CONTINUOUS FUNCTIONS
4.1CONTINUITY: IDEAS, BASIC TERMINOLOGY, PROPERTIES
The idea of continuity of a function ƒ at a point x is familiar. Intuitively, as y gets close to x, ƒ(y) gets close to ƒ(x). This has a direct translation in terms of sequences.
4.1.1Definition
Let ƒ be a function defined on the metric space Ω and taking values in the metric space Ω′; the standard notation is ƒ : Ω → Ω′. The set Ω is called the domain of f, and Ω′ the codomain of f; ƒ is sometimes called a mapping. (If it aids the understanding, you may assume Ω = Ω′ = R.) If x ∈ Ω, we say that ƒ is continuous at x if for all sequences x1, x2, ... in Ω with xn → x, we have f(xn) →ƒ(x).
There is another way of describing continuity, the so-called epsilon-delta approach.
4.1.2THEOREM. The function ƒ is continuous at x if and only if for every ∈ > 0 there is a δ > 0 such that whenever1 y is a point of Ω with d(x, y) < δ, we have d (f(x), f(y)) < ∈.
Thus, if we wish to force ƒ(y) to be “close” to ƒ(x), that is, within distance ∈, we can accomplish this by takings “close” to x, in other words, within distance δ. (We have followed common practice and used the same letter d for the metrics on Ω and Ω′. This should cause no confusion since in each expression involving d it is clear which space is being considered).
Proof. We may write the epsilon-delta condition described in Theorem 4.1.2 as follows (the symbol ⇒ denotes “implies”):
The negation of this statement is
Thus, if (1) is false then we can find some ∈ > 0 such that no matter which δ > 0 is selected, there is a point y within distance δ of x such that f(y) is at least distance ∈ from ƒ(x). Since any choice of δ is allowed, we can take δ = 1/n, n = 1, 2, … ; label the corresponding point y as xn. Then d(x, xn) < 1/n, and consequently xn → x. But d(f(x), f(xn)) > ∈ for all n, so f(xn) 
ƒ(x). We conclude that ƒ is discontinuous at x.
Now assume that (1) holds, and let xn → x. Given ∈ > 0, choose δ > 0 so that (1) is satisfied. Since xn → x, we have d(x, xn) < δ eventually, so by (1), d(f(x), f(xn)) < ∈ eventually. It follows that f(xn) →f(x). ■
It is possible to show directly that a familiar function such as a polynomial is continuous (see Problems 1–3), but the easiest way to do this is to show that the function is differentiable. In the next chapter we study differentiation, and show that differentiability implies continuity.
A basic continuity result involves composition of functions, which we now describe. If f: Ω → Ω′ and g: Ω′ → Ω″, we define the composition of f and g as the function h: Ω → Ω″ given by
Thus, given x we compute ƒ(x) ∈ Ω′ and then apply g to ƒ(x) to get h(x) (see Fig. 4.1.1). The standard notation for composition is g ∘ ƒ, sometimes read “ƒ followed by g.” ■
4.1.3THEOREM. The composition of continuous functions is continuous; that is, if f is continuous at x and g is continuous at ƒ(x), then g ∘ f is continuous at x.
Proof. If xn → x, then f(xn) → f(x); hence, g(f(xn)) →g(f(x)). ■
We have described continuity at a particular point of Ω; “continuity on Ω” simply means continuity at each point. However, the global concept of continuity on the entire space (as opposed to the local idea of continuity at a single point) has an important description in terms of open and closed sets. Before discussing this, one more bit of terminology is needed.
4.1.4Definition
If ƒ : Ω → Ω′ and A is a subset of Ω′, the preimage (or inverse image) of A under ƒ is
In other words,
Figure 4.1.1 Composition of Functions
Thus ƒ−1(A) is the set of points mapped by ƒ into A. See Fig. 4.1.2 for an explicit calculation in a simple case.
Preimages behave very well with respect to unions, intersections, and complements, as the following result shows.
