

Theorem 3.12. (Density of the irrationals in R) If a and b are real numbers with

a < b, then there exists an irrational number z with a < z < b.

Proof. By the last theorem, with a replaced by a√2and b replaced by b√

2, ∃r ∈ Q

s.t. a√2< r < b√

2. Multiply by

√2. So a < w < b, where w = r

√2. Also, w


is irrational (for if w were rational then√2 = w/r would be rational, since the

quotient of two rationals is obviously rational. But we proved in Proposition 3.4

that√2 is irrational). �

Theorem 3.13. If S is a nonempty set of real numbers, then S is an interval iff

S has the property that if x, z are in S and if x < y < z then y is in S. (Recall that

an interval is a set of one of the 9 types described in the third ‘bullet’ in Chapter

2.)

Proof. (⇒) Do this for each of the 9 types, every one of these is easy. For example,if S = (a, b] and x, z ∈ S and x < y < z, then a < x < y < z ≤ b so y ∈ S.

(⇐) If S is a nonempty set of real numbers, then either S is bounded above orit is not, and either S is bounded below or it is not. And if S is bounded above

(resp. bounded below) then either it has a maximum (resp. minimum) or it does

not. Thinking about the above one sees that there are exactly nine possibilities:

either (1) S is bounded above and below and has a minimum and a maximum,

(2) S is bounded above and below and has a minimum but not a maximum, (3)

S is bounded above and below and has a maximum but not a minimum, (4) S is

bounded above and below and has neither a maximum nor a minimum, (5) S is

bounded below but not above and has a minimum, (6) S is bounded below but

not above and does not have a minimum, (7) S is bounded above but not below

and has a maximum, (8) S is bounded above but not below and does not have a

maximum, (9) S is not bounded above nor is bounded below.

Now show that each of these 9 cases leads to one of the 9 types described in

the third ‘bullet’ in Chapter 2. I will do two as examples, the other seven are

similar. For example, in case (3), let a = inf(S) and b = sup(S) = max(S), then

S ⊂ (a, b]. Let y ∈ (a, b]. By the ‘Principle’ on page 18, there exists an x ∈ S witha ≤ x < y ≤ b. So y ∈ S since x, b ∈ S. So every y ∈ (a, b] is in S. Therefore(a, b] ⊂ S, so S = (a, b], the third kind of interval.

In case (8), let b = sup(S), then S ⊂ (−∞, b). Let y ∈ (−∞, b). Since S is notbounded below, y is not a lower bound for S, so there exists x ∈ S with x < y. Bythe ‘Principle’ on page 18, there exists a z ∈ S with y < z ≤ b. So y ∈ S sincex, z ∈ S. So every y ∈ (−∞, b) is in S. Therefore (−∞, b) ⊂ S, so S = (−∞, b),the eighth kind of interval. �

Homework 2.

• Text 5th Ed Section 3.2 number 3a,b,c, Section 3.3 numbers 1a-c, 3degi(prove these), 4e,i (prove these), 5, 6, 7, 8 [4th (and 3rd) Ed numbers:

12.1a-c, 12.3d,e,g,i (prove these), 12.4e,i (prove these), 12.5, 12.6, 12.7,

12.8.]


• (A) Show that every nonempty subset of R which is bounded below hasa greatest lower bound or inf (this falls out of the proof of Question 7(b)

above (or 12.7 (b)) with k = −1).• Also do case (4) and (9) in the proof of Theorem 3.13, in both directions((⇒) and (⇐)).

3.4. Open and closed sets. In calculus you met some important definitions,

something like the following:

• A neighborhood of x is an open interval centered at x. For example, (1, 3)is a neighborhood of 2.

Every neighborhood of x is of the form

N(x, ǫ) = (x− ǫ, x+ ǫ) = {y ∈ R : |x− y| < ǫ},

for some ǫ > 0. [Picture drawn in class.]

We occasionally say that ǫ is the radius of N(x, ǫ).

• A deleted neighborhood of x is a neighborhood of x with the midpoint xremoved:

N∗(x, ǫ) = (x− ǫ, x) ∪ (x, x + ǫ) = N(x, ǫ)\{x} = {y ∈ R : 0 < |x− y| < ǫ}.

[Picture drawn in class.]

• An interior point of a set S is a point x ∈ S such that S contains someneighborhood of x. That is, ∃ǫ > 0 s.t. N(x, ǫ) ⊂ S. Or equivalently, ∃ǫ > 0s.t. (x− ǫ, x+ ǫ) ⊂ S.

• The interior of a nonempty set S is the set of all interior points of S. Wewrite this set as S◦ or Int(S). This may be the empty set (if S has no

interior points).

For example if A = [1, 3) then A◦ = (1, 3).

• A boundary point of a nonempty set S is a number x such that everyneighborhood of x contains at least one point in S and at least one point

not in S. That is, ∀ǫ > 0, the neighborhood N(x, ǫ) of x contains at leastone point in S and at least one point not in S. Or equivalently, ∀ǫ > 0,N(x, ǫ) ∩ S 6= ∅ and N(x, ǫ) ∩ Sc 6= ∅. (Here Sc is the complement of S.)

• The boundary of a nonempty set S is the set of all boundary points of S.We write this set as ∂S or Bd(S). This may be the empty set.

• A nonempty set S is called open if it contains none of its boundary points.That is, S ∩ Bd(S) = ∅. We also declare the empty set to be open.

• A nonempty set S is called closed if it contains all of its boundary points.That is, Bd(S) ⊂ S. We also declare the empty set to be closed.

• The sets R and ∅ are both open and closed.• A set can be neither open nor closed, for example [1, 3) :


Examples.

(a) A = [1, 3). Let us prove that Int(A) = (1, 3), and Bd(A) = {1, 3}. Fromthis it follows that A is neither open nor closed (since A contains some

boundary points but not other boundary points.

