Haddad W.M. Nonlinear Dynamical Systems and Control: A Lyapunov-Based Approach

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DYNAMICAL SYSTEMS AND DIFFERENTIAL EQUATIONS 25

Deﬁnition 2.15. Let S ⊆ R

. The set Q ⊆ S is ope n (respectively,

closed) relative to S if there exists an open (respectively, closed) set R ⊆ R

such that Q = S ∩ R.

Note that since S = S ∩R

, it follows that any set is open relative to

itself.

Example 2.8. Let Q = (0, 1] and S = (−1, 1]. Note that Q is open

relative to S. In particular, for all α > 1, (0, 1] = (−1, 1] ∩(0, α). Similarly,

(0, 1] is closed relative to (0, 2) since (0, 1] = (0, 2) ∩ [α, 1] for all α < 0. △

A sequence of scalars {x

}

∞

n=0

⊂ R is said to converge to a scalar x ∈ R

if for every ε > 0, there exists N = N(ε) such that |x

− x| < ε for every

n > N . If a sequence {x

}

∞

n=0

converges to some x ∈ R, we say x is the

limit of {x

}

∞

n=0

, that is, lim

n→∞

= x. A sequence {x

}

∞

n=0

is a Cauchy

sequence if for every ε > 0, there exists N = N(ε) such that |x

− x

| < ε

for all n, m > N.

A scalar sequence {x

}

∞

n=0

⊂ R is said to be bounded above (respec-

tively, bounded below) if there exists α ∈ R such that x

≤ α (respectively,

≥ α) for all n ∈ Z

. A scalar sequence {x

}

∞

n=0

is said to be

nonincreasing (respectively, nondecreasing) if x

n+1

≤ x

(respectively,

n+1

≥ x

) for all n ∈ Z

Example 2.9. Consider th e sequence {

}

∞

n=1

= {1,

, . . .}. To see

that this sequence converges to x = 0, let ε > 0. In this case, |1/n − 0| =

|1/n| = 1/n < ε if and only if n > 1/ε. Hence, letting N = N (ε) > 1/ε, it

follows that |1/n − 0| < ε for every n > N . △

If a ﬁnite limit x ∈ R does not exist for a given scalar sequence, then

the sequence is said to be divergent. In p articular, th e sequ en ce {n}

∞

n=0

diverges as n → ∞. In addition, the s equence {(−1)

}

∞

n=0

, though bounded,

is also divergent since it yields the oscillating sequence {−1, 1, −1, 1, . . .},

and h en ce does not converge to a ﬁnite limit.

The supremum of a nonempty set S ⊂ R of scalars, denoted by sup S,

is deﬁ ned to be the smallest scalar x such that x ≥ y f or all y ∈ S. If no

such scalar exists, th en sup S

△

= ∞. Similarly, the inﬁmum of S, denoted by

inf S, is deﬁned to be the largest scalar x such th at x ≤ y for all y ∈ S. If no

such scalar exists, then inf S

△

= −∞. Given a scalar sequence {x

}

∞

n=0

⊂ R,

the supremum of the sequence, denoted by sup

, is deﬁned as sup{x

n = 1, 2, . . .}. Similarly, the inﬁmum of the sequen ce, denoted by inf

is deﬁ ned as inf{x

: n = 1, 2, . . .}. Finally, given a sequence {x

}

∞

n=0

, let

= inf{x

: n ≥ m} and z

= sup{x

: n ≥ m}. Since the sequences

}

∞

m=0

and {z

}

∞

m=0

are nondecreasing and nonincreasing, respectively,

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26 CHAPTER 2

it follows from Proposition 2.6 that if these sequences are bounded, th en

they have ﬁnite limits. The limit of {y

}

∞

m=0

is denoted by lim inf

m→∞

and is called the limit inferior of x

, and the limit of {z

}

∞

m=0

is denoted

by lim sup

m→∞

and is called the limit superior of x

. If {y

}

∞

m=0

and

}

∞

m=0

are unbounded, then lim inf

m→∞

= −∞ and lim sup

m→∞

∞.

