We now begin the study of the most important class of functions that arises in real analysis:
the class of continuous functions. The term "continuous" has been used since the time of Newton to refer to the motion of bodies or to describe an unbroken curve, but it was not made precise until the nineteenth century. Work of Bernhard Bolzano in 1 8 17 and Augustin-Louis Cauchy in 1 821 identified continuity as a very significant property of functions and proposed definitions, but since the concept is tied to that of limit, it was the careful work of Karl Weierstrass in the 1 870s that brought proper understanding to the idea of continuity.
We will first define the notions of continuity at a point and continuity on a set, and then show that various combinations of continuous functions give rise to continuous functions.
Then in Section 5.3 we establish the fundamental properties that make continuous functions so important. For instance, we will prove that a continuous function on a closed bounded interval must attain a maximum and a minimum value. We also prove that a continuous function must take on every value intermediate to any two values it attains. These properties and others are not possessed by general functions, as various examples illustrate, and thus they distinguish continuous functions as a very special class of functions.
In Section 5.4 we introduce the very important notion of unifonn continuity. The distinction between continuity and uniform continuity is somewhat subtle and was not fully appreciated until the work of Weierstrass and the mathematicians of his era, but it proved to be very significant in applications. We present one application to the idea of approximating continuous functions by more elementary functions (such as polynomials).
Karl Weierstrass
Karl Weierstrass (=Weierstraj3) ( 1 8 1 5-1 897) was born in Westphalia, Germany. His father, a customs officer in a salt works, insisted that he study law and public finance at the University of Bonn, but he had more interest in drinking and fencing, and left Bonn without receiving a diploma. He then enrolled in the Academy of Munster where he studied mathematics with Christoph Gudermann. From 1 841 to 1 854 he taught at various gymnasia in Prussia. Despite the fact that he had no contact with the mathematical world during this time, he worked hard on mathematical research and was able to publish a few papers, one of which attracted considerable attention. Indeed,
the University of Konigsberg gave him an honorary doctoral degree for this work in 1 855. The next year, he secured positions at the Industrial Institute of Berlin and the University of Berlin. He remained at Berlin until his death.
A methodical and painstaking scholar, Weierstrass distrusted intuition and worked to put everything on a firm and logical foundation. He did fundamental work on the foundations of arithmetic and analysis, on complex analysis, the calculus of variations, and algebraic geometry.
Due to his meticulous preparation, he was an extremely popular lecturer; it was not unusual for him to speak about advanced mathematical topics to audiences of more than 250. Among his auditors are counted Georg Cantor, Sonya Kovalevsky, Gosta Mittag-Leffler, Max Planck, Otto Holder, David Hilbert, and Oskar Bolza (who had many American doctoral students). Through his writings and his lectures, Weierstrass had a profound influence on contemporary mathematics.
124
5 . 1 CONTINUOUS FUNCTIONS 125 The notion of a "gauge" is introduced in Section 5.5 and is used to provide an alternative method of proving the fundamental properties of continuous functions. The main signifi
cance of this concept, however, is in the area of integration theory where gauges are essential in defining the generalized Riemann integral. This will be discussed in Chapter 10.
Monotone functions are an important class of functions with strong continuity properties and they are discussed in Section 5.6.
Section 5.1 Continuous Functions
In this section, which is very similar to Section 4. 1 , we will define what it means to say that a function is continuous at a point, or on a set. This notion of continuity is one of the central concepts of mathematical analysis, and it will be used in almost all of the following material in this book. Consequently, it is essential that the reader master it.