4.1.5THEOREM. Let ƒ : Ω → Ω′, and let {Αi} be an arbitrary family (there may be uncountably many i) of subsets of Ω′. Then
and
Also, if A1 ⊆A2, then
Proof
The final statement is a consequence of the definition of preimage. ■
We may now relate continuity to open and closed sets.
4.1.6THEOREM. Let ƒ : Ω → Ω′, where Ω and Ω′ are metric spaces. The function ƒ is continuous on Ω if and only if for each open set V ⊆ Ω′ the preimage f−1(C) is an open subset of Ω. Equivalently, for each closed set C ⊆ Ω′, the preimage f−1(C) is a closed subset of Ω.
Proof. Assume ƒ continuous. Let x belong to ƒ−1(V), where V is open in Ω′. Then ƒ(x) ∈ V, so or some ∈ > 0, B∈ (f (x)) ⊆ V. If δ > 0 is as given by the statement that ƒ is continuous at x (see Theorem 4.1.2), then
Thus, y ∈ f−1(V) proving that Bδ(x) ⊆ f−1(V). Therefore ƒ−1(V) is open.
Conversely, assume V open implies ƒ−1(V) open. If x ∈ Ω, we show that ƒ is continuous at x. Given ∈ > 0, f(x) ∈ B∈(f(x), which is an open set V. Thus, x ∈ ƒ−1(V), which is open by hypothesis, so Bδ(x) ⊆ ƒ−1 (V) for some δ > 0. Consequently,
in other words,
By Theorem 4.1.2, ƒ is continuous at x.
Finally, we must show that the preimage of each closed set is closed if and only if the preimage of each open set is open. Suppose that for each closed C ⊆ Ω′, ƒ−1(C) is closed, and assume V is an open subset of Ω′. Then Vc is closed, so ƒ−1 (Vc) is closed. But by Theorem 4.1.5,
Thus, [ƒ−1 (V)]c is closed, so ƒ−1(V) is open. Conversely, if the preimage of each open set is open and C is a closed subset of Ω′, then Cc is open, and hence ƒ−1(Cc) = [ƒ−1(C)]c is open. Therefore ƒ−1(C) is closed. ■
A companion to the preimage idea is that of the direct image, defined as follows.
4.1.7Definition
If ƒ : Ω → Ω′ and A is a subset of Ω, the image (or direct image) of A under ƒ is
thus, ƒ(A) is the set of all values ƒ(x) as x ranges over A.
Direct images are not as well behaved as preimages (see Problem 4), but they have some desirable properties.
4.1.8THEOREM. Let f : Ω → Ω′. Then
(a)If A ⊆ Ω, then A ⊆ f−1[f(A)].
(b)If B ⊆ Ω′, then ƒ [f−1(B)] ⊆ B.
(c)If A ⊆ C ⊆ Ω, then f (A) ⊆ f(C).
(d)If {Ai} is an arbitrary family of subsets of Ω, then
(a)If x ∈ A, then f(x) ∈ f (A); hence, x ∈ f−1[f(A)]. Note that A can be a proper subset of ƒ−1[f (A)]; see Fig. 4.1.3.
(b)If y ∈ f[ƒ−1(B)], then y = f(x) for some x ∈ƒ−1(B). But then ƒ(x) ∈ B; that is, y ∈ B. Note that f [f−1(B)] can be a proper subset of B; see Fig 4.1.4.
(c)If y = ƒ(x) for some x ∈ 4, then x also belongs to C, so y ∈ f (C).
(d)We have y = ƒ(x) for some 
iff for at least one i, y = f(x) for some x ∈ Ai, iff 
■
Problems for Section 4.1
In Problems 1–3, all numbers are real and all functions are from R to R.
1.If xn → x and yn → y, show that
(a)xn +yn → x + y.
(b)xn − yn → x − y.
(c)xn yn → xy.
(d)xn/yn → x/y (if y ≠ 0).