An equivalent deﬁnition for the limit superior of an inﬁnite sequence

of real numbers {x

}

∞

n=0

is the existence of a real number S s uch that for

every ε > 0 there exists an integer N such that for all n > N, x

< S + ε,

and for every ε > 0 and M > 0 there exists an n > M such that x

> S −ε.

Then S is the limit superior of {x

}

∞

n=0

. The limit inferior of x

is simply

lim inf

n→∞

= −lim sup

n→∞

(−x

). Finally, note that {x

}

∞

n=0

converges

if and only if lim sup

n→∞

= lim inf

n→∞

= lim

n→∞

. (See Problem

2.18.)

Proposition 2.6. Let {x

}

∞

n=0

⊂ R be a nonin creasing or nondecreas-

ing scalar sequence. If {x

}

∞

n=0

is bounded, then {x

}

∞

n=0

converges to a

ﬁnite real number.

Proof. Let {x

}

∞

n=0

⊂ R be a nondecreasing scalar sequence that is

bounded above. Then, by the completeness axiom,

the supremum α

△

sup{x

: n ∈ Z

} exists and is ﬁnite. Now, let ε > 0 and choose N ∈ Z

such that α −ε < x

≤ α. Since x

≤ x

for n ≥ N and x

≤ α for all n ∈

, it follows that α−ε < x

≤ α for all n ≥ N, and hence, lim

n→∞

= α.

If, alternatively, {x

}

∞

n=0

⊂ R is nonincreasing with inﬁmum β

△

= inf{x

n ∈ Z

}, then {−x

}

∞

n=0

is nondecreasing with supremum −β. Hence,

β = −(−β) = −lim

n→∞

(−x

) = lim

n→∞

Example 2.10. Consider the scalar sequence S = {

}

∞

n=1

⊂ R.

Clearly, S is bounded below, and inf S = 0. However, 0 6∈ S. Alternatively,

sup S = 1 and 1 ∈ S. Hence, the maximum element of th e set S is attained

and is given by sup S, whereas the minimum element of the set S does not

exist. Finally, note that lim inf

n→∞

= lim su p

n→∞

= lim

n→∞

= 0. △

Norms can be used to deﬁne convergent sequences in R

. For the

next deﬁnition, a sequence of vectors {x

, x

, . . .} is deﬁned as an ord ered

multiset with countably inﬁnite elements.

Deﬁnition 2.16. A sequence of vectors {x

}

∞

n=0

⊂ R

converges to

x ∈ R

if lim

n→∞

kx − x

k = 0. In this case, we write lim

n→∞

= x or,

The completeness axiom states that every nonempty set S of real numbers which is bounded

above has a supremum, that is, there exists a real number α s uch that α = sup S. As a consequence

of this axiom it follows that every nonempty set of real numbers which is bounded below has an

inﬁmum.

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DYNAMICAL SYSTEMS AND DIFFERENTIAL EQUATIONS 27

equivalently, x

→ x as n → ∞. The sequence {x

}

∞

n=0

⊂ R

is called a

Cauchy sequence if, for every ε > 0, there exists an integer N = N(ε) such

that kx

− x

k < ε, whenever n, m > N.

Note that convergence guarantees that for every ε > 0, there exists

N = N(ε) such that kx

− xk < ε for all n > N . The next pr oposition

shows that if a sequence converges, its limit is unique.

Proposition 2.7. Let {x

}

∞

n=0

⊂ R

be a convergent sequence. Then

its limit is unique.

Proof. Suppose, ad absurdum, x

→ x and x

→ y as n → ∞, where

x 6= y. Then it follows from the triangle inequality for norms that

kx −yk = kx − x

+ x

− yk ≤ kx −x

k+ kx

− yk.

Hence, as n → ∞, kx − yk → 0. Thus, x = y.

It is always possible to construct a subsequence from any sequence

by selecting arbitrary elements from that sequence. However, in this

construction the order of the terms must be preserved; that is, no interchange

of terms is permissible. More precisely, given a sequen ce {x

}

∞

n=1

⊂ R

and

a sequence of positive integers {n

}

∞

k=1

such that n

< n

< . . ., then the

sequence {x

}

∞

k=1

is a subsequ en ce of {x

}

∞

n=1

. Sequences in general do not

converge, but they often h ave subsequences that do. T he next proposition

gives a characterization of a closure point in terms of sequences.