5.1.1 Definition Let A t;::; R letf : A -> R and let c E A. We say thatfis continuous at c if, given any number 8 > 0, there exists 8 > 0 such that if x is any point of A satisfying lx � Iff cl
<
fails to be continuous at c, then we say that f 8, then lf(x) � f(c) l< 8.
is discontinuous at c.As with the definition of limit, the definition of continuity at a point can be formulated very nicely in terms of neighborhoods. This is done in the next result. We leave the verification as an important exercise for the reader. See Figure 5. 1 . 1 .
c
VB(c)
Figure 5.1.1 Given Vr. (f(c) ), a neighborhood V8 (c) is to be determined
5.1.2 Theorem A function f : A -> IR is continuous at a point c E A if and only if given any �::-neighborhood V" (!(c) ) of f(c) there exists a a-neighborhood V8 ( c) of c such that if x is any point of A n V8 (c), then f(x) belongs to V, (f(c)), that is,
f(A n Va (c)) t;::; V, (f(c)) .
Remarks (1) If c E A is a cluster point of A, then a comparison of Definitions 4. 1 .4 and 5. 1 . 1 show that f is continuous at c if and only if
( 1 ) f(c) = x � c limf(x).
Thus, if cis a cluster point of A, then three conditions must hold forjto be continuous at c: (iii) these two values must be equal. (ii) the limit off at c must exist in (i) fmust be defined at c (so thatf(c) makes sense),
lR(so that lim
X---->C f(x)makes sense), and
(2)
If c
EA is not a cluster point of A, then there exists a neighborhood V8(c) of c such that A n V8(c) = {c}. Thus we conclude that a functionfis automatically continuous at a point c
EA that is not a cluster point of A. Such points are often called "isolated points" of A. They are oflittle practical interest to us, since they have no relation to a limiting process. Since continuity is automatic for such points, we generally test for continuity only at cluster points. Thus we regard condition (1) as being characteristic for continuity at c.
A
slight modification of the proof of Theorem 4.1.8 for limits yields the following sequential version of continuity at a point.
5.1.3 Sequential Criterion for Continuity
A
function f: A
---> lR is continuous at the point c EA
if and only if for every sequence (xn) inA
that converges to c, the sequence(f(xn)) converges to
f(c).
The following Discontinuity Criterion is a consequence of the last theorem. It should be compared with the Divergence Criterion 4.1.9(a) with
L= f(c). Its proof should be written out in detail by the reader.
5.1.4 Discontinuity Criterion Let
A
� JR, letf : A
---> R and let c EA.
Thenf
is discontinuous at c if and only if there exists a sequence (xn) inA
such that (xn) converges toc,
but the sequence (f(xn)) does not converge to f(c).So far we have discussed continuity at a
point.To talk about the continuity of a function on a
set,we will simply require that the function be continuous at each point of the set. We state this formally in the next definition.
5.1.5 Definition
Let A
� lRand let f : A
---> RIf
Bis a subset of
A,we say that f is
continuous on the set B
iff is continuous at every point of
B.5.1.6 Examples (a)
It was seen in Example 4.1.7(a) that if c The constant function f(x) :=
E lR,then lim
bis continuous on
f(x) = b.Sincef(c) =
R b,we have lim
X...--J. L' f(x)= f(c), and thus f is continuous at ev'er:y point c
E RTherefore f is continuous on
R(b) g(x)
It was seen in Example 4.1.7(b) that if c :=
xis continuous on
R E lR,then we have lim
g= c. Since
g(c)= c, then
gis continuous at every point c
E RThus
gis continuo�sSn
R(c) h(x)
It was seen in Example 4.1.7(c) that if c := x2 is continuous on
R E lR,then we have limh = c2. Since
h(c)
= c2, then h is continuous at every point c
E RThus h is contintots on
R (d)It was seen in Example 4.1.7(d) that if c
({i(
x) := ljx is continuous on A := {
x E lR : E x >A, then we have lim (/! = ljc. Since 0}.
x----j.c
({i(
c) = 1 j c, this shows that(/! is continuous at every point c
EA. Thus(/! is continuous on A.
5 . 1 CONTINUOUS FUNCTIONS 127 (e) <p(x) : = 1 /x is not continuous at x = 0.