Usually the way we proceed with sets like these is as follows. Note that

[1, 3) divides R into 5 disjoint pieces: numbers < 1, the number 1, numbers

in (1, 3), the number 3, and numbers > 3. Any number x < 1 cannot

be a boundary point of [1, 3), since if you set ǫ = 1 − x then the interval(x− ǫ, x+ ǫ) = (x− ǫ, 1) is contained in Ac, so contains no points in A. Onthe other hand, the number 1 is a boundary point of [1, 3), since for every

ǫ > 0, N(1, ǫ) contains points in A (like 1), and points not in A, like the

average of 1 and 1 − ǫ, namely 1 − ǫ2(assuming that ǫ < 1). Note that 1

is not an interior point of [1, 3) since N(1, ǫ) is not contained in A if ǫ > 0.

Similarly, the number 3 is not an interior point of [1, 3). However, 3 is a

boundary point of [1, 3), since for every ǫ > 0, N(3, ǫ) contains points in A

(like the average 3− ǫ2of 3 and 3− ǫ if ǫ < 1), and points not in A, like 3.

The final case is if 1 < x < 3: here let ǫ = min{x− 1, 3− x}. If y ∈ N(x, ǫ)then

1 = x− (x− 1) ≤ x− ǫ < y < x+ ǫ ≤ x+ 3− x = 3.

So y ∈ A = [1, 3). That is, N(x, ǫ) ⊂ A, and so x is an interior pointof A. Notice such x are not boundary points since for ǫ as above, the

interval N(x, ǫ) contains no points from Ac. Since Int(A) ⊂ A, we musthave Int(A) = (1, 3). And Bd(A) = {1, 3}.

(b) E = [0, 1]. Proceeding exactly as in the previous example one sees that

Int(E) = (0, 1) and Bd(E) = {0, 1}. Hence this set is closed and not open,since it contains all of its boundary points.

(c) G = (−1, 2) ∪ (2, 5). Proceeding exactly as in (a) (but notice that here Gdivides R into 7 disjoint pieces), one sees that Int(G) = G = (−1, 2)∪(2, 5),and that Bd(G) = {−1, 2, 5}. For example, every interval centered at 2contains points not in G (like 2), and points in G.

This set is open, since it contains none of its boundary points. It is not

closed, since A does not contain its boundary points.

(d) I = [2,∞). Proceeding exactly as in (a) (but notice that here I dividesR into 3 disjoint pieces), one sees that Int(I) = (2,∞) and Bd(I) = {2}.Hence this set is closed and not open, since it contains all of its boundary

points.


(e) Every number in R is an interior point of R, and is not a boundary point

of R (since no interval can contain a point in the complement of R). So

Int(R) = R and Bd(R) = ∅. Thus R is open and closed.(f) Let B = {1, 1

2, 13, 14, · · · }. Proceed exactly as in (a), but notice that here B

divides R into infinitely many pieces, which I would label as follows: the

numbers < 0, the number 0, the numbers > 1, the numbers in B, and the

open intervals whose endpoints are consecutive numbers in B like (12, 13).

Proceeding exactly as in (a), we see that B has no interior points, and that

every point in B is a boundary point, and that the only point not in B which

could possibly be a boundary point is 0. And indeed 0 is a boundary point

of B since given ǫ > 0, the interval (−ǫ, ǫ) centered at 0 contains pointsnot in B (like 0), and points in B. Indeed by the Archimidean property

Theorem 3.8 (c), there exists an n ∈ N with 0 < 1n< ǫ.

Thus Int(B) = ∅ and Bd(B) = B ∪ {0}. Thus B is neither open norclosed, since it contains some of its boundary points but not others.

(g) Let C = {0} ∪ {1, 12, 13, 14, · · · }. Proceeding exactly as in (f) we see that

Int(C) = ∅ and Bd(C) = C. Thus C is closed but not open, since itcontains all of many its boundary points.

Lemma 3.14. (a) Let x ∈ S. Then either x ∈ Int(S) or x ∈ Bd(S) but notboth. So Int(S) = S \ Bd(S).

(b) Bd(S) = Bd(Sc), where Sc = R\S.

Proof. (a) If x ∈ Int(S) then ∃ǫ > 0 s.t. N(x, ǫ) ⊂ S. Hence N(x, ǫ) ∩ Sc = ∅, sox cannot be a boundary point. So a point cannot be both an interior point and a

boundary point.

Now suppose that x ∈ S, but x is not an interior point. Thus ∀ǫ > 0, N(x, ǫ) 6⊂ S.That is, N(x, ǫ) contains a point in Sc. Also, N(x, ǫ) contains a point in S, namely

x. Thus, x is a boundary point of S.

(b) x ∈ Bd(Sc) ⇔ (∀ǫ > 0, N(x, ǫ) contains at least one point in Sc and atleast one point in S) ⇔ x ∈ Bd(S). �

Lemma 3.15. For a nonempty set S in R, the following are equivalent:

(a) S is open.

(b) S = Int(S).

(c) For every point x ∈ S, there exists ǫ > 0 s.t. N(x, ǫ) ⊂ S.(d) Sc is closed.

Proof. (a) ⇔ (b) S is open iff it does not contain any boundary points. This isequivalent, by (a) of the last Lemma, to saying that all points in S are interior

points; that is, iff S = Int(S).


(b) ⇔ (c) Note that (c) is just saying that every point in S is an interior pointof S, or equivalently that S = Int(S).

(a) ⇔ (d) S is open iff S ∩Bd(S) = ∅ iff Bd(S) ⊂ Sc. This is equivalent, by (b)of the last Lemma, to saying that Bd(Sc) ⊂ Sc, which is the definition of Sc beingclosed. �

Corollary 3.16. A nonempty set E is closed iff Ec is open.

Proof. Apply (a) ⇔ (d) in Lemma 3.15 to S = Ec. �

Digression on sets: If {Si : i ∈ I} is a collection of sets, then we write∪i∈I Si for their union, namely {x : ∃i ∈ I s.t. x ∈ Si}. We write ∩i∈I Si fortheir intersection, namely {x : x ∈ Si ∀i ∈ I}. If I is finite we usually writeS1 ∪ S2 ∪ · · · ∪ Sn or ∪ni=1 Si for the union; and write S1 ∩ S2 ∩ · · · ∩ Sn or ∩ni=1 Sifor the intersection.