Proposition 2.8. Let D ⊆ R

. Then x ∈ R

is a closure point of D if

and only if there exists a sequence {x

}

∞

n=1

⊆ D such that x = lim

n→∞

Proof. Let x ∈ R

be a closure point of D. Then, for all n ∈ Z

there exists x

∈ D such that kx − x

k < 1/n. Hence, x

→ x as n → ∞.

Conversely, suppose {x

}

∞

n=1

⊆ D is such that x = lim

n→∞

and let

ε > 0. Then, there exists k ∈ Z

such that kx − x

k < ε for n > k. Thus,

k+1

∈ D ∩B

(x), which implies that D ∩B

(x) is not empty. Hence, x is a

closure point of D.

The following theorem is du e to Bolzano and Weierstrass and forms

the cornerstone of mathematical analysis.

Theorem 2.3 (Bolzano-Weierstrass). Let S ⊂ R

be a bounded set

that contains inﬁnitely many points. Then there exists at least one point

p ∈ R

that is an accumulation point of S.

Proof. See [371].

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28 CHAPTER 2

The next resu lt shows an equivalence between sequential compactness

and compactness; th at is, every sequence in a compact set has a convergent

subsequence and conversely, every compact set is sequentially compact. This

result is known as the Bolzano-Lebesgue theorem.

Theorem 2.4 (Bolzano-Lebesgue). Let D

⊂ R

. For every sequence

}

∞

n=1

⊆ D

there exists a convergent subsequence {x

}

∞

k=1

⊆ {x

}

∞

n=1

such that lim

k→∞

∈ D

if and only if D

is compact.

Proof. To show necessity, let D

be compact, let {x

}

∞

n=1

⊆ D

, and

let Q = {x

, x

, . . .}. Note that Q is an inﬁnite set contained in D

. Now,

since D

is bounded, Q is bound ed , and hence, by the Bolzano-Weierstrass

theorem (Theorem 2.3), Q has an accumulation point x. Since D

is closed,

x is also an accumulation point of D

, and hen ce, x ∈ D

. Thus, for each k ∈

we can ﬁ nd an n

> n

k−1

such that x

∈ B

1/k

(x). Hence, lim

k→∞

x ∈ D

To show suﬃciency, assume that every sequence contained in D

has a

convergent sub s equence and suppose, ad absurdum, that D

is not bounded.

Then there exists a sequence {x

}

∞

n=1

⊆ D

such that kx

k ≥ n, n ∈ Z

and there exists a convergent subsequence {x

}

∞

k=1

⊆ {x

}

∞

n=1

such that

lim

k→∞

= x ∈ D

. Now, for every k ∈ Z

such that n

> 1 + kxk,

−xk ≥ kx

k− kxk ≥ n

− kxk > 1,

which is a contradiction. Hence, D

is bounded.

Next, suppose, ad absurdum, that D

is not closed. Let {x

}

∞

n=1

⊆

be such that lim

n→∞

= x 6∈ D

. By assumption, there exists a

subsequence {x

}

∞

k=1

⊆ {x

}

∞

n=1

such that lim

k→∞

= y ∈ D

. Now,

since lim

n→∞

= x it follows that for every ε > 0, there exists N ∈ Z

such that kx

−xk < ε, n ≥ N. Furthermore, there exists k ∈ Z

such that

≥ N , wh ich implies that kx

−xk < ε. Hence, lim

k→∞

= x. Now, it

follows fr om the uniqueness of limits of sequences (see Proposition 2.7) that

x = y, which is a contradiction. Hence, D

is compact.

The next lemma is needed to show that a sequence in R

converges if

and only if it is a Cauchy sequ en ce.

Lemma 2.1. A Cauchy sequence in R

is bounded.

Proof. Let {x

}

∞

n=0

⊂ R

be a Cauchy sequen ce and let N ∈ Z

such that kx

− x

k < 1 for all n > N. Now, for n > N,

k = kx

− x

+ x

k ≤ kx

− x

k+ kx

k < 1 + kx

k, (2.41)

which proves the result.

