POMA Solution (Chapter 4)

Zhanxiong Xu

Department of Statistics, Penn State University

Summer 2015

Chapter 4

Continuity

1. Suppose f is a real function defined on R1 which satisfies

limh→0

[f(x+ h)− f(x− h)] = 0

for every x ∈ R1. Does this imply that f is continuous?

Proof. No, it doesn’t. Consider f(x) = I{0}(x), x ∈ R1. Clearly, the assumption in

the problem is satisfied. However, f is not continuous at 0, since |f(x)− f(0)| = 1 no

matter how small the x is.

2. If f is a continuous mapping of a metric space X into a metric space Y , prove that

f(E) ⊂ f(E)

for every set E ⊂ X. Show, by an example, f(E) can be a proper set of f(E).

Proof. Let y ∈ f(E), then ∃x ∈ E such that y = f(x). Since x ∈ E, there exists a

sequence {xn} ⊂ E such that xn → x as n → ∞. By continuity of f , this implies

f(xn) → f(x) = y. Since f(xn) ∈ E for every n, this means y ∈ f(E). Thus

f(E) ⊂ f(E).

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For the example, let X = Y = (0,+∞), E = {1, 2, 3, . . .} = N ⊂ X. Define f(x) =

1x , x ∈ (0,+∞), which is continuous in X. For this case, we have f(E) = f(E) =

{1, 12 ,13 , . . .}, while f(E) = {0, 1, 12 ,

13 , . . .}. Hence f(E) $ f(E).

3. Let f be a continuous real function on a metric space X. Let Z(f) (the zero set of f)

be the set of all p ∈ X at which f(p) = 0. Prove that Z(f) is closed.

Proof. Note that Z(f) = {p ∈ X : f(p) = 0} = f−1({0}) and {0} is a closed set in

R1. Therefore by Corollary 4.2.3., Z(f) is closed, as f is continuous on X.

4. Let f and g be continuous mappings of a metric space X into a metric space Y , and

let E be a dense subset of X. Prove that f(E) is dense in f(X). If g(p) = f(p) for all

p ∈ E, prove that g(p) = f(p) for all p ∈ X. (In other words, a continuous mapping

is determined by its values on a dense subset of its domain.)

Proof. Take y ∈ f(X), then ∃x ∈ X such that y = f(x). If x is a point of E, then

y ∈ f(E). Otherwise x /∈ E is a limit point of E, as E is dense in X. Let ε be any given

positive, since f is continuous at x, there exists δ > 0 such that dY (f(x′), f(x)) < ε

for all x′ ∈ X such that dX(x, x′) < δ. Since x is a limit point of E, there exists

x0 ∈ E such that dX(x0, x) < δ, hence dY (f(x0), f(x)) < ε, which means y = f(x)

is a limit point of f(E) since f(x0) ∈ f(E). Since y is arbitrary, we showed f(E) is

dense in f(X).

Let ε be any given positive. For any p ∈ X\E, p must a limit point of E. Since both

f and g are continuous at p, there exists δ > 0 such that dY (f(p), f(q)) < ε/2 and

dY (g(p), g(q)) < ε/2 whenever dX(p, q) < δ. Because E is dense in X, we can take an

q0 ∈ E with dX(p, q0) < δ. Then we have, by triangle inequality

dY (f(p), g(p)) ≤ dY (f(p), f(q0)) + dY (f(q0), g(p))

= dY (f(p), f(q0)) + dY (g(q0), g(p))

<ε

2+ε

2= ε.

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The equality on the second line above holds since by assumption, f(q0) = g(q0). Since

ε > 0 can be arbitrarily small, it is necessary to have dY (f(p), g(p)) = 0, that is,

f(p) = g(p). Therefore, f(p) = g(p) for all p ∈ X.

5. If f is a real continuous function defined on a closed set E ⊂ R1, prove that there exist

continuous real functions g on R1 such that g(x) = f(x) for all x ∈ E. (Such functions

g are called continuous extension of f from E to R1.) Show that the result becomes

false if the word “closed” is omitted. Extend the result to vector-valued functions.