De Morgan’s laws say:

(∪i∈I Si)c = ∩i∈I Sci ,and

(∩i∈I Si)c = ∪i∈I Sci .Here is a proof of De Morgan’s laws: x ∈ (∩i∈I Si)c iff x /∈ ∩i∈I Si, iff ∃i ∈ I s.t. x /∈Si, iff ∃i ∈ I s.t. x ∈ Sci , iff x ∈ ∪i∈I Sci .

Also x ∈ (∪i∈I Si)c iff x /∈ ∪i∈I Si, iff x /∈ Si ∀i ∈ I, iff x ∈ Sci ∀i ∈ I, iffx ∈ ∩i∈I Sci .

Theorem 3.17. (a) The union of any collection of open sets is open.

(b) The intersection of a finite number of open sets is open.

Proof. We can avoid trivialities and suppose that all sets here are nonempty.

(a) Let {Ui : i ∈ I} be a collection of open sets. Let V = ∪i∈I Ui. We willapply (c) in Lemma 3.15 to show that V is an open set. Let x ∈ V = ∪i∈I Ui. So∃j ∈ I s.t. x ∈ Uj . Since Uj is open, ∃ǫ > 0 s.t. N(x, ǫ) ⊂ Uj ⊂ ∪i∈I Ui = V .Hence by (c) in Lemma 3.15, V is an open set.

(b) Let U1, U2, · · · , Un be a finite number of open sets. Let U = U1∩U2∩· · ·∩Un.We will apply (c) in Lemma 3.15 to show that U is open. If x ∈ U = U1∩U2∩· · ·∩Unthen x ∈ Ui for all i = 1, 2, · · · , n. Since Ui is open, ∃ǫi > 0 s.t. N(x, ǫi) ⊂ Ui forall i = 1, 2, · · · , n. Let ǫ = min{ǫ1, . . . , ǫn}. Then N(x, ǫ) ⊂ N(x, ǫi) ⊂ Ui, ∀i =1, . . . , n. Thus N(x, ǫ) ⊂ U1 ∩ U2 ∩ · · · ∩ Un = U . Hence by (c) in Lemma 3.15, Uis open. �

Example. An = (−1/n, 1/n) is an open set, but⋂∞

n=1 An = {0} is not open.This is why a restriction to finite collections in Theorem 3.17 (b) is necessary.

Corollary 3.18. (a) The intersection of any collection of closed sets is closed.


(b) The union of any finite collection of closed sets is closed.

Proof. (a) Let {Di : i ∈ I} be a collection of closed sets. Then Dci is open, so byTheorem 3.17 (a), ∪i∈I Dci is open. Hence (∪i∈I Dci )c is closed by Lemma 3.15 (d).But by De Morgan’s laws, (∪i∈I Dci )c = ∩i∈I Di. So ∩i∈I Di is closed.

(b) Similar. If D1, D2, · · · , Dn are closed sets, then Dci is open, so ∩ni=1 Dci isopen by Theorem 3.17 (b). Hence (∩ni=1 Dci )c is closed. But by De Morgan’s laws,(∩ni=1 Dci )c = ∪ni=1 Di. So the latter is closed. �

• Note that any singleton set {x} is closed. Hence any finite set {x1x2, · · · , xn}is closed (here n ∈ N), being the union of a finite number of closed sets.

Lemma 3.19. Let S be a nonempty set of real numbers.

(1) If S is bounded above (resp. below) then sup(S) (resp. inf(S)) is a boundary

point of S.

(2) If S is closed and bounded above (resp. below), then S has a maximum

(resp. minimum). [5]

Proof. (1) We just do the ‘bounded above’ case, the other is similar. Let β = supS,

and let ǫ > 0 be given. Since β is an upper bound of S, the interval (β, β+ǫ) contains

no points of S, and hence contains points of Sc. By the Principle on page 18, the

interval (β−ǫ, β] must contain a point in S. Thus the interval (β−ǫ, β+ǫ) containsboth points of Sc and points of S. So β ∈ Bd(S).

(2) By (1), sup(S) (resp. inf(S)) is in Bdy(S) ⊂ S. So S has a maximum (resp.minimum). �

Definition. The closure S̄ of a nonempty set S is defined to be S̄ = S ∪Bd(S).The textbook writes cl S for S̄. We define the closure of ∅ to be ∅.

Example. (0, 1] = (0, 1] ∪ {0, 1} = [0, 1].

• In the homework you will prove that the closure S̄ of S is the smallestclosed set containing S. You will also show that S̄ = Int(S) ∪ Bd(S) (notethat we proved earlier that Int(S) ∩ Bd(S) = ∅).

Lemma 3.20. If S is a nonempty set of real numbers then x ∈ S̄ iff every neigh-borhood of x contains at least one point in S. This is also equivalent to saying that

every open set containing x contains at least one point in S.

Proof. Recall that S̄ = S∪Bd(S). If x ∈ S then the conditions on both sides of thefirst ‘iff’ are clearly true. On the other hand, if x /∈ S then every neighborhood of xcontains at least one point in Sc, namely x, and so in this case: x ∈ S̄ iff x ∈ Bd(S)iff every neighborhood of x contains at least one point in S.


The (⇐) direction of the last assertion from the first ‘iff’, since any neighborhoodis open. For the other direction, if U is an open set containing x, then by Lemma

3.15 (c), U contains a neighborhood of x, which contains at least one point in S. �

Lemma 3.21. Let S be a subset of R. Then S̄ is a closed set.

Proof. By Corollary 3.16 it is enough to show that S̄c is open. We will show that

S̄c is open using Lemma 3.15 (c) with S replaced by S̄c. So let x ∈ S̄c, if we canshow that x has a neighborhood N which is contained in S̄c then we will be done

(by that Lemma). By the contrapositive of Lemma 3.20, there is a neighborhood N

of x which is contained in Sc. If y ∈ N , then since N is open, by the contrapositiveof the last assertion in Lemma 3.20 with x replaced by y, we see that y /∈ S̄. SoN ⊂ S̄c. Hence by Lemma 3.15 (c), S̄c is open. �

A number x ∈ R is called an accumulation point of S if every deleted neighbor-hood of x contains at least one point of S. [Picture drawn in class.] Equivalently,

∀ǫ > 0, N∗(x, ǫ) ∩ S 6= ∅, where N∗(x, ǫ) = N(x, ǫ)\{x} We write S′ for the set ofaccumulation points of S.