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DYNAMICAL SYSTEMS AND DIFFERENTIAL EQUATIONS 29

Theorem 2.5. Let {x

}

∞

n=0

⊂ R

. Then {x

}

∞

n=0

is convergent if and

only if {x

}

∞

n=0

is a Cauchy sequence.

Proof. Assume {x

}

∞

n=0

is convergent, and hence, x

→ x as n → ∞

for some x ∈ R

. Given ε > 0, let N ∈ Z

be such that kx

− xk < ε/2

whenever n > N. Now, if m > N, then kx

− xk < ε/2. If n > N and

m > N it follows that

− x

k = kx

− x + x − x

≤ kx

− xk + kx − x

= ε, (2.42)

which s hows that {x

}

∞

n=0

is a Cauchy sequence. Conversely, assume that

}

∞

n=0

is a Cauchy sequen ce. In this case, it f ollows f rom Lemma 2.1 that

△

= {x

}

∞

n=0

is bounded, and hence, by the Bolzano-Weierstrass theorem

(Theorem 2.3), S has an accumulation point p ∈ R

. Now, since {x

}

∞

n=0

is Cauchy, given ε > 0, there exists N such that kx

−x

k < ε/2 whenever

n, m > N. Since p is an accumulation point of S it follows that the open

ball B

ε/2

(p) contains a point x

with m > N. Hence, if m > N then it

follows that

− pk = kx

− x

+ x

− pk

≤ kx

− x

k + kx

− pk

< ε/2 + ε/2

= ε, (2.43)

and h en ce, x

→ p as n → ∞.

It follows from Proposition 2.8 that the deﬁnition of convergence can

be used to characterize closed sets.

Proposition 2.9. Let D ⊆ R

. Then D is closed if and only if every

convergent sequence {x

}

∞

n=0

⊆ D has its limit point in D.

Proof. The resu lt is a direct consequence of Proposition 2.8.

The next results show that the ﬁnite intersection and inﬁnite union of

open sets is open, whereas the ﬁnite u nion and inﬁnite intersection of closed

sets is closed.

Theorem 2.6. The f ollowing statements hold:

i) L et S

, . . . , S

be open subsets of R

. Then S = S

∩ ··· ∩ S

is an

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30 CHAPTER 2

open s et.

ii) Let {S

: i ∈ I} be an indexed family of open sets contained in R

with the index set I being ﬁnite or inﬁnite. Then S =

i∈I

is an

open s et.

Proof. i) The result is obvious if S = Ø. Assume S 6= Ø, let x ∈ S,

and note that x ∈ S

for all i = 1, . . . , n. Since for all i = 1, . . . , n, S

is an

open set, there exists ε

> 0, i = 1, . . . , n, such that B

(x) ⊆ S

, i = 1, . . . , n.

Now, taking ε = min ε

, it follows that B

(x) ⊆ S

, i = 1, . . . , n, and hence,

(x) ⊆ S, which proves S is open.

ii) Let x ∈ S and note that since S =

i∈I

it follows that x ∈ S

for some j ∈ I. Since S

is open, there exists ε > 0 such that B

(x) ⊆ S

which implies that B

(x) ⊆ S. Thus, x ∈

◦

, and hence, for every x ∈ S,

x ∈

◦

; that is,

◦

= S, which proves that S is open.

Theorem 2.7. The f ollowing statements hold:

i) L et S

, . . . , S

be closed subsets of R

. Then S = S

∪ ··· ∪ S

is a

closed set.

ii) Let {S

: i ∈ I} be an indexed family of closed sets contained in R

with the ind ex set I being ﬁnite or inﬁnite. T hen S =

i∈I

is a

closed set.