Indeed, if <p(x) = 1 /x for x > 0, then <p is not defined for x = 0, so it cannot be continuous there. Alternatively, it was seen in Example 4. 1 . 1 0( a) that lim <p does not exist
x----rO
in lR, so <p cannot be continuous at x = 0.
(f) The signum function sgn is not continuous at 0.
The signum function was defined in Example 4. 1 . 1 O(b ), where it was also shown that lim sgn(x) does not exist in R Therefore sgn is not continuous at x = 0 (even though sgn 0
x---+0
is defined). It is an exercise to show that sgn is continuous at every point c "I 0.
Note In the next two examples, we introduce functions that played a significant role in the development of real analysis. Discontinuities are emphasized and it is not possible to graph either of them satisfactorily. The intuitive idea of drawing a curve in the plane to represent a function simply does not apply, and plotting a handful of points gives only a hint of their character. In the nineteenth century, these functions clearly demonstrated the need for a precise and rigorous treatment of the basic concepts of analysis. They will reappear in later sections.
(g) Let A : = IR and let f be Dirichlet's "discontinuous function" defined by f(x) : =
{ �
if x is rational,if x is irrational.
We claim thatfis not continuous at any point of!R'.. (This function was introduced in 1 829 by P. G. L. Dirichlet.)
Indeed, if c is a rational number, let (xn) be a sequence of irrational numbers that converges to c. (Corollary 2.4.9 to the Density Theorem 2.4.8 assures us that such a sequence does exist.) Since f(xn ) = 0 for all n E N, we have lim(f(xn ) ) = 0, while f( c) = I . Therefore f is not continuous at the rational number c.
On the other hand, if b is an irrational number, let ( Yn) be a sequence of rational numbers that converge to b. (The Density Theorem 2.4.8 assures us that such a sequence does exist.) Since f(Yn) = I for all n E N, we have lim(.f"( yn ) ) = I, while f(b) = 0.
Therefore f is not continuous at the irrational number b.
Since every real number is either rational or irrational, we deduce that f is not continuous at any point in R
0
. .
. .
" . . . .
0.5
. . . . . '
1
,•. .
... . . . .
1 . 5
Figure 5.1.2 Thomae's function
2
(h) Let A := {x E � : x > 0} . For any irrational number x > 0 we define h(x) := 0.
For a rational number in A of the form mjn, with natural numbers m, n having no common factors except 1 , we define h(mjn) := 1 /n. (We also define h(O) := 1 .)
We claim that h is continuous at every irrational number in A, and is discontinuous at every rational number in A. (This function was introduced in 1 875 by K. J. Thomae.)
Indeed, if a > 0 is rational, let (xn) be a sequence of irrational numbers in A that converges to a. Then lim(h(xn ) ) = 0, while h(a) > 0. Hence h is discontinuous at a.
On the other hand, if b is an irrational number and £ > 0, then (by the Archimedean Property) there is a natural number n0 such that 1 /no < D. There are only a finite number of rationals with denominator less than no in the interval (b -1 , b + 1 ) . (Why?) Hence 8 > 0 can be chosen so small that the neighborhood (b - 8, b + 8) contains no rational numbers with denominator less than n0. It then follows that for lx -bl < 8, x E A, we have lh(x) -h(b) l = lh(x) l :::; 1 /no < D. Thus h is continuous at the irrational number b.
Consequently, we deduce that Thomae's function h is continuous precisely at the
irrational points in A. (See Figure 5 . 1 .2.) D
5.1.7 Remarks (a) Sometimes a function f : A -+ � is not continuous at a point c because it is not defined at this point. However, if the functionfhas a limit L at the point c and if we define F on A U { c} --+ � by
F(x) : =
{ J
(x) for x = c, for x E A ,then F i s continuous at c. To see this, one needs to check that lim F = L, but this follows (why?), since lim f = L.