Lemma 3.22. x is an accumulation point of a set S iff every neighborhood of x

contains infinitely many points of S.

Proof. (⇐) If a neighborhood N of x contains infinitely many points of S, thenclearly the deleted neighborhood N \ {x} contains at least one point of S.

(⇒) By way of contradiction, suppose that x is an accumulation point of theset S, and that N is a neighborhood of x containing only finitely many points from

S, say n points where n ∈ N. Let ǫ > 0 be the shortest distance from x to any ofthese n points (excluding x). Then the smaller deleted neighborhood N∗(x, ǫ/2)

contains no points from S, contradicting the definition of an accumulation point.

So every neighborhood of x contains infinitely many points from S. �

Accumulation points are closely related to boundary points. However an accu-

mulation point need not be a boundary point, and vice versa. What is true is that

if x /∈ S then x is a boundary point of S iff x is an accumulation point of S. Thisis because if x /∈ S then as in the first paragraph of the proof of Lemma 3.20,x ∈ Bd(S) iff every neighborhood of x contains at least one point in S iff everydeleted neighborhood of x contains at least one point in S (recall that x is not in

S) iff x ∈ S′.Examples.

(1) Let A = (0, 1) then A′ = [0, 1].

(2) Let B = N then B′ = ∅.(3) Let D = {1, 1

2, 13, 14, · · · }, then D′ = {0}.


[Pictures drawn in class. These are done similarly to the method we used for

boundary/interior points.]

Proposition 3.23. If S is a nonempty set of real numbers then S̄ = S ∪ S′.

Proof. We need to show that S ∪ S′ = Bd(S) ∪ S. If x ∈ S then x is contained inboth sides of the equation. If x /∈ S then by the observation above the example,x ∈ Bd(S) iff x ∈ S′. This does it. �

Corollary 3.24. If S is a nonempty set of real numbers then S is closed iff S̄ = S,

and iff S′ ⊂ S.

Proof. We will use twice the basic principle for sets A and B that A ⊂ B iffB = A ∪B. (Draw a Venn diagram to see this.)

Then S is closed iff Bd(S) ⊂ S iff S = Bd(S) ∪ S = S̄, by the last paragraph.By Proposition 3.23 this is equivalent to S = S ∪ S′, which happens (by the lastparagraph) iff S′ ⊂ S. �

Definition. An isolated point of a set S is a point x ∈ S which is not in S′.That is, x ∈ S and there exists ǫ > 0 with N∗(x, ǫ) ∩ S = ∅. [Picture drawn inclass.]

Homework 4 Exercise: Show that any isolated point of S is in Bd(S).

Homework 3: See course webpage.

3.5. Compact sets. It is easy to see that a set S is bounded iff it is contained

in some interval of the form [a, b], and this is equivalent to saying that there is a

number K > 0 such that |x| ≤ K for all x ∈ S. (To see the latter is equivalent, takean interval of the form [a, b] containing S and stretch it to the left and to the right,

making it bigger, so that it is centered at 0, so is of the form [−K,K].) [Picturesdrawn in class.]

Theorem 3.25. (The Heine-Borel theorem) If S is a nonempty set of real numbers

then the following are equivalent:

(1) S is closed and bounded.

(2) S has the following property: whenever C is a collection of open sets whoseunion contains S, then actually S is contained in the union of a finite

number of the sets in C.

We will prove this theorem later in class (Appendix to this chapter).

A set satisfying condition (1) or (2) in the theorem is called compact. We will

work only with (1) in this section, so you can take a compact set to simply be a

closed and bounded set. We also say that ∅ is compact.


Example. The set [0, 1] is closed and bounded, hence is compact.

Proposition 3.26. If S is a nonempty compact set of real numbers then S has a

maximum and a minimum.

Proof. This follows from Lemma 3.19 (2). �

Proposition 3.27. The intersection of any collection of compact sets is compact.

The union of two compact sets is compact.

Proof. Let C be an intersection of a collection of compact sets. Since compact sets

are closed, and since an intersection of closed sets is closed, we see that C is closed.

Since any one set D in this collection is bounded, and since C ⊂ D, we see that Cis bounded. So C is compact.

Let C be a union of two compact sets A and B. Since compact sets are closed,

and since a union of closed sets is closed, we see that C is closed. If K is an upper

bound for A andM is an upper bound for B then max{K,M} is an upper bound forA∪B. Similarly, A∪B has a lower bound. So C is bounded. So C is compact. �

Theorem 3.28. (The nested intervals theorem) If C1 ⊃ C2 ⊃ C3 ⊃ · · · are nestedcompact intervals (that is, each Ck is of the form [a, b] for numbers a, b ∈ R), then∩∞n=1 Cn is not empty. Equivalently: ∃x such that x ∈ Cn for every n ∈ N.

Proof. Let Cn = [an, bn], where an ≤ bn, ∀n ∈ N. So[a1, b1] ⊇ [a2, b2] ⊇ · · · ⊇ [an, bn] ⊇ . . . .

Let S = {a1, a2, . . . }, which is bounded above (say, by b1, since an ≤ bn ≤ b1 forall n. ). Let α = sup(S). Claim: α ∈

∞⋂

n=1

[an, bn]. BWOC, if α /∈∞⋂

n=1

[an, bn], then

∃m ∈ N s.t. α /∈ [am, bm]. Since α ≥ am, we must have α > bm. But bm is anupper bound for S. Indeed, if x ∈ S, then x = an for some n. If n ≥ m, thenan ≤ bn ≤ bm, and if n ≤ m, then an ≤ am ≤ bm. In either case, x = an ≤ bm, sothat bm is an upper bound for S. Since α > bm, this contradicts the fact that α

was the least upper bound. �

Theorem 3.29. (The Bolzano-Weierstrass theorem for sets) Every infinite bounded

set in R has an accumulation point.