2.Let ƒ and g be continuous at x. Show that ƒ + g, ƒ – g, fg, and ƒ/g are continuous at x. (p = ƒg is the product of ƒ and g, i.e., p(y) = f(y)g(y), and q = f/g is the quotient of ƒ and g, i.e., q(y) = f(y)/g(y). Assume g(x) ≠ 0, so that q(y) will be defined for y sufficiently close to x.)
Figure 4.1.3 A Can Be a Proper Subset of f−1 [f(A)]
3.Show that a polynomial (ƒ(x) = a0 + a1x + ··· + anxn) is continuous everywhere.
4.If f: Ω → Ω′ and {Ai} is an arbitrary family of subsets of Ω, is it true that
5.Let A be the set of real numbers x such that sin x = c, where c is a constant between −1 and +1. Show that A is closed. Can you identify a theorem of which this is a special case?
4.2CONTINUITY AND COMPACTNESS
Continuous functions on compact sets have a number of special properties; in particular, they are bounded and attain a maximum and a minimum value. These results are consequences of the following fundamental theorem.
4.2.1THEOREM. Let f : K → Ω′, where K is a compact subset of the metric space Ω. If ƒ is continuous on K, the image f(K), that is, {f(x) : x ∈ K}, is a compact subset of Ω′. For short, the continuous image of a compact set is compact.
Proof. Let 
, where the Gi are open subsets of Ω′. Then
By compactness, there is a finite subcovering, say
But then
Since ƒ [ƒ−1(Gj)] ⊆ Gj by Theorem 4.1.8(b), we have a finite subcovering of the original open covering of ƒ(K). ■
4.2.2COROLLARY. If f is a continuous function on the compact set K, then ƒ is bounded (in other words, f(K) is bounded set). If f is real-valued, then ƒ attains a maximum and a minimum on K.
Proof. By Theorem 4.2.1, f(K) is compact, and therefore closed and bounded by Theorem 2.2.4. In the real case, let s be the sup of ƒ(if), and let t = inf f(K). By Theorem 2.4.3, s and t belong to f(K). If f(x) = s and f(y) = t, then ƒ attains a maximum at s and a minimum at t. ■
If the hypothesis of compactness is dropped, it is easy to produce continuous functions having no maximum or minimum value. For example, consider ƒ(x) = x on (0, 1). Of course, in this case there is a natural extension of ƒ to the compact set [0, 1], and ƒ has a maximum and a minimum on the larger domain. However, no such extension is possible for g(x) = (1/x) sin(1/x), 0 < x < 1. The question of extension of continuous functions will be studied later in this section; it is closely connected with the idea of uniform continuity, which we now discuss.
We first look at the epsilon-delta description of continuity (see Theorem 4.1.2) for the function ƒ(x) = 1/x, x > 0. Given ∈ > 0 and x > 0, we wish to find δ > 0 such that if |x – y| < δ then |f (x) – ƒ(y)| < ∈; that is, |x – y | < xy ∈. Our first try is 
, which is less than xy ∈ if x < 2y. Now if 
(so that 
; remember that x, y > 0 here), then
Thus, our final choice of δ is min
.
Our aim is not to develop techniques for finding a δ for a given ∈, but to point out that δ may depend on x as well as ∈. To force 1/y to be within distance ∈ of 1/x, we must choosey closer to x if x is near 0 than if x is far from 0. This can be verified by looking at a picture of 1/x; it is harder to get 1/y close to 1/x near 0 because the derivative (the slope of the tangent to the curve) is – 1/x2, which approaches – ∞ as x → 0. If we can use the same δ for all x in a given set, the function is said to be uniformly continuous on that set.
4.2.3Definition
Let ƒ : Ω → Ω′, where Ω and Ω′ are metric spaces, and let E be a subset of Ω. We say that ƒ is uniformly continuous on E if for every ∈ > 0 we can find δ > 0, where δ depends on ∈ but not on x, such that whenever x and y are points of E with d(x, y) < δ, we have d(f(x), f(y)) < ∈. For example, let ƒ(x) = x2, 0 ≤ x ≤ 1. Then given ∈ > 0, the δ for x = 1 works for all x in [0, 1]. The key idea is that the maximum slope occurs at x = 1 (see Fig. 4.2.1).