Proof. i) Let x ∈ R

be an accumulation point of S. Then, f or every

ε > 0, the set S ∩ (B

(x) \ {x}) is not empty or, equivalently, there exists

a sequence {x

}

∞

n=1

⊆ S such that x = lim

n→∞

. Now, since S is a ﬁnite

collection of closed sets, at least one, say S

, contains an inﬁnite subsequence

}

∞

k=1

⊆ {x

}

∞

n=1

such that x = lim

k→∞

. (This follows from the fact

that if x = lim

n→∞

, th en x = lim

k→∞

for every subsequence {x

}

∞

k=1

of {x

}

∞

n=1

.) Then, x ∈ R

is an accumulation point of S

, and, since S

closed, x ∈ S

. Hence, x ∈ S.

ii) Let x ∈ R

be an accumulation point of S so that, for every ε > 0,

the set S ∩ (B

(x) \ {x}) is not empty. Then every open ball centered at

x contains an inﬁnite number of points of S, and hence, since for every

i ∈ I, S ⊆ S

, contains an inﬁnite number of points of S

. Hence, x is an

accumulation point of S

, and hence, x ∈ S

for each i ∈ I since the sets S

i ∈ I, are all closed. Thus, x ∈ S, and hence, S is closed.

Deﬁnition 2.17. Let S ⊆ R

. Then S is connected if there do not

exist open sets O

and O

in R

such that S ⊂ O

∪ O

, S ∩ O

6= Ø,

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DYNAMICAL SYSTEMS AND DIFFERENTIAL EQUATIONS 31

S ∩ O

6= Ø, and S ∩ O

∩ O

= Ø. S is arcwise connected if for every two

points x, y ∈ S there exists a continuous function g : [0, 1] → S such that

g(0) = x and g(1) = y. A connected component of the set S is a connected

subset of S that is not properly contained in any connected subset of S.

Note that S is connected if and only if it is not the union of two disjoint,

nonempty, relatively open subsets of S. Recall that S is a connected subset

of R if and only if S is either an interval or a single point. S is arcwise

connected if any two points in S can be joined by a continuous cur ve that

lies in S. Arcwise connectedness is an important notion in dynamical sys tem

theory since the state of a dynamical system taking values in R

is typically

a continuous function of time. Hence, in order to control the system state

from any point x ∈ D ⊆ R

to any other point in D, the su bset of the state

space D must be arcwise connected.

Example 2.11. The set of rational numbers Q ⊂ R is disconnected.

To see this, note that Q = X ∩ Y, where X = {x ∈ Q : x <

√

3} and

Y = {y ∈ Q : y >

√

3}. Since X = (−∞,

√

3) ∩ Q and Y = (

√

3, ∞) ∩ Q,

X and Y are nonempty, disjoint, and open relative to Q. Hence, Q is

disconnected. △

Deﬁnition 2.18. Let S ⊆ R

. Then S is convex if µx + (1 − µ)y ∈ S

for all 0 ≤ µ ≤ 1 and x, y ∈ S. The convex hull of S, denoted by co S, is

the intersection of all convex sets containing S, that is, the smallest convex

set that contains S.

Proposition 2.10. Let B

(x) ⊂ R

. Then B

(x) is convex.

Proof. Without loss of generality, consider the open ball with radius

1 and center 0, that is, B

(0) = {x ∈ R

: kxk < 1}. Now, note that if

, y

∈ B

(0), then kx

k < 1 and ky

k < 1. Next, let µ ∈ [0, 1] and note

that

kµx

+ (1 − µ)y

k ≤ kµx

k + k(1 −µ)y

= µkx

k + (1 −µ)ky

< 1, x

, y

∈ B

(0). (2.44)

Hence, µx

+ (1 − µ)y

∈ B

(0) for all x

, y

∈ B

(0) and µ ∈ [0, 1].

Proposition 2.11. Let C ⊆ R

be a convex set. Then C and

◦

are

convex.

Proof. If C = Ø, then C is convex. Let x

, y

∈ C and choose µ =

(0, 1). Now, for a given ε > 0, let x, y ∈ C be such that kx − x

k < ε and

ky −y

k < ε. In this case, kµx + (1 −µ)y −µx

−(1 −µ)y

k < ε, and hen ce,

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32 CHAPTER 2

kz −z

k < ε, where z

△

= µx + (1 −µ)y and z

△

= µx

+ (1 −µ)y

, and z ∈ C.

Since ε > 0 is arbitrary, it follows that z

is a closure point of C, and hence,

C is convex.