Proof. Since S is bounded, S ⊂ [−K,K], for a number K > 0. Divide [−K,K] intotwo equal compact intervals I1 and J1. Either I1 ∩ S or J1 ∩ S has infinitely manypoints (or else their union, which is S has finitely many points, a contradiction).

If I1 ∩ S is finite, switch the names of I1 and J1. Now I1 ∩ S has infinitely manypoints. Divide I1 into two equal intervals I2 and J2. Either I2 ∩ S or J2 ∩ S hasinfinitely many points (or else their union, which is I1 ∩S has finitely many points,


a contradiction). If I2 ∩ S is finite, switch the names of I2 and J2. Now I2 ∩ S hasinfinitely many points. Divide I2 into two equal intervals I3 and J3. Continue in

this way, producing a sequence of nested closed intervals I1 ⊇ I2 ⊇ I3 . . . such thatIn ∩ S has infinitely many elements ∀n ∈ N. Write In = [an, bn].

By the nested intervals theorem,∞⋂

n=1

In 6= ∅. Let x ∈∞⋂

n=1

In. Claim: x is an

accumulation point of S. To see this, given any ǫ > 0, ∃n ∈ N s.t. 1n< ǫ

2K(using

(c) of the Archimedian property). Now n ≤ 2n by mathematical induction, so1

2n≤ 1

n< ǫ

2K. Hence 2K

2n< ǫ. Since the length of [an, bn] is

2K2n

, we have

bn = an +2K

2n< an + ǫ ≤ x+ ǫ,

(recall that x ∈ [an, bn]). Similarly,

x− ǫ ≤ bn − ǫ < bn −2K

2n= an.

So [an, bn] ⊂ (x − ǫ, x + ǫ). Since [an, bn] contains infinitely many points in S, sodoes (x − ǫ, x + ǫ). Since ǫ > 0 was arbitrary, x is an accumulation point of S byLemma 3.22. �

Homework 3/4. (Go by the webpage for the precise numbers and due dates)

(1) Show that the interior of a set is open. Also show that Int(S) is the largest

open set contained in S.

(2) Show that S̄ is the smallest closed set containing S.

(3) Homework 3.5 Exercise 1 from the textbook 5th Ed (Exercise 14.1 in 4th

Edition).

(4) Explain why each of the following sets are not compact: (a) [1, 3), (b) N,

(c) {1/n : n ∈ N}, (d) {x ∈ Q : 0 ≤ x ≤ 2}.(5) Which of the 9 types of intervals are compact?

(6) Show that any isolated point of S is in Bdy(S).

(7) Show that the closure of a bounded set is bounded.

(8) Read the proofs above in this Section which we skipped in class.

END OF WORK FOR TEST 2

Appendix to Chapter 3. Proof of the Heine-Borel theorem.

THE PROOF BELOW IS NOT EXAMINABLE

We define an open cover of a set S to be a collection C of open sets whose unioncontains S. A finite subcover of S from C is a finite collection {C1, C2, · · · , Cn} of


the sets in C, whose union contains S. That is, Ck ∈ C for all k, and S ⊂ ∪nk=1 Ck.Note that if C1 ⊂ C2 ⊂ · · · ⊂ Cn then Cn = ∪nk=1 Ck, so S ⊂ Cn.

Example. Let S = [0, 1] and let

C = {(− 150

,1

50), (

1

2,3

2), (

1

4,3

4), (

1

8,3

8), (

1

16,3

16), (

1

32,3

32), (

1

64,3

64), (

1

128,

3

128), · · · }.

It is easy to see that in this last example, C is an infinite collection of open setswhose union contains S. That is, C is an open cover of S. But in fact the first 7sets listed in C have a union which already contains S. These 7 sets comprise afinite subcover. This example also illustrates the condition in (2) of the Heine-Borel

Theorem.

Theorem 3.30. (The Heine-Borel theorem) If S is a nonempty set of real numbers

then the following are equivalent:

(1) S is closed and bounded.

(2) S has the following property: whenever C is a collection of open sets whoseunion contains S, then actually S is contained in the union of a finite

number of the sets in C. (In English: every open cover of S has a finitesubcover.)

Proof. (2) ⇒ (1) Let C = {(−n, n) : n ∈ N}. By (2), this collection has a finitesubcover, and by the line ending the first paragraph of this appendix, S ⊂ (−n, n)for some n ∈ N. So S is bounded.

If x /∈ S, let D = {[x − 1n, x + 1

n]c : n ∈ N}. This is an open cover of R \{x},

and hence of S. By (2), this collection has a finite subcover, and by the line ending

the first paragraph of this appendix, S ⊂ [x − 1n, x + 1

n]c for some n ∈ N. Thus

(x− 1n, x+ 1

n) ⊂ Sc. Hence Sc is open by Lemma 3.15 (c), so S is closed.

(1) ⇒ (2) BWOC, suppose that S is closed and bounded, and that C is an opencover of S with no finite subcover. We follow the proof of Theorem 3.29 closely.