Uniform continuity is a global concept, referring to the behavior of ƒ on an entire set, as opposed to the local concept of continuity at a particular point.
In some cases, a continuous function is automatically uniformly continuous, as the following result shows.
4.2.4THEOREM. If f is continuous on the compact set K, then ƒ is uniformly continuous on K.
Proof. Fix ∈ > 0. If x ∈ K, we find δ(x) > 0 such that whenever y ∈ K and d(x, y) < δ(x), we have d(f(x), f(y)) < ∈/2. The intuition behind the argument is as follows. Since K is covered by the balls Βδ(x)(x), compactness yields a finite subcovering. If x and y are close enough, they are likely to be in the same ball of the sub-covering, say Bδ(xi)(xi). Then d (x, xi) < δ(xi), d(xi, y) < δ(xi), so that d((x), f (xi)) < ∈/2, d(f(xi), f(y)) < ∈/2. Consequently, d(f(x), f(y)) < ∈. But if x and y are not in the same ball, there is a technical difficulty, which we resolve by using balls of radius δ(x)/2.
Figure 4.2.1 Uniform Continuity
To formalize, note that since x ∈ Bδ(x)/2(x), we have
By compactness, there is a finite subcovering, say
Let 
. If x, y ∈ K and d(x, y) < δ, then x belongs to some Bδ(xi)/2(xi), so 
, and hence d(f(x), f(xi)) < ∈/2. Furthermore,
and therefore d (ƒ(xi), ƒ(y)) < ∈/2. Thus,
(The key idea is that if d (x, y ) < δ, then x and y are forced to lie in the same ball Bδ(xi)(xi), so the previous argument works; see Fig. 4.2.2.)
Now consider again the function ƒ(x) = 1/x, x > 0. Let D = (0, ∞), Ω = [0, ∞), Ω′ = R. Then ƒ : D → Ω′, and D is dense in Ω; that is, 
. (By Theorem 1.5.1 this means that every point of Ω can be expressed as a limit of a sequence of points in D. The most familiar result of this type is that the rationals are dense in the reals.) There is no way to extend ƒ to a continuous mapping of Ω into Ω′, for if g were such an extension then g (0) would have to be + ∞, which is not in Ω′. (To see this, let x → 0; then g(x) must approach g(0). But g(x) = f(x) = 1/x for x > 0.) In fact, this shows that ƒ cannot be uniformly continuous, as a consequence of the following result.
Figure 4.2.2 Proof of Theorem 4.2.4
4.2.5THEOREM. Let f be a uniformly continuous mapping from D to Ω′, where D is a dense subset of Ω and Ω′ is a complete metric space. Then ƒ has a unique extension to a continuous function g: Ω → Ω′. Furthermore, g is uniformly continuous on Ω.
Proof If x ∈ Ω, there is a sequence of points xn ∈ D with xn → x. Since {xn} converges, it is a Cauchy sequence (see Section 2.4.4); we claim that {ƒ(xn)} is a Cauchy sequence in Ω′. For if ∈ > 0, the uniform continuity of ƒ provides δ > 0 such that for all x′, y′ ∈ D, d{x′, y′) < δ ⇒ d(f(x′), f(y′)) < ∈. Since d(xn, xm) < δ for n and m sufficiently large, we have d(f(xn), f(xm)) < ∈ for large n and m, as desired. By completeness of Ω′, ƒ(xn) approaches a limit L; we define g(x) = L. We must verify that L does not depend on the particular sequence {xn } so that g is well defined.
Suppose xn → x, yn → x with f(xn) → L and f(yn) → M. Then
The first and third terms on the right approach 0 by hypothesis, and the second term approaches 0 because x1, y2, x2, y2, x3, y3, ... is a Cauchy sequence. Thus, d (L, M ) = 0, so L = M and g is well defined.