◦

= Ø, then

◦

is convex. Let x

, y

∈

◦

and choose µ ∈ (0, 1). Since

, y

∈

◦

, there exists ε > 0 such that B

) ⊂ C and B

) ∈ C. Now,

since z

+ u = µ(x

+ u) + (1 − µ)(y

+ u), it follows that z

+ u ∈ C for all

kuk < ε. Hence, z

is an interior point of C, and hence,

◦

is convex.

Note that in the light of Proposition 2.11 the closed ball B

(x) is also

convex. The next result shows that the intersection of an arbitrary number

of convex sets is convex.

Proposition 2.12. Let S be an arbitrary collection of convex sets.

Then Q =

C∈S

C is convex.

Proof. If Q = Ø, then the result is immediate. Now, assume x

, x

∈

Q and choose µ ∈ [0, 1]. Then x

, x

∈ C for all C ∈ S, and since C is convex,

µx

+ (1 − µ)x

∈ C for all C ∈ S. Thus, µx

+ (1 − µ)x

∈ Q, and hence,

Q is convex.

2.4 Analysis in R

If X and Y are sets, then a function f (·) (or f) that maps X into Y is a

rule f : X → Y that assigns a unique element f (x) of Y to each element

x in X. Equivalently, a function f can be viewed as a subset F of X × Y

such that if (x

, y

) ∈ F, (x

, y

) ∈ F, and x

= x

, then y

= y

. This

connection is denoted by F = graph(f). The set X is called the domain

of f while the set Y is called the codomain of f . If X

⊆ X, then it is

convenient to deﬁne f(X

)

△

= {f(x) : x ∈ X

}. The set f(X) is called the

range of f. If, in addition, Z is a set and g : Y → Z, then g ◦ f : X → Z

is deﬁned by (g ◦ f)(x)

△

= g(f(x)). If x

, x

∈ X and f (x

) = f (x

) implies

that x

= x

, then f is one-to-one (or injective); if f(X) = Y, then f is

onto (or surjective). If f is one-to-one and onto then f is bijective. The

function I

: X → X deﬁned by I

(x) = x, x ∈ X, is called the identity on

X. The function f : X → Y is called left invertible if there exists a function

g : Y → X (called a left inverse of f) such that g ◦ f = I

, while f is right

invertible if there exists a function h : Y → X (called a right inverse of f )

such th at f ◦ h = I

. In addition, the function f : X → Y is invertible if

there exists f

−1

: Y → X such that f

−1

◦f = I

and f ◦ f

−1

= I

A function f : X → Y is said to be into if and only if f (X) ⊂ Y.

For example, if X = Y = Z, then f (x) = x

is an into f unction since f(x)

NonlinearBook10pt November 20, 2007

DYNAMICAL SYSTEMS AND DIFFERENTIAL EQUATIONS 33

contains only integers that can be expressed as x

for x ∈ Z, and hence,

f(X) ⊂ Y = Z. Alternatively, if f(x) = 2x, then f is injective since for

every x

, x

∈ Z such that x

6= x

, f (x

) 6= f(x

). For X = R and Y = R

the function f(x) = x

is surjective since f (X) = Y = R

. Finally, for

X = Y = Z, the function f(x) = −x is bijective since for every x

, x

∈ Z

such that x

6= x

, f(x

) = −x

6= f (x

) = −x

, and hence, f is injective.

In addition, since for every −x ∈ Z, x ∈ Z it follows that f(X) = Y = Z,

and h en ce, f is also surjective.

Two functions f and g are equal, th at is, f = g, if and only if they

have the same domain, the same codomain, and for all x in the common

domain, f(x) = g(x). For example, given f : R → R

deﬁned by f (x) = |x|

and g : R → R

deﬁned by g(x) =

√

, then f (x) = g(x) for all x ∈ R.

The notion of bijection is used to determine set equivalence. I n

particular, given two sets X and Y we say th at X and Y are equivalent,

denoted as X ∼ Y, if and only if there exists a bijective m apping f : X → Y.

Since f is one-to-one and onto, it follows that there exists a one-to-one

correspondence between the elements of the equivalent sets X and Y. We

say a set X is ﬁnite if and only if X = Ø or X is equivalent to the set

{1, . . . , n} for some n ∈ Z

; otherwise X is inﬁnite. If a set X is equivalent

to Z

, then X is called countably inﬁnite (or denumerable). A set X is

countable if and only if it is either ﬁnite or countably inﬁnite.