X--J.C
X--J.C
(b) If a function g : A -+ � does not have a limit at c, then there is no way that we can obtain a function G : A U { c} -+ � that is continuous at c by defining
G(x) : =
{�
(x) for x = c, for x E A .To see this, observe that if lim G exists and equals C, then lim g must also exist and
equal C. X----+C X----t C
5.1.8 Examples (a) The function g(x) := sin ( 1 /x) for x -1 0 (see Figure 4. 1 .3) does not have a limit at x = 0 (see Example 4. 1 . 1 0( c)). Thus there is no value that we can assign at x = 0 to obtain a continuous extension of g at x = 0.
(b) Letf(x) := x sin ( 1 /x) for x -1 0. (See Figure 5. 1 .3.) It was seen in Example 4.2.8(f) that lim (x sin( l /x)) = 0. Therefore it follows from Remark 5. 1 .7(a) that if we define x�o F : � -+ � by
{
0F ( ) X · = · x sin( l /x) then F is continuous at x = 0.
for x = 0, for x -1 0,
D
y
5.1 CONTINUOUS FUNCTIONS 129
Figure 5.1.3 Graph of f(x) = x sin ( l /x) (x f' 0) Exercises for Section 5.1
1 . Prove the Sequential Criterion 5. 1 .3.
2. Establish the Discontinuity Criterion 5. 1 .4.
3. Let a < h < c. Suppose that f is continuous on [a, h i , that g is continuous on [b, c] , and that f(b) = g(b ). Define h on [a, c] by h(x) := f(x) for x E [a, b] and h(x) := g(x) for x E [b, c].
Prove that h is continuous on [a, c] .
4. If x E R, we define [x] to be the greatest integer n E Z such that n ::; x. (Thus, for example, [8.3] = 8, [n] = 3, [ - n] = -4.) The function x >--? [x] is called the greatest integer function.
Determine the points of continuity of the following functions:
(a) f(x) := [x], (b) g(x) := x [x],
(c) h(x) := [sin x], (d) k(x) := [ 1 /x] (x f' 0) .
5. Let f be defined for all x E R, x i 2, by f(x) = (x2 + x - 6)/(x - 2). Can f be defined at x = 2 in such a way that f is continuous at this point?
6. Let A C::: R and letf : A ---. R be continuous at a point c E A. Show that for any D > 0, there exists a neighborhood V8 (c) of c such that if x, y E A n V0 (c) , then lf(x) -f(y) l < D.
7. Letf : R ---. R be continuous at c and letf(c) > 0. Show that there exists a neighborhood V8 (c)
of c such that if x E V8 (c), then f(x) > 0.
8. Let f : R .:...., R be continuous on R and let S := {x E R : f(x) = 0} be the "zero set" off If (xn) is in S and x = lim(x11), show that x E S.
9. Let A C::: B C::: R, let { : B ---. R and let g be the restriction of f to A (that is, g(x) = f(x) for
X E A).
(a) Iff is continuous at c E A, show that g is continuous at c.
(b) Show by example that if g is continuous at c, it need not follow that f is continuous at c.
1 0. Show that the absolute value function f(x) := lxl is continuous at every point c E R.
I I . Let K > 0 and let f : R ---. R satisfy the condition lf(x) - f(y) l ::; Klx -Yl for all x, y E R.
Show that f is continuous at every point c E R.
1 2. Suppose thatf : R ---. R is continuous on R and thatf(r) = 0 for every rational number r . Prove that f(x) = 0 for all x E R.
1 3. Define g : R ---. R by g(x) := 2x for x rational, and g(x) := x + 3 for x irrational. Find all points at which g is continuous.
1 4. Let A : = ( 0, oo) and let k : A __. JR. be defined as follows. For x E A, x irrational, we define k(x) = 0; for x E A rational and of the form x = mjn with natural numbers m, n having no common factors except I , we define k(x) := n. Prove that k is unbounded on every open interval in A. Conclude that k is not continuous at any point of A. (See Example 5 . 1 .6(h).)