Note that g is actually an extension of f. If x ∈ D, take xn = x for all n; then f(xn) ≡ ƒ(x), so g(x) = ƒ(x).
Given ∈ > 0, let δ > 0 be such that if x, y ∈ D with d(x, y) < δ, we have d(f(x), f(y)) < ∈/3. Now consider points x, y 
with d (x, y) < δ. Suppose we have sequences {xn}, {yn} in D with xn → x, yn → y so that f(xn) → g(x), f(yn) → g(y). Then
If d(x, y) < δ, then d(xn, yn) < δ eventually, so d(f(xn), f(yn)) < ∈/3 for all sufficiently large n. It follows that d (g (x), g (y)) < 3(∈/3) = ∈, proving that g is uniformly continuous on Ω.
To prove uniqueness, suppose h1 and h2 are continuous on Ω, and h1 = h2 on D. If x ∈ Ω, let xn, ∈ D, xn → x. Since h1(xn) = h2(xn) for all n, we have h1(x) = h2(x) by continuity. Thus, h1 = h2 on Ω. ■
Problems for Section 4.2
1.Let f(x) = x In x, x > 0. Can ƒ be defined at 0 so as to be continuous on the extended domain [0, ∞)?
2.Let ƒ : E → R, where E is a bounded subset of Rp. If ƒ is uniformly continuous on E, use Theorem 4.2.5 to show that the image f(E) is bounded.
3.Do Problem 2 without invoking Theorem 4.2.5. (Use the definition of uniform continuity to show that if ∈ > 0, then E, hence ƒ(E ), can be covered by a finite number of open balls of diameter less than ∈.)
4.Let {xn} be a sequence in the compact subset K of the metric space Ω. If x ∈ Ω and xn does not converge to x, show that {xn} has a subsequence converging to a limit y ∈ K with y ≠ x.
5.Apply Problem 4 to prove the following result. Let f : K → Ω′, where K is a compact susbet of the metric space Ω. If ƒ is continuous on K and one-to-one, i.e., f(x) = f(y) implies x = y, show that ƒ has a continuous inverse on ƒ(K). In other words, if x1, x2, … ∈ K, x ∈ K, and f(xn) → ƒ(x) > then xn → x.
6.Give an example of a continuous function ƒ and a closed set A such that ƒ(A) is not closed.
4.3TYPES OF DISCONTINUITIES
It is useful to study the behavior of a function at a point where it is discontinuous. The most familiar situation occurs in the real case when ƒ “jumps” at a point x. As t approaches x from the right (notation t → x+), f(t) approaches a limit denoted by ƒ(x+), and as t approaches x from the left (notation t → x– ), ƒ(t) approaches a limit ƒ(x–). If there is a jump, then f (x–) ≠ (x+).
Formally, ƒ is said to have a simple discontinuity or a discontinuity of the first kind at x if ƒ is discontinuous at x but ƒ(x+) and ƒ(x–) both exist. Within discontinuities of the first kind we have the following subcategories:
removable or point discontinuity: f (x–) = f (x+) finite
jump discontinuity: ƒ(x–) ≠ ƒ(x+), both finite
infinite discontinuity: ƒ(x –), ƒ(x +) not both finite
(see Fig. 4.3.1).
For example, if
then ƒ has a removable discontinuity at x = 0. If
then ƒ has a jump discontinuity at x = 0. If
then ƒ has an infinite discontinuity at x = 0.