Hint: Let the graph of g be a straight line on each of the segments which constitute

the complement of E (compare Exercise 29, Chap. 2). The result remains true if R1

is replaced by any metric space, but the proof is not so simple.

Proof. We will prove the statement for vector-valued functions directly.

Since E is closed, Ec is an open subset of R1, by Exercise 29, Chap, 2. Ec is the

union of an at most countable collection of disjoint segments. Write Ec =⋃∞i=1(ai, bi)

(if Ec is the finite union of disjoint segments, then from some i, (ai, bi) = ∅). Since

{ai} ⊂ E, {bi} ⊂ E, f(ai),f(bi) are all defined. For every i ∈ {1, 2, 3, . . .}, define

g(x) = f(ai) +x− aibi − ai

(f(bi)− f(ai)), x ∈ (ai, bi).

and g(x) = f(x) if x ∈ E. We shall show g is continuous everywhere on R1. Obviously,

g is continuous at every point of E◦ and Ec, so it suffices to show g is continuous at

every ai and bi. Given ε > 0, since ai ∈ E and f is continuous at ai, there exists

δ1 > 0 such that |f(x)− f(ai)| < ε whenever 0 < ai − x < δ1. On the other hand, for

x ∈ (ai, bi),

|g(x)− g(ai)| =|f(bi)− f(ai)|

bi − ai(x− ai).

Thus if taking δ2 = min(

bi−ai|f(bi)−f(ai)|ε, bi − ai

), then 0 < x− ai < δ2 implies |g(x)−

g(ai)| < ε. Now let δ = min(δ1, δ2), then for all x ∈ (ai − δ, ai + δ), we have |g(x) −

g(ai)| < ε, i.e., g is continuous at ai. Similarly, it can be shown g is continuous at bi.

Therefore, g is a continuous extension of f from E to R1.

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If the word “closed” is omitted, the result becomes false. For example, f(x) = 1/x is

continuous in E = (0,+∞), where E is an open set. It doesn’t admit any continuous

extension g such that g is continuous on R1 and g(x) = f(x) for x ∈ E. If such g

existed, then g would be bounded on the interval [−1, 1] (Theorem 4.15), which is

clearly impossible.

6. If f is defined on E, the graph of f is the set of points (x, f(x)), for x ∈ E. In

particular, if E is a set of real numbers, and f is real-valued. the graph of f is a

subset of the plane.

Suppose E is compact, and prove that f is continuous on E if and only if its graph is

compact.

Proof. Denote by Γ(f) = {(x, f(x)) : x ∈ E} ⊂ X × R1 the graph of f . Define

π to be the mapping of E × R1 onto E, i.e., π(x, y) = x, for (x, y) ∈ E × R1, the

projection function of E×R1 onto E. Let dE be the metric on E, to proceed, we have

to assign some metric d := dE×R1 on E×R1 on E×R1 (since we must talk about the

compactness of Γ(f), which is a subset of E × R1. Under current scope, we have to

introduce some metric). One clear choice is

d((x1, y1), (x2, y2)) = dE(x1, x2) + |y1 − y2|.

Under this metric, the function π defined above is clearly continuous on E × R1.

Given ε > 0, for every x ∈ E, since f is continuous on E, there exists δ > 0 such that

|f(y)− f(x)| < ε/2 whenever dE(y, x) < δ. Take δ0 = min(δ, ε/2), then for all y ∈ E

such that dE(y, x) < δ0, it follows that

d((x, f(x)), (y, f(y))) = dE(x, y) + |f(y)− f(x)| < δ0 +ε

2≤ ε

2+ε

2= ε,

therefore the function F : E → E × R1, defined by F (x) = (x, f(x)) is continuous on

E. And clearly F (E) = Γ(f). Since E is compact, by Theorem 4.14, Γ(f) = F (E) is

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compact.

Conversely, suppose Γ(f) is a compact subset of E × R1. To show f is continuous on

E, by Corollary of Theorem 4.8, it suffices to show that for every closed set C in R1,

f−1(C) is closed in E. It can be shown that

f−1(C) = π((E × C) ∩ Γ(f)).

Indeed, if p ∈ f−1(C), then f(p) ∈ C, therefore (p, f(p)) ∈ (E × C) ∩ Γ(f), thus

p = π(p, f(p)) ∈ π((E × C) ∩ Γ(f)). On the other hand, if p ∈ π((E × C) ∩ Γ(f)),

then there exists (x, y) ∈ (E × C) ∩ Γ(f) such that p = π(x, y). Since (x, y) ∈ Γ(f),

y = f(x); since (x, y) ∈ E ×C, y ∈ C. That is f(x) ∈ C. Consequently, p = π(x, y) =

x ∈ f−1(C). Now since both E and C are closed, it follows that E × C is a closed

subset of E × R11 . Since Γ(f) is compact in E × R1, it follows that (E × C) ∩ Γ(f),

as a closed subset of the compact set Γ(f), is a compact set in E ⊂ R1. Since π is

continuous on E × R1, by Theorem 4.14, f−1(C) = π((E × C) ∩ Γ(f)) is compact, a

fortiori, closed. Therefore, f is continuous on E!

7. Let E ⊂ X and if f is a function defined on X, the restriction of f to E is the function g

whose domain of definition is E, such that g(p) = f(p) for p ∈ E. Define f and g on R2

by: f(0, 0) = g(0, 0) = 0, f(x, y) = xy2/(x2 + y4), g(x, y) = xy2/(x2 + y6), if (x, y) 6=

(0, 0). Prove that f is bounded on R2, that g is unbounded in every neighborhood of

(0, 0), and that f is not continuous at (0, 0); nevertheless, the restrictions of both f

and g to every straight line in R2 are continuous.

Proof. By algebraic-geometric mean inequality, |xy2| ≤ 12 (x2 + y4) for all (x, y) ∈ R2,

therefore, |f(x, y)| ≤ 12 if (x, y) 6= (0, 0), and since f(0, 0) = 0, f is bounded on R2.

For t 6= 0, g(t3, t) = t5/(t6 + t6) = 12t → ∞ as t → 0, hence g is unbounded in every

neighborhood of (0, 0).

1Given ε > 0, let (x0, y0) ∈ (E × C)′, then ∃(x0, y0) 6= (x, y) ∈ E × C, such that d((x, x0), (y, y0)) =dE(x, x0) + |y − y0| < ε, which implies dE(x, x0) < ε and |y − y0| < ε, i.e., x0 ∈ E and y0 ∈ C. Since Eand C are closed, it follows that x0 ∈ E and y0 ∈ C, hence (x0, y0) ∈ E × C. Therefore, E × C is closed inE × R1.

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Since f(t2, t) = t4/(t4 + t4) = 1/2 doens’t converge to f(0, 0) as t → 0, f is not

continuous at (0, 0).

Let E be any straight line in R2. If E doesn’t contain (0, 0), then by the remarks

of Example 4.11, f(x, y) is continuous on E. If E contains (0, 0), as before, f is

continuous at every point of E that is not (0, 0). At (0, 0) ∈ E, since every point

(x, y) ∈ E can be written as x = k1t, y = k2t, t ∈ R1, where (k1, k2) 6= (0, 0) is fixed,

it follows that

|f(x, y)− f(0, 0)| =∣∣∣∣ k1k

22t

3

k21t2 + k42t

4

∣∣∣∣ =|k1|k22tk21 + k42t

2→ 0, as t→ 0.

Hence f is continuous at (0, 0), when it is restricted to E. Similarly, it can be shown

that the restriction of g to every straight line in R2 is continuous.

8. Let f be a real uniformly continuous function on the bounded set E in R1. Prove that

f is bounded on E.

Show that the conclusion is false if boundedness of E is omitted from the hypothesis.

Proof. Since E is a bounded set in R1, its closure E is compact, as it is a closed

and bounded subset in R1. Since f is uniformly continuous on E, for ε = 1, there

exists δ > 0 such that |f(x) − f(y)| < 1 whenever |x − y| < δ, x ∈ E, y ∈ E. It is

easily seen {Nδ(x) : x ∈ E} form an open cover of E. Because E is compact, there

exists a finite subcover Nδ(x1), . . . , Nδ(xm), x1, . . . , xm ∈ E of E. Fix x1 ∈ E, and set

M = max{|f(xi)−f(x1)| : i = 2, . . . ,m}. For every x ∈ E, there exists i ∈ {1, . . . ,m}

such that x ∈ Nδ(xi), or |x− xi| < δ, for which

|f(x)− f(x1)| ≤ |f(x)− f(xi)|+ |f(xi)− f(x1)| < 1 +M.

In other words, |f(x)| < 1 + |f(x1)|+M . That is, f is bounded on E.

If E is not bounded, the conclusion is false. For example, let E = R1, f(x) = x, x ∈ E.

Obviously, f is uniformly continuous on E but unbounded.

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9. Show that the requirement in the definition of uniform continuity can be rephrased as

follows, in terms of diameters of sets: To every ε > 0 there exists a δ > 0 such that

diam f(E) < ε for all E ⊂ X with diam E < δ.

Proof. Call the definition made in this exercise “Definition 4.18′”. Let f be defined

as in Definition 4.18.

Definition 4.18 implies Definition 4.18′: Given ε > 0, suppose Definition 4.18 holds

(with ε/2 in place of ε). Let E be any subset of X with diam E < δ. For any two

points p and q in E, we have dX(p, q) ≤ diam E < δ, hence dY (f(p), f(q)) < ε/2,

which implies diam f(E) ≤ ε/2 < ε, Definition 4.18′ follows.

Definition 4.18′ implies Definition 4.18: Given ε > 0, suppose Definition 4.18′ holds.

Let p and q be two points in X such that dX(p, q) < δ. Set E = {p, q}, then

diam E = dX(p, q) < δ. By Definition 4.18′, dY (f(p), f(q)) = diam f(E) < ε. There-

fore Definition 4.18 follows.

10. Complete the details of the following alternative proof of Theorem 4.19: If f is not

uniformly continuous, then for some ε > 0 there are sequences {pn}, {qn} in X

such that dX(pn, qn) → 0 but dY (f(pn), f(qn)) > ε. Use Theorem 2.37 to obtain a

contradiction.

Proof. If f is not uniformly continuous, then for some ε > 0 there are sequences {pn},

{qn} in X such that dX(pn, qn)→ 0 but dY (f(pn), f(qn)) > ε. It can be seen that both

{pn} and {qn} have infinitely many distinctive terms, otherwise we have pn ≡ p and

qn ≡ q for all n ≥ N , for some integer N . Then dX(pn, qn) → 0 implies dX(p, q) = 0

hence p = q, and it thus impossible to have ε < dY (f(pn), f(qn)) = dY (f(p), f(q)) =

dY (f(p), f(p)) = 0. Since {pn} is an infinite sequence and X is compact, by Theorem

2.37, {pn} has a limit point p0 in X, that is, there exists a subsequence {pnk} of {pn}

such that d(pnk, p0)→ 0 as k →∞. Consider {qnk

}, if it is infinite, again by Theorem

2.37, {qnk} has a limit point q0 in X, i.e., there exists a subsequence {qnki

} of {qnk}

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such that d(qnki, q0) → 0 as i → ∞ (if {qnk

} has finite range, q0 can be taken as the

eventual term, with the aforementioned property still holds.). Since

dX(p0, q0) ≤ dX(p0, pnki) + dX(pnki

, qnki) + dX(qnki

, q0)→ 0 as i→∞,

it follows that p0 = q0. For the ε > 0 made in assumption, since dX(pnki, p0) → 0,

dX(qnki, p0)→ 0 as i→∞ and f is continuous at p0, for sufficiently large i, we have

dY (f(pnki), f(p0)) < ε/2 and dY (f(qnki

), f(p0)) < ε/2. Consequently,

dY (f(pnki), f(qnki

)) ≤ dY (f(pnki), f(p0)) + dY (f(qnki

), f(p0)) < ε/2 + ε/2 = ε,

which contradicts with the assumption that dY (f(pnki), f(qnki

)) > ε. Therefore f

must be uniformly continuous on X.

11. Suppose f is a uniformly continuous mapping of a metric space X into a metric space

Y and prove that {f(xn)} is a Cauchy sequence in Y for every Cauchy sequence {xn}

in X. Use this result to give an alternative proof of the theorem stated in Exercise 13.

Proof. Let {xn} be a Cauchy sequence in X. Given ε > 0, since f is continuous on

X, there exists δ > 0 such that dY (f(p), f(q)) < ε for all p ∈ X, q ∈ X such that

dX(p, q) < δ. To this δ > 0, since {xn} is Cauchy in X, there exists N ∈ N such that

dX(xn, xm) < δ for n,m ≥ N . Therefore, for all n,m ≥ N , we have

dY (f(xn), f(xm)) < ε.

That is, {f(xn)} is a Cauchy sequence in Y .

12. A uniformly continuous function of a uniformly continuous function is uniformly con-

tinuous.

State this more precisely and prove it.

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Restatement: Suppose f is a mapping of a metric space X into a metric space Y and

f is uniformly continuous on X. g is a mapping of Y into another metric space Z and

g is uniformly continuous on f(X), the range of f . Then the composition mapping of

f and g, h = g ◦ f , as a mapping of X into Z, is uniformly continuous on X.

Proof. Given ε > 0, since g is uniformly continuous on f(X), there exists δ1 > 0 such

that dZ(g(p), g(q)) < ε for all p ∈ f(X), q ∈ f(X) and dY (p, q) < δ1. For this δ1 > 0,

since f is uniformly continuous on X, there exists δ2 > 0 such that dY (f(x), f(y)) < δ1

such that x ∈ X, y ∈ X and dX(x, y) < δ2. It then follows that, for all x, y ∈ X such

that dX(x, y) < δ2, since f(x), f(y) ∈ f(X) and dY (f(x), f(y)) < δ1, we have

dZ(h(x), h(y)) = dZ(g(f(x)), g(f(y))) < ε.

That proves h is uniformly continuous on X.

13. Let E be a dense subset of a metric space X, and let f be a uniformly continuous

real function defined on E. Prove that f has a continuous extension from E to X

(see Exercise 5 for terminology). (Uniqueness follows from Exercise 4.) Hint: For

each p ∈ X and each positive integer n, let Vn(p) be the set of all q ∈ E with

d(p, q) < 1/n. Use Exercise 9 to show that the intersection of the closures of the

sets f(V1(p)), f(V2(p)), . . ., consists of a single point, say g(p), of R1. Prove that the

function g so defined on X is the desired extension of f .

Could the range space R1 be replaced by Rk? By any compact metric space? By any

complete metric space? By any metric space?

Proof. For each p ∈ X, let Vn(p) be defined as in the hint. Clearly, we have V1(p) ⊃

V2(p) ⊃ V3(p) ⊃ · · · , which implies f(V1(p)) ⊃ f(V2(p)) ⊃ f(V3(p)) ⊃ · · · . Given

ε > 0, since f is uniformly continuous on E, by Exercise 9, there exists δ > 0

such that diam f(F ) < ε for every F ⊂ E with diam F < δ. For sufficiently

large n ∈ N so that 2n < δ/2, for any q1, q2 ∈ Vn(p), since d(q1, q2) ≤ d(q1, p) +

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d(p, q2) < 2/n < δ/2, it follows that diam Vn(p) ≤ δ/2 < δ, hence diam f(Vn(p)) =

diam f(Vn(p)) < ε. This shows limn→∞ diam f(Vn(p)) = 0. Note the metric space

containing f(Vn(p)), n = 1, 2, . . ., is R1, whichis complete, hence by the result of Ex-

ercise 21, Chap. 3,⋂∞

1 f(Vn(p)) consists of a single point, which we shall denote it by

g(p). We need to show the function g so defined on X is the desired extension of f .

First we need to show g(p) = f(p), for every p ∈ E. By construction, if p ∈ E, we

have p ∈⋂∞

1 Vn(p), therefore f(p) ∈⋂∞

1 f(Vn(p)) = {g(p)}, hence g(p) = f(p).

Next we need to show g is continuous at every point p ∈ X. Given ε > 0, fix p ∈ X,

since f is uniformly continuous on E and g(x) = f(x) for all x ∈ E, there exists δ > 0

such that |g(x) − g(y)| = |f(x) − f(y)| < ε/3 for all x ∈ E, y ∈ E, d(x, y) < δ. Since

g(p) ∈⋂∞

1 f(Vn(p)), there exists p0 ∈ Vn(p) with 1/n < δ/3 such that |g(p)−g(p0)| <

ε/3. For every q ∈ X such that d(p, q) < δ/3, since g(q) ∈⋂∞

1 f(Vn(q)), there exists

q0 ∈ Vn(q) (the n is taken to be the same as before) such that |g(q) − g(q0)| < ε/3.

Notice that since

d(p0, q0) ≤ d(p0, p) + d(p, q) + d(q, q0) <1

n+δ

3+

1

n< δ,

and p0 ∈ E, q0 ∈ E, it follows that |g(p0) − g(q0)| < ε/3. Therefore, for every q ∈ X

such that d(p, q) < δ/3, we have:

|g(p)− g(q)| ≤ |g(p)− g(p0)|+ |g(p0)− g(q0)|+ |g(q0)− g(q)| < ε

3+ε

3+ε

3= ε.

That is, g is continuous at p.

A careful examination of the above proof reveals that the only property we used about

R1 is its completeness (because of that, Exercise 21, Chap. 3 can be applied.). So the

range space can be replaced by Rk, by any compact metric space (as it is then a

complete metric space), or by any complete metric space.

The result, however, is not true for an arbitrary metric space. Let X = R1, E = Q

which is dense in X. Clearly, f(x) = x is uniformly continuous on E. If we restricted

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the range space of f to Q, then f would not have a continuous extension from E

to X (it is required that the extension also has the range in f(E)). For if such

continuous extension g existed, take α ∈ R1\Q, since g is continuous at α, it follows

that g(α) = α ∈ R1\Q, contradiction.

14. Let I = [0, 1] be the closed unit interval. Suppose f is a continuous mapping of I into

I. Prove that f(x) = x for at least one x ∈ I.

Proof. Define g(x) = f(x) − x, x ∈ I, since f maps I into I, it is easily seen g(0) =

f(0) − 0 ≥ 0, g(1) = f(1) − 1 ≤ 0. If g(0) = 0 or g(1) = 0, then we have f(0) = 0

or f(1) = 1, the proof is complete. Otherwise we have g(0) > 0 > g(1), since g is

continuous on I, by Theorem 4.23, there exists x ∈ (0, 1) such that g(x) = 0, which

implies f(x) = x.

15. Call a mapping of X into Y open if f(V ) is an open set in Y whenever V is an open

set in X.

Prove that every continuous open mapping of R1 into R1 is monotonic.

Proof. Suppose f is a continuous open mapping of R1 into R1. If f were not monotonic,

there would exist a < c < b such that f(a) < f(c), f(c) > f(b) (the case f(a) >

f(c), f(c) < f(b) is similar.). This means on the interval [a, b], f attains its maximum

in (a, b), say x0 such that f(x0) = supx∈[a,b] f(x). Since (a, b) is an open interval

and f is an open mapping, we have f((a, b)) is an open set in R1. Therefore, there

exists δ > 0, such that (f(x0) − δ, f(x0) + δ) ⊂ f((a, b)), therefore, f(x0) + δ ≤

sup f((a, b)) ≤ f(x0), leading to δ ≤ 0, contradiction. Hence the assumption is false,

and every continuous mapping of R1 into R1 is monotonic.

16. Let [x] denote the largest integer contained in x, that is, [x] is the integer such that

x − 1 < [x] ≤ x; and let (x) = x − [x] denote the fractional part of x. What

discontinuities do the functions [x] and (x) have?

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Proof. [x] has simple discontinuities at Z, it is monotonic, and the jump size is 1 at

each jump point.

(x) also has simple discontinuities at Z, however, it is not monotonic (but periodic

and bounded).

Note that both [x] and (x) are right continuous.

17. Let f be a real function defined on (a, b). Prove that the set of points at which

f has a simple discontinuity is at most countable. Hint: Let E be the set on which

f(x−) < f(x+). With each point x of E, associate a triple (p, q, r) of rational numbers

such that

(a) f(x−) < p < f(x+),

(b) a < q < t < x implies f(t) < p,

(c) x < t < r < b implies f(t) > p.

The set of all such triples is countable. Show that each triple is associated with at most

one point of E. Deal similarly with the other possible types of simple discontinuities.

Proof. Let E and (p, q, r) be defined as in the hint.

We first show that for each x ∈ E, there exists (p, q, r) ∈ A such properties (a), (b), (c)

hold. Since x ∈ E, f(x−) < f(x+), hence we can take a rational p ∈ (f(x−), f(x+)).

Having this p be taken, by the definition of f(x−), there exists δ > 0 such that

|f(t)− f(x−)| < p− f(x−) for all t ∈ (x− δ, x) and a < x− δ. Let q be any rational

in (x − δ, x), then for all q < t < x, it holds that f(t) < f(x−) + p − f(x−) = p,

therefore, q satisfies property (b). With the same argument, we can show that there

exists r ∈ Q such that property (c) holds. We thus associated each x ∈ E with a

rational triple (p, q, r) satisfying (a), (b), (c).

Next we show that such (p, q, r) is uniquely associated with element in E. Let (p, q, r)

be associated with some x ∈ E. Suppose that there were another point y ∈ E, y 6= x

so that y is also associated with the same triple (p, q, r), without losing of generality,

12

we may assume y > x. Clearly y ∈ (x, r), take α ∈ (x, y), on one hand, since

a < q < α < y, by (b), we have f(α) < p, on the other hand, since x < α < r < b, by

(c), we have f(α) > p, contradiction. Therefore y cannot differ from x.

Denote the set of all (p, q, r) triples by A, since A ⊂ Q × Q × Q, which is countable,

thus A is at most countable. Additionally, by the above argument, the mapping of E

into A is injective, therefore E is at most countable.

For the cases of other types of simple discontinuities, the proof is almost identical.

18. Every rational x can be written in the form x = m/n, where n > 0, and m and n are

integers without any common divisors. When x = 0, we take n = 1. Consider the

function f defined on R1 by

f(x) =

0 (x irrational),

1n (x = m

n ).

Prove that f is continuous at every irrational point, and that f has a simple discon-

tinuity at every rational point.

Proof. In fact, we can show that for every x0 ∈ R1, limx→x0f(x) = 0.

Given ε > 0, take q0 ∈ N sufficiently large such that 1/q0 < ε. Notice that there are

only finitely many rationals pq with the denominator q ∈ (0, q0] fall in the segment

(x0 − 1, x0 + 1). Therefore there exists δ > 0, such that for every rational pq ∈

(x0 − δ, x0 + δ), its denominator q > q0. Now for every x such that 0 < |x− x0| < δ,

if x is irrational, then |f(x)− 0| = 0 < ε; if x is rational, |f(x)− 0| = 1/q < 1/q0 < ε.

That is, limx→x0f(x) = 0.

Now if x0 is irrational, then limx→x0f(x) = 0 = f(x0), thus f is continuous at x0. If

x0 is rational, then f(x0) > 0 = limx→x0f(x), thus f has a simple discontinuity (of

second kind, i.e., f(x0+) = f(x0−) 6= f(x0)) at x0.

19. Suppose f is a real function with domain R1 which has the intermediate value property:

13

If f(a) < c < f(b), then f(x) = c for some x between a and b. Suppose also, for every

rational r, that the set of all x with f(x) = r is closed. Prove that f is continuous.

Hint: If xn → x0 but f(xn) > r > f(x0) for some r and all n, then f(tn) = r for some

tn between x0 and xn; thus tn → x0. Find a contradiction. (N. J. Fine, Amer. Math.

Monthly, vol. 73, 1966, p. 782.)

Proof. If f were not continuous in R1, then there would exist x0 at which f is not

continuous. Therefore ∃ε > 0 and a sequence {xn} that converges to x0 such that

|f(xn) − f(x0)| ≥ ε. It then follows that there are either infinitely many n satisfy

f(xn) − f(x0) ≥ ε or infinitely many n satisfy f(x0) − f(xn) ≥ ε. Let’s assume

the former case holds (the latter case would equally lead to contradiction, using the

same argument as below), for which we denote the subsequence also by {xn} and

f(xn) ≥ ε + f(x0) holds for all n ∈ N. Therefore there exists some rational r such

that f(xn) > r > f(x0) for all n. By the intermediate value property of f , for each

n, there exists some tn between x0 and xn such that f(tn) = r. Since xn → x0 as

n → ∞, we have tn → x0 as n → ∞. Let E := {x ∈ R1 : f(x) = r}. Since {tn} ⊂ E

and tn → x0, x0 ∈ E. The closedness of E then implies that x0 ∈ E, hence f(x0) = r,

contradicts with that f(x0) < r.

Therefore, f is continuous in R1.

20. If E is a nonempty subset of a metric space X, define the distance from x ∈ X to E

by

ρE(x) = infz∈E

d(x, z).

(a) Prove that ρE(x) = 0 if and only if x ∈ E.

(b) Prove that ρE is a uniformly continuous function on X, by showing that

|ρE(x)− ρE(y)| ≤ d(x, y)

for all x ∈ X, y ∈ X.

14

Proof.

(a) Suppose x ∈ E. Given ε > 0, there exists y ∈ E such that d(x, y) < ε. Hence

ρE(x) = infz∈E d(x, z) ≤ d(x, y) < ε. Since ε can be arbitrarily small, ρE(x) = 0.

Conversely, suppose ρE(x) = 0. Given ε > 0, since ρE(x) is the greatest lower

bound of {d(x, z) : z ∈ E}, there exists z0 ∈ E such that d(x, z0) < ρE(x)+ε = ε.

That is, x ∈ E.

(b) Since for any z ∈ E, ρE(x) ≤ d(x, y) + d(y, z), it follows that ρE(x) ≤ d(x, y) +

ρE(y). In the same manner, we have ρE(y) ≤ d(x, y) + ρE(x). Therefore,

|ρE(x)− ρE(y)| ≤ d(x, y),

hence ρE is a uniformly continuous function on X.

21. Suppose K and F are disjoint sets in a metric space X, K is compact, F is closed.

Prove that there exists δ > 0 such that d(p, q) > δ if p ∈ K, q ∈ F . Hint: ρF is a

continuous positive function on K.

Show that the conclusion may fail for two disjoint closed sets if neither is compact.

Proof. By part (a) of Exercise 20, ρF (x) > 0 for every x ∈ K (otherwise K ∩F 6= ∅).

By part (b) of Exercise 20, ρE is continuous on K. Since K is compact, by Theorem

4.16, there exists a point p0 ∈ K, such that 0 < ρF (p0) = infz∈K ρF (z). Set δ =

ρF (p0)/2 > 0, then if p ∈ K, q ∈ F ,

d(p, q) ≥ ρF (p) ≥ ρF (p0) > δ.

The proof is complete.

For the counterexample, let X = R1, K = N, F = {n + 1n : n ≥ 2}. Clearly, both K

and F are closed and K ∩ F = ∅. However, there doesn’t exist any δ > 