As we said there, S ⊂ [−K,K], for a number K > 0. Let D = {Sc} ∪ C. This isan open cover of [−K,K], since C covers the part of [−K,K] which is in S, and Sccovers the part of [−K,K] not in S. Then D has no finite subcover of [−K,K] (forif it did then C would have a finite subcover of S). Divide [−K,K] into two equalhalves. At least one of these halves, call it I1, cannot be covered by a finite number

of the sets in D (or else the union of the two halves, which is [−K,K], could becovered by a finite number of the sets in D, a contradiction). Divide I1 into twoequal halves. At least one of these halves, call it I2, cannot be covered by a finite

number of the sets in D (or else the union of the two halves, which is I1, could be


covered by a finite number of the sets in D, a contradiction). Divide I2 into twoequal halves. Continue in this way, producing a sequence of nested closed intervals

I1 ⊇ I2 ⊇ I3 . . . such that In cannot be covered by a finite number of the sets inD, ∀n ∈ N. As in the proof we are following, ∃x ∈

∞⋂

n=1

In. Since D covers [−K,K],choose a set U in D with x ∈ U . By Lemma 3.15 (c), ∃ǫ > 0 with (x− ǫ, x+ ǫ) ⊂ U .As in the proof we are following, ∃n ∈ N s.t. In ⊂ (x − ǫ, x + ǫ) ⊂ U . So In iscontained in one set from D. This contradicts the fact that In cannot be coveredby a finite number of the sets in D. This contradiction shows that C does have afinite subcover, as desired in (2). �


4. Sequences

4.1. Convergent sequences. We studied sequences in Calculus 2. In this course

our sequences are infinite sequences s1, s2, s3, · · · of real numbers. We often writethe sequence as (sn)

∞n=1 or simply as (sn). The nth term is sn.

• Recall that we say that a sequence (sn) converges to a real number s if∀ǫ > 0, ∃Ns.t. |sn − s| < ǫ ∀n ≥ N .

• Of course saying that |sn−s| < ǫ is the same as saying that s−ǫ < sn < s+ǫ,which is the same as saying that sn ∈ N(s, ǫ), in the notation of Chapter3.

• If (sn) converges to s then we say that s is the limit of (sn) and writes = limn sn, or s = limn→∞ sn, or sn → s as n → ∞, or simply sn → s.

• If (sn) does not converge to any real number then we say that it diverges.• The simplest example of a sequence is a constant sequence. If C is a con-stant, let s1 = C, s2 = C, s3 = C, · · · . It is clear that this sequence haslimit C too, since for any ǫ > 0, |sn − C| = 0 < ǫ ∀n ≥ 1.

• Another simple example from calculus is the sequence ( 1n). This converges

to 0 since given ǫ > 0, choose N > 1ǫ, then n ≥ N ⇒ | 1

n− 0| = 1

n≤ 1

N< ǫ.

• A sequence (sn) is called bounded if the set {sn : n ∈ N} is a bounded set.That is, there are numbers m and M such that m ≤ sn ≤ M for all n ∈ N.This is the same as saying that {sn : n ∈ N} ⊂ [m,M ]. It is easy to seethat this is equivalent to: there exists a number K ≥ 0 such that |sn| ≤ Kfor all n ∈ N. (See the first lines of the last Section.)

Fact 1. Any convergent sequence is bounded.

Proof: Suppose that sn → s as n → ∞. Taking ǫ = 1 in the definition ofconvergence gives that there exists a number N ∈ N such that |sn−s| < 1 whenevern ≥ N . Thus

|sn| = |sn − s+ s| ≤ |sn − s|+ |s| < 1 + |s|

whenever n ≥ N . Now let M = max{|s1|, |s2|, · · · , |sN |, 1+ |s|}. We have |sn| ≤ Mif n = 1, 2, · · · , N , and |sn| ≤ M if n ≥ N . So (sn) is bounded.

• A sequence (an) is called nonnegative if an ≥ 0 for all n ∈ N. To say thata nonnegative sequence converges to zero is simply to say that:

∀ǫ > 0, ∃Ns.t. an < ǫ ∀n ≥ N.

This is because |an − s| = |an − 0| = |an| = an in this case.• The reason we introduce nonnegative sequences, is that the convergenceof a general sequence can be restated as another nonnegative sequence

converging to 0. In fact:


Fact 2. If (sn) is a general sequence then:

limn

sn = s ⇐⇒ limn

(sn − s) = 0 ⇐⇒ limn

|sn − s| = 0.

That is, the sequence (sn) converges to s if and only if the nonnegative sequence

(|sn − s|) converges to 0.To prove this, simply look at the ǫ-definition of limn sn = s, and compare it

with the ǫ-definition of limn |sn − s| = 0.• Next we will prove some facts for nonnegative sequences, which we willuse later, together with Fact 2, to prove many other facts about general

sequences.

Fact 3. If (an) and (bn) are nonnegative sequences, with limn an = 0 and

limn bn = 0, and if C ≥ 0, then

limn

an + bn = limn

Can = 0.

Also, if limn an = 0 and if (bn) is any bounded sequence, then limn anbn = 0.

Proof: Let ǫ > 0 be given. Then ∃N1 such that an < ǫ whenever n ≥ N1. Infact, replacing ǫ by ǫ/2, we can say that ∃N1 such that an < ǫ/2 whenever n ≥ N1.Similarly, ∃N2 such that bn < ǫ/2 whenever n ≥ N2. Let N = max{N1, N2}, sothat if n ≥ N then n ≥ N1 and n ≥ N2. Thus if n ≥ N then an+bn < ǫ/2+ǫ/2 = ǫ.We have proved that limn an + bn = 0.

To prove limn Can = 0, we can assume that C > 0, because if C = 0 this

result is obvious. If ǫ > 0 is given, then replacing ǫ by ǫ/C, we can say that ∃N3such that an < ǫ/C whenever n ≥ N3. Thus Can < ǫ whenever n ≥ N3. Hencelimn Can = 0.

The proof of the final assertion is similar to the last paragraph. For if |bn| ≤ Cfor all n, then choosing N3 as above, we have |anbn − 0| = an|bn| ≤ Can < C ǫC = ǫwhenever n ≥ N3. Hence limn anbn = 0.

• Now we turn away from nonnegative sequences, and prove some thingsabout general sequences.

Fact 4. If (sn) and (tn) are sequences with sn ≤ tn for every n ≥ 1. Iflimn sn = s and limn tn = t, then s ≤ t.

Proof: By contradiction. Suppose that t < s. Let ǫ = s−t2. Then ǫ > 0, in fact

ǫ is half the distance from t to s. Thus t + ǫ = s− ǫ. Since limn tn = t, ∃N1 suchthat |tn − t| < ǫ whenever n ≥ N1. Thus −ǫ < tn − t < ǫ, so that tn < t+ ǫ. Sincelimn sn = s, ∃N2 such that |sn − s| < ǫ whenever n ≥ N2. Thus −ǫ < sn − s < ǫ,so that sn > s − ǫ. Let N = max{N1, N2}, so that if n ≥ N then n ≥ N1 andn ≥ N2. Thus if n ≥ N then

tn < t+ ǫ = s− ǫ < sn.