To avoid confusion, we must say precisely what we mean by a statement of the form limt→x f(t) =A. The basic idea is that ƒ(x) is not considered in the limit statement. One way of describing the statement is to say that for all sequences tn converging to x, with tn never equal to x, we have ƒ(tn) → A. (A = ±∞ is allowed in this case.) In epsilon-delta terms (assuming A finite), if ∈ > 0 there is a δ > 0 such that if t ≠ x and d(t, x) < δ, we have d(f(t), A) < ∈ (Problem 3). Similarly, in the case where ƒ is defined on an interval of reals, limt→ x + f(t) = B (that is, f (x+) = B) means that whenever tn → x, with tn > x for all n, we have ƒ(tn) → B. Equivalently, if ∈ > 0, we can find δ > 0 such that if x < t < x + <δ, then d(f(t), B) < ∈. In the definition of limt→x f(t) = C (that is, f(x–) = C), we require that tn < x for all n, or x – δ < t < x in the epsilon-delta statement. It follows that limt→x f(t) = A if and only if ƒ(x+) = f(x–) = A; we need not have f(x)= A, and in fact ƒ might not even be defined at x (see Fig. 4.3.2). Similarly, if ƒ has a jump discontinuity at x, ƒ(x) need not coincide with either ƒ(x+) or ƒ(x–). The statement that the limits ƒ(x+) and ƒ(x–) exist and coincide with ƒ(x) is equivalent to continuity of ƒ at x ; a formal proof can be given by using the epsilon-delta descriptions of ƒ(x+) and ƒ(x–) (Problem 4).
It is not hard to give an explicit example of a function that has a non-simple discontinuity (also called discontinuity of the second kind). Consider ƒ(x) = sin(1/x), x ≠ 0; ƒ(0) = 0. The function ƒ is continuous everywhere except at 0, where there is a nonsimple discontinuity. For supposes xn = 1/nπ, n = 1, 2,…; then xn → 0+, and ƒ(xn) = 0 for all n; hence ƒ(xn) → 0. Now if yn = 2/(4n + 1)π, n = 1, 2,…, then xn → 0+, and f(yn) = 1 for n; hence ƒ(yn) → 1. Thus, f(t) has no limit as t → 0+ (similarly, ƒ(t) has no limit as t → 0– either), and therefore 0 is a nonsimple discontinuity of ƒ. As x gets close to 0, sin(1/x) oscillates with ever-increasing frequency.
In fact we can produce an explicit example of a function that has nothing but discontinuities of the second kind: g(x) = 1 for x rational, g(x) = 0 for x irrational. Since any real number x can be expressed as the limit of a sequence of rationals or equally well as the limit of a sequence of irrationals, every x is a nonsimple discontinuity of g.
Figure 4.3.2 limt→x f(t) = A
In certain cases, discontinuities of the second kind are excluded.
4.3.1THEOREM. Let f : R → R, and assume f monotone (that is, either increasing on R or decreasing on R). Then all discontinuities off are jumps, and ƒ has at most countably many discontinuities.
Proof. Assume ƒ increasing (the decreasing case is handled similarly, or by applying the result for increasing functions to –ƒ ). Intuitively, if tn decreases to a limit x, then ƒ(tn) converges by Theorem 2.4.6, so we expect that f(x+) exists. But how do we know that limn ƒ(tn) is the same for all sequences {tn}? Also, what about nonmonotone {tn}? To avoid such difficulties, we take a different approach. If x ∈ R, let L = inf {f(t) : t > x}. Given ∈ > 0, by Theorem 2.4.3 there is a real number t0 > x such that L ≤ f (t0) < L + ∈. If tn > x, tn → x, then x < tn < t0 for all sufficiently large n. Since ƒ is increasing, L ≤ f(tn) ≤ f (t0) < L + ∈ for large n, and therefore f(tn) → + L. Thus, the limit f(x+) exists. Similarly, let M = sup{f(t) : t < x} and choose 
such that 
. If tn < x, tn, →x, then 
for large n, so 
. Therefore, f(tn) → M, and the limit f(x–) exists. Thus, all discontinuities are simple, and since ƒ is real-valued L and M (i.e., f(x+) and f(x–)) are finite. A removable discontinuity cannot occur because ƒ is monotone, and it follows that all discontinuities are jumps.
To show that there are at most countably many discontinuities, note that each jump determines a nonempty open interval (ƒ(x–), ƒ(x+)), and, by monotonicity, the open intervals are disjoint. In each interval we may choose a rational number (see Fig. 4.3.3). Since there are only countably many rationals available, there can only be countably many jumps. ■
Theorem 4.3.1 applies equally well to a function defined on an interval I ⊆ R; the proof is the same.