Example 2.12. The set Z of all integers is countable. In particular,

the f unction f : Z → Z

deﬁned by

f(n) =



2n + 1, n ∈ Z

−2n, n ∈ Z

−

(2.45)

is bijective, and hence, Z is a countable set. The set E = {n ∈ Z

n is even} of all positive even numbers is countable. Speciﬁcally, the

function f : Z

→ E deﬁned by f (n) = 2n is bijective. Finally, the set

P = {2, 4, 8, . . . , 2

, . . .} of powers of 2 is countable as shown by the obvious

one-to-one correspondence of f : Z

→ P given by f (n) = 2

. △

Deﬁnition 2.19. Let D ⊆ R

and f : D → R

. Then the image of

X ⊂ D under f is the set

f(X) = {y : y = f (x) for some x ∈ X}.

The inverse image of Y ⊂ R

under f is the set

−1

(Y) = {x ∈ D : f (x) ∈ Y}.

Note that f

−1

(Y) may be empty even when Y 6= Ø. In particular, an

NonlinearBook10pt November 20, 2007

34 CHAPTER 2

inverse function f

−1

does not exist for the function f : R → R

deﬁned by

y = f (x) = x

since x = ±

√

y, and hence, there always exist two image

points

√

y and −

√

y for every y > 0. This violates the uniquen ess property

of a function. However, f : R

→ R

and g : R

−

→ R

deﬁned by

y = f(x) = g(x) = x

, f

−1

= R

→ R

and g

−1

: R

→ R

−

exist and are

given by x =

√

y and x = −

√

y, respectively.

The next result su mmarizes some of the important properties of images

and inverse images of functions.

Theorem 2. 8. Let X and Y be sets, let f : X → Y, let A, A

, and

be s ubsets of X, and let B

and B

be s ubsets of Y. Then the following

statements hold:

i) I f A

⊆ A, then f (A

) ⊆ f(A).

ii) f (A

∪ A

) = f(A

) ∪f(A

iii) f

−1

∪ B

) = f

−1

) ∪f

−1

iv) f

−1

∩ B

) = f

−1

) ∩f

−1

Proof. i) Let y ∈ f (A

). I n this case, there exists x ∈ A

such that

y = f(x). Since A

⊆ A, x ∈ A. Hence, f(x) = y ∈ f(A), wh ich proves

f(A

) ⊆ f(A).

ii) Let y ∈ f (A

∪ A

) and note that y = f (x), where x ∈ A

x ∈ A

. Hence, y ∈ f(A

) or y ∈ f(A

) or, equivalently, y ∈ f (A

) ∪f (A

Conversely, let y ∈ f (A

) ∪f(A

) and note that y = f(x), where x ∈ A

x ∈ A

. Thus, x ∈ A

∪ A

, and hence, y = f(x) ∈ f (A

∪ A

iii) Let x ∈ f

−1

∪ B

) and note th at in this case f(x) ∈ B

∪

. Hence, f (x) ∈ B

or f(x) ∈ B

. This implies that x ∈ f

−1

) or

x ∈ f

−1

) or, equivalently, x ∈ f

−1

) ∪ f

−1

). Conversely, let x ∈

−1

) ∪ f

−1

). In this case, x ∈ f

−1

) or x ∈ f

−1

), and hence,

f(x) ∈ B

or f(x) ∈ B

. Thus, f (x) ∈ B

∪ B

, which implies that x ∈

−1

∪ B

iv) Let x ∈ f

−1

∩ B

) and note that in this case f(x) ∈ B

∩ B

Hence, f(x) ∈ B

and f (x) ∈ B

. This implies that x ∈ f

−1

) and

x ∈ f

−1

) or, equivalently, x ∈ f

−1

) ∩ f

−1

). Conversely, let x ∈

−1

) ∩ f

−1

). In this case, x ∈ f

−1

) and x ∈ f

−1

), and hence,

f(x) ∈ B

and f(x) ∈ B

. Thus, f(x) ∈ B

∩ B

, which implies that

x ∈ f

−1

∩ B