This contradicts the fact that sn ≤ tn. So s ≤ t.

Fact 5: The ‘squeezing’ or ‘pinching rule’. Suppose that (sn), (xn), and

(tn) are sequences with sn ≤ xn ≤ tn, for every n ≥ 1. If limn sn = s andlimn tn = s, then limn xn = s.

Proof: Let ǫ > 0 be given. Since limn sn = s, as in the previous proof, ∃N2such that sn > s− ǫ whenever n ≥ N2. Similarly, ∃N1 such that tn < t+ ǫ = s+ ǫwhenever n ≥ N1. Let N = max{N1, N2}, so that if n ≥ N then

s− ǫ < sn ≤ xn ≤ tn < s+ ǫ.

That s− ǫ < xn < s+ ǫ simply says that |xn − s| < ǫ. Thus limn xn = s.

Fact 6, sometimes call ’squeezing’ too, since it is a corollary of Fact 5: Let

(sn) be a sequence, let s be a number, and suppose that |sn− s| ≤ an for all n ≥ 1,where (an) is a sequence with limit 0. Then limn sn = s.

Proof: We have 0 ≤ |sn − s| ≤ an. Now an → 0 as n → ∞, so by ‘squeezing’(Fact 5), limn |sn − s| = 0. By Fact 2, limn sn = s.

Fact 7: The limit of a sequence, or even if the sequence converges, has nothing to

do with its first few terms. Thus, for example, the sequence 500, 1000, 1500, 2000, 1, 12, 13, 14, 15, · · · ,

converges to 0. Because of this principle, many of the last few Facts above, which

are stated with an ‘n ≥ 1’, would also be true with ‘n ≥ 1’ replaced by ‘n ≥ 2’, or‘n ≥ 25’, or indeed ‘n ≥ K’ for any fixed number K ∈ N.

Fact 8: If limn tn = t, and if t 6= 0, then there exists a number N with |tn| > |t|2for all n ≥ N .

Proof: Let ǫ = |t|2. Since limn tn = t, ∃N with |tn − t| < |t|2 whenever n ≥ N .

Hence for such n,

|t| = |t− tn + tn| ≤ |t− tn|+ |tn| <|t|2

+ |tn|,

and so |tn| > |t| − |t|2 =|t|2, whenever n ≥ N .

Fact 9: If limn sn = s and limn tn = t, then:

(1) limn sn + tn = s+ t;

(2) limn sn − tn = s− t;(3) limn sntn = st;

(4) limn Csn = Cs, if C is a constant;

(5) limnsntn

= st, if t 6= 0;

(6) limn |sn| = |s| ;(7) limn

√tn =

√t, if tn ≥ 0 for all n ∈ N.


Proof: By Fact 2, we have |sn − s| → 0, and |tn − t| → 0, as n → ∞. To prove(1), by Fact 6 it is enough to show that |(sn + tn)− (s+ t)| ≤ an, where an → 0 asn → ∞. But

|(sn + tn)− (s+ t)| = |(sn − s) + (tn − t)| ≤ |sn − s|+ |tn − t| → 0,

using the triangle inequality, and Fact 3 in the very last step. So we have proved

(1). The proof of (3) is similar, by Fact 6, it is enough to show that |sntn−st| ≤ an,where an → 0, as n → ∞. But

|sntn−st| = |sntn−snt+snt−st| ≤ |sntn−snt|+|snt−st| = |sn(tn−t)|+|t(sn−s)| = |sn||tn−t|+|t||sn−s|.

Now |sn − s| → 0, so |t||sn − s| → 0, as n → ∞, by Fact 3. On the other hand,since (sn) is convergent, it is bounded, by Fact 1. Thus (|sn|) is bounded. By thefinal assertion of Fact 3, |sn||tn − t| → 0 as n → ∞. By the first assertion of Fact3, we now see that |sn||tn − t| + |t||sn − s| → 0 as n → ∞. Since |sntn − st| ≤|sn||tn − t|+ |t||sn − s|, by Fact 6 we deduce that sntn → st as n → ∞. So we haveproved (3).

(4) follows from (3), if we set tn = C for all n. Applying (4) with C = −1 showsthat limn(−tn) = −t. Using this with (1), gives

limn(sn − tn) = lim

n(sn + (−tn)) = lim

nsn + lim

n(−tn) = s− t.

This proves (2).

For (5), we use a similar strategy to (1) and (3). Note that∣

∣

∣

∣

sntn

− st

∣

∣

∣

∣

=

∣

∣

∣

∣

snt− tnstnt

∣

∣

∣

∣

=|snt− st+ st− tns|

|tn||t|≤ |snt− st|+ |st− tns||tn||t|

=|sn − s||t|+ |s||t− tn|

|tn||t|.

By Fact 8, ∃N s.t. |tn| > |t|/2 for n ≥ N . Thus for n ≥ N ,∣

∣

∣

∣

sntn

− st

∣

∣

∣

∣

≤ |sn − s||t|+ |s||t− tn||tn||t|≤ 2|t|2 (|sn − s||t|+ |s||t− tn|).

Since |sn − s| → 0 and |t − tn| → 0, as n → ∞, by Fact 3 it follows that |sn −s||t| + |s||t − tn| → 0 too. By Fact 3 again, 2|t|2 (|sn − s||t| + |s||t − tn|) → 0 asn → ∞. Thus we conclude from Fact 6 (in conjunction with Fact 7), that sn

tn→ s

t

as n → ∞.(6) follows by squeezing too, since we have using Theorem 3.3 (f) above, that

||sn| − |s|| ≤ |sn − s| → 0. So by Fact 6, |sn| → |s|.Finally, to prove (7), we note first that since tn ≥ 0 for all n, we have t ≥ 0 by

Fact 4. We consider two cases. Case 1: t = 0. In this case, since limn tn = 0, given

ǫ > 0 there is a number N such that tn < ǫ2 whenever n ≥ N . We have simply

replaced ǫ by its square, but since ǫ > 0 was arbitrary, that is permissable. Taking

the square root,√tn < ǫ whenever n ≥ N . Hence limn

√tn = 0. Case 2: t > 0. In


this case,√tn +

√t ≥

√t > 0. Then

|√tn −

√t| =

∣

∣

∣

∣

(√tn −

√t)(

√tn +

√t)√

tn +√t

∣

∣

∣

∣

≤ 1√t|tn − t|.