Perhaps unexpectedly, a monotone function of more than one variable can have uncountably many discontinuities (see Problem 1).
In calculus, you became familiar with the idea that a continuous function is “smooth”; in other words, its graph can be drawn without lifting pencil from paper. As a consequence, if, say, ƒ(a) = 3 and ƒ(b) = 10, then as x ranges from a to b, ƒ will take on all values between 3 and 10. We now prove a theorem to this effect.
Figure 4.3.3 Proof of Theorem 4.3.1
4.3.2Intermediate Value Theorem
Let ƒ be a continuous, real-valued function on [a, b]. If f (a) < c < f(b), there is an x ∈ (a, b) such that f (x) = c.
Proof. The argument is rather tricky; we look at the last time that ƒ takes a value less than c. Formally, let A = {t ∈ [a, b] : ƒ(t ) < c} and let x = sup A. (Note that a ∈ A, so A is not empty and therefore has a sup. Also, A ⊆ [a, b], so x ∈ [a, b]; see Fig. 4.3.4.) We show that ƒ(x) = c by eliminating the other possibilities ƒ(x) < c and ƒ(x) > c. By Theorem 2.4.3 there is a sequence of points tn ∈ A with tn → sup A = x. Then f(tn) < c for all n, so, by continuity, ƒ(x) ≤ c. Thus, it remains only to dispose of the case ƒ(x) < c.
If f(x) < c, then x < b, and, by continuity, if y is taken greater than x and sufficiently close to x, we have f(y) < c. Thus, y ∈ A, contradicting the fact that x is an upper bound of A. ■
Problems for Section 4.3
1.Let ƒ(x, y) = 1 if x, y ≥ 0; f(x, y) = 0 elsewhere (x,y real). Show that ƒ is increasing in the sense that if x1 ≤ x2 and y1 ≤ y2, then ƒ(x1, y1) ≤ ƒ(x2, y2), but ƒ has uncountably many discontinuities.
2.Classify the discontinuities of the function given by
3.Show that the “sequence” and “epsilon-delta” characterizations of limt→x ƒ(t) = A are equivalent (assumed finite).
4.If f: R → R, show that f is continuous at x iff f(x+) and f(x–) exist and coincide with f(x).
4.4THE CANTOR SET
This is a convenient place to introduce the Cantor set, an interesting source of examples and counterexamples. We take Ω to be the closed unit interval [0, 1], and remove the “middle third” V1; let E1 be the set that remains:
Thus, x ∈ E1 if and only if x can be expressed in ternary form with first digit 0 or 2. (Note that in ternary, 
Although 
has two ternary representations, it can be expressed using a first digit 0, so there is no ambiguity.) We now remove the middle thirds of the intervals of Ε1 and let E2 be the set that remains:
Thus, x ∈ E2 if and only if x can be expressed in ternary form with the first two digits 0 or 2. (For example, 
iff x = .20.…)
We continue this process inductively. To make sure everyone understands what is going on, we write out one more step:
At step n we obtain a set En, the union of 2n intervals each of which is of length 3–n, such that x ∈ En if and only if x can be expressed in ternary form using 0’s and 2’s in the first n digits. Since 
the sets En form a decreasing sequence. The Cantor set C is defined by
Thus, x ∈ C if and only if x can be expressed in ternary form using only digits 0 and 2. Since each En is closed, C is a closed set.
The following properties of C may be established.
(a)C is uncountable.
(b) The length of C is 0. (We have not defined the length of an arbitrary set of reals; this is done in measure theory. However, if we are given ∈ > 0, C can be shown to be a subset of a finite union of intervals whose total length is less than ∈. Thus, if length is defined in any reasonable way, C must have length 0. )
(c) Every point of C is a limit point of C. (A set with this property is sometimes called “perfect,” although this terminology seems rather extravagant.)