1 5 . Let j' : (0, I ) __.JR. be bounded but such that lillJ f does not exist. Show that there are two sequences (xn) and (Yn) in (0, I ) with lim(xn) �0 = lim(yn) , but such that (f(xn)) and (f(yn) ) exist but are not equal.
Section 5.2 Combinations of Continuous Functions
Let A <;;;: IR and letf and g be functions that are defined on A to IR and let b E R In Definition 4.2.3 we defined the sum, difference, product, and multiple functions denoted by f + g, J - g, Jg, hf. In addition, if h : A --+ IR is such that h(x) =/= 0 for all x E A, then
we defined the quotient function denoted by f I h.
The next result is similar to Theorem 4.2.4, from which it follows.
5.2.1 Theorem Let A <;;;: R let f and g be functions on A to R and let b E R Suppose that c E A and that f and g are continuous at c.
(a) Then f + g, f - g, fg, and bf are continuous at c.
(b) If h : A --+ IR is continuous at c E A and if h(x) =/= 0 for all x E A, then the quotient
!I h is continuous at c.
Proof. If c E A is not a cluster point of A, then the conclusion is automatic. Hence we assume that c is a cluster point of A .
(a) Since f and g are continuous at c, then
f(c) = lim f and g(c) = lim g.
x�c x�c
Hence it follows from Theorem 4.2.4(a) that
(f + g) (c) = J(c) + g(c) = lim (f + g) .
X--+('
Therefore f + g is continuous at c. The remaining assertions in part (a) are proved in a similar fashion.
(b) Since c E A, then h(c) =/= 0. But since h(c) = lim h, it follows from Theorem 4.2.4(b)
X---+C
that
[_ (c) = f(c) =h h(c) ��lim h /= limx�c
(
£h)
·X..--.--+C
Therefore f/ h is continuous at c. Q.E.D.
The next result is an immediate consequence of Theorem 5.2. 1 , applied to every point of A. However, since it is an extremely important result, we shall state it formally.
5.2.2 Theorem Let A <;;;: IR, let f and g he continuous on A to IR, and let b E R
(a) The functions f + g, f - g, fg, and bf are continuous on A.
(b) If h : A --+ IR is continuous on A and h(x) =/= 0 for x E A, then the quotient !I h is continuous on A .
5.2 COMBINATIONS OF CONTINUOUS FUNCTIONS 131 Remark To define quotients, it is sometimes more convenient to proceed as follows. If cp : A --> R let A 1 := { x E A : cp(x) y'c 0} . We can define the quotient! jcp on the set A 1 by ( 1 )
([_)
cp (x) : = f(x) cp (x) for x E A 1 •If cp is continuous at a point c E A 1 , it is clear that the restriction cp1 of cp to A 1 is also continuous at c. Therefore it follows from Theorem 5.2. 1 (b) applied to cp1 that f/cp1 is continuous at c E A. Since (! jcp) (x) = (! /cp1 ) (x) for x E A 1 it follows that f/cp is continuous at c E A 1 . Similarly, if f and cp are continuous on A, then the function f / cp, defined on A 1 by (1 ), is continuous on A 1 .
5.2.3 Examples (a) Polynomial functions.
Ifp is a polynomial function, so thatp(x) = anx" + an_1xn-l + · · · + a 1 x + ao for all x E R then it follows from Example 4.2.5(f) that p(c) = lim p for any c E JR. Thus a polynomial function is continuous on R x ... c
(b) Rational functions.
If p and q are polynomial functions on R then there are at most a finite number
a 1 , • • • , am of real roots of q. If x ¢:. { a 1 , • • • , am } then q(x) y'c 0 so that we can define the rational function r by
. p(x) r(x) : =q(x)
It was seen in Example 4.2.5(g) that if q( c) y'c 0, then p(c) . p(x) . r(c) = -( ) = hm -( ) = bm r(x) .
q C X->C q X X->C
In other words, r is continuous at c. Since c is any real number that is not a root of q, we infer that a rational function is continuous at every real number for which it is defined.