Since |tn − t| → 0, by Fact 3, we have 1√t |tn − t| → 0. So by Fact 6,√tn →

√t as

n → ∞.

Example: Show using an ǫ argument that limnn2+2nn3−5 = 0.

Solution: To do this, we ‘make the top smaller than something simpler’: observe

that n2 +2n ≤ n2 + 2n2 = 3n2. Also, we ‘make the bottom bigger than somethingsimpler’: note if n ≥ 3 then n3 − 5 ≥ n3

2(since this last inequality is the same as

2n3 − 10 ≥ n3, which is the same as n3 ≥ 10). Thus∣

∣

∣

∣

n2 + 2n

n3 − 5

∣

∣

∣

∣

≤ 3n2

n3

2

6n2

n3=

6

n.

We want this to be < ǫ, which will happen if 6n< ǫ, or equivalently, n > 6

ǫ. So let

N be any number > 6ǫ, which is also bigger than 3.

Neat solution: If ǫ > 0 is given, choose N to be any number > max{ 6ǫ, 3}. If

n ≥ N then n > 6ǫ, so that 6

n< ǫ. Hence, and using the arguments for the centered

equation above, we have∣

∣

∣

n2 + 2n

n3 − 5∣

∣

∣≤ n

2 + 2n2

n3

2

=6

n< ǫ.

Hence we have verified the ǫ-definition of limnn2+2nn3−5 = 0.

Example: Show using Fact 6 that limnn2+2nn3−5 = 0.

Solution: As above, if n ≥ 3 then∣

∣

∣

∣

n2 + 2n

n3 − 5 − 0∣

∣

∣

∣

≤ 6n

→ 0 as n → ∞.

By Fact 6, limnn2+2nn3−5 = 0.

Example: Show using an ǫ argument that limn4n2−35n2+2n

= 45.

Solution: We look at the difference

4n2 − 35n2 + 2n

− 45=

20n2 − 15− 20n2 − 8n5(5n2 + 2n)

=−8n− 15

5(5n2 + 2n),

and so∣

∣

∣

4n2 − 35n2 + 2n

− 45

∣

∣

∣=

8n+ 15

5(5n2 + 2n)<

8n+ 15

5(5n2)≤ 8n+ 15n

5(5n2)=

23n

25n2=

23

25n.

We want this to be < ǫ, which will happen if 2325n

< ǫ, or equivalently, n > 2325ǫ

. So

let N be any number > 2325ǫ

.


Neat solution: If ǫ > 0 is given, choose N to be any number > 2325ǫ

. If n ≥ Nthen n > 23

25ǫ, so that 23

25n< ǫ. Hence, and using the centered equations above, we

have∣

∣

∣

4n2 − 35n2 + 2n

− 45

∣

∣

∣<

23

25n< ǫ.

Hence we have verified the ǫ-definition of limn4n2−35n2+2n

= 45.

Example: Show using Fact 6 that limn4n2−35n2+2n

= 45.

Solution: As above, we have∣

∣

∣

4n2 − 35n2 + 2n

− 45

∣

∣

∣<

23

25n→ 0,

as n → ∞. So by Fact 6, limn 4n2−3

5n2+2n= 4

5.

Infinite limits. If (sn) is a sequence, then we write limn sn = +∞ if ∀M >0, ∃N s.t. sn > M ∀n ≥ N . We write limn sn = −∞ if ∀M > 0, ∃N s.t.sn < −M ∀n ≥ N . Such sequences must be unbounded, and hence divergent (bythe contrapositive to Fact 1).

Example. Show that limn n3 = +∞.

Solution: Given M > 0, choose N > 3√M . If n ≥ N then n3 ≥ N3 > M . Hence

limn n3 = +∞ (we’ve just checked the definition of the latter).

Proposition 4.1. Suppose that (sn) and (tn) are sequences such that sn ≤ tn, ∀n.(a) If limn sn = +∞, then limn tn = +∞.(b) If limn tn = −∞, then limn sn = −∞.

Proof. We just prove (a), (b) being similar. Given M > 0, ∃N s.t. sn > M forn ≥ N . But tn ≥ sn, ∀n. Thus tn > M for n ≥ N . So limn tn = +∞. �

Theorem 4.2. Let (sn) be a sequence of positive numbers (so sn > 0, ∀n). Thenlimn sn = +∞ ⇔ limn 1sn = 0.

Proof. Suppose that lim sn = +∞. Given ǫ > 0, let M = 1ǫ > 0. Then ∃N s.t.sn > M =

1

ǫif n ≥ N . Since sn > 0, we have 1sn = |

1

sn− 0| < ǫ for n ≥ N .

Conversely, suppose that limn1

sn= 0. Suppose that M > 0 is given. Taking

ǫ = 1M, ∃N s.t. 1

sn< 1

Mif n ≥ N . Hence sn > M if n ≥ N . So limn sn = +∞. �

Example. Show that limnn3−5n2+2n

= ∞.Solution: We proved a couple of pages back that limn

n2+2nn3−5 = 0. So by the last

theorem, limnn3−5n2+2n

= ∞.

Theorem 4.3. If S is a nonempty set in R then

16 MATH 3333–INTERMEDIATE ANALYSIS–BLECHER NOTESdblecher/3333note2.pdf16 MATH...

Documents

Transcript of 16 MATH 3333–INTERMEDIATE ANALYSIS–BLECHER NOTESdblecher/3333note2.pdf16 MATH...