(d) C is totally disconnected; that is, if x0, x1 ∈ C with x0 < x1, there is a point x2 ∈ (x0, x1) with x2 ∉ C.
(e) C is nowhere dense; that is, the closure of C (which coincides with C since C is closed) has no interior; in other words, there are no open subsets of C except the empty set.
(a)The points of C are in one-to-one correspondence with the points in [0, 1] that can be represented in binary form using the digits 0 and 1—in other words, with all points of [0, 1].
(b)Note that C ⊆ En, and the length of En is 
(c)If x = .a1a2a3 …∈ C (in ternary form using digits 0 and 2), let 
, where we take 
if an + 1 = 2, and 
if an + 1 = 0. Then xn ∈ C, xn ≠ x, and xn → x. Thus, x is a limit point of C.
(d)Examine the ternary expansions of x0 and x1. Since x0 < x1, for some n we have
Let x2 = .a …an11 … Then x2 ∉ C and x0 < x2 < x1.
(e)If(a, b) ⊆ C, let x0, x1 ∈ (a, b) with x0 < x1. By (d) we can find x2 ∈ (x0, x1) with x2 ∉ C, contradicting x2 ∈ (a, b) ⊆ C. ■
The Cantor set can be used to construct continuous functions with rather unusual properties. Here is one example. If A is a nonempty subset of the metric space Ω and x ∈ Ω, minimizing the distance d(x, y ) as y ranges over A would seem to produce a reasonable notion of distance from a point to a set. Formally, we define
Now
thus
Take the inf over z ∈ A to obtain d(x, A) ≤ d(x, y) + d(y, A); by symmetry,
Thus, the function defined by f(x) = d(x, A), x ∈ Ω, is uniformly continuous. If A is closed and x ∉ A, then (since Ac is open), Br (x) ⊆ Ac for some r > 0. If y ∈ A, then d(x, y) ≥ r; hence, d(x, A) ≥ r > 0. (Of course, if x ∈ A, then d(x, A) = 0 whether or not A is closed).
Now let A be the Cantor set C and take Ω to be [0, 1]. Since C is closed, f(x) = 0 if and only if x ∈ C. By Theorem 4.4.1(a), ƒ has uncountably many zeros, and by Theorem 4.4.1(d), ƒ is never identically 0 on any open subinterval of [0, 1].
Problems for Section 4.4
1.Let Ω be a metric space with distance function d. Let f(x) = d(x, y), x ∈ Ω, where y is any fixed element of Ω. Show that ƒ is uniformly continuous on Ω.
2.Indicate how to modify the construction of the Cantor set so that a set of positive length is obtained.
3.If A is compact (and nonempty) and x is arbitrary, show that there is a point y0 ∈ A such that
4.In Problem 3, show that “compact” can be replaced by “closed” if Ω = Rp.
REVIEW PROBLEMS FOR CHAPTER 4
1.Give an example of a function ƒ: R → R that has a nonsimple discontinuity at x = 2 but no other discontinuities.
2.Give an example of a set A and a point x ∉ A such that d(x, A) = 0.
3.Let f(x) = x if x is rational; ƒ(x) = 0 if x is irrational. Is ƒ discontinuous everywhere?
4.Let f: A → R, where A is the set of rational numbers x such that 0 < x < 7. If ƒ is uniformly continuous on A, show that ƒ is bounded.
5.If f(x) = x3 + 2x + 2, show that ƒ(x) = 0 for some x in (–1, 0).
6.Let ƒ(x) = In x; is ƒ uniformly continuous on (0, 1]?
7.Let A be the set of all real numbers x such that ex2 is not an integer. Show that A is open.
8.Give an example of a countably infinite nowhere dense subset of R.
9.Let ƒ(x) = 1 if x ≥ 0; ƒ(x) = 0 if x < 0. Find a closed set C ⊆ R such that ƒ–1(C) is not closed.
10.Classify the discontinuities of the function given by
1 The “whenever” phrase is translated as “for every y in Ω such that d(x, y) < 6,” etc.