(c) We shall show that the sine function sin is continuous on R
To do so we make use of the following properties of the sine and cosine functions.
(See Section 8.4.) For all x, y, z E IR'. we have:
lsin zl s; lzl , Ieos zl s; 1 ,
sin x - sin y = 2 sin [!( x -y) J cos [!( x + y) J . Hence if c E IR'., then we have
I sin x - sin cl s; 2 ·! lx -cl · 1 = lx -ci -
Therefore sin is continuous at c. Since c E IR'. is arbitrary, it follows that sin is continuous on R
(d) The cosine function is continuous on R
We make use of the following properties of the sine and cosine functions. For all x, y, z E IR'. we have:
I sin zl s; lzl , I sin zl s; 1 ,
cos x - cos y = -2 sin [!( x + y) J sin [!( x -y)].
Hence if c E R then we have
Ieos x - cos cl s; 2 · 1 · !lc - xl = lx -c1 -
Therefore cos is continuous at c. Since c E JR. is arbitrary, it follows that cos is continuous on R (Alternatively, we could use the relation cos x = sin(x + n /2).)
(e) The functions tan, cot, sec, esc are continuous where they are defined.
For example, the cotangent function is defined by cos x cot x : = -.sm-x
provided sin x -1 0 (that is, provided x -1 nn, n E Z). Since sin and cos are continuous on JR., it follows (see the Remark before Example 5.2.3) that the function cot is continuous on its domain. The other trigonometric functions are treated similarly. D
5.2.4 Theorem Let A � R letf : A ---> JR., and let IJI be defined by IJI (x) : = lf(x) l for
X E A.
(a) Iff is continuous at a point c E A, then III is continuous at c.
(b) Iff is continuous on A, then III is continuous on A.
Proof. This is an immediate consequence of Exercise 4.2. 14. Q.E.D.
5.2.5 Theorem Let A � R letf : A ---> JR., and letf(x) ?: Ofor all x E A. We let ,jJ be definedfor x E A by (VJ) (x) := Jl(X5.
(a) Iff is continuous at a point c E A, then ,jJ is continuous at c.
(b) Iff is continuous on A, then ,jJ is continuous on A.
Proof. This is an immediate consequence of Exercise 4.2. 1 5. Q.E.D.
Composition of Continuous Functions ________________ _
We now show that if the function! : A ---> JR. is continuous at a point c and if g : B ---> JR. is continuous at b = f(c), then the composition g o f is continuous at c. In order to assure that g of is defined on all of A, we also need to assume that f(A) � B.
5.2.6 Theorem Let A, B � JR. and let f : A ---> JR. and g : B ---> JR. be functions such that f(A) � B. Iff is continuous at a point c E A and g is continuous at b = f( c) E B, then the
composition g of : A ---> JR. is continuous at c.
Proof. Let W be an £-neighborhood of g(b). Since g is continuous at b, there is a 8-neighborhood V of b = f(c) such that if y E B n V then g(y) E W. Sincefis continuous at c, there is a y-neighborhood U of c such that if x E A n U, then f(x) E V. (See Figure 5.2. 1 .) Since f(A) � B, it follows that if x E A n U, then f(x) E B n V so that g o f(x) = g(f(x)) E W. But since W is an arbitrary £-neighborhood of g(b), this implies
that g o f is continuous at c. Q.E.D.
5.2.7 Theorem Let A , B � JR., let f : A ---> JR. be continuous on A, and let g : B ---> JR. be continuous on B. lff(A) � B, then the composite function g o f : A ---> JR. is continuous on A.
Proof. The theorem follows immediately from the preceding result, if f and g are continuous at every point of A and B, respectively. Q.E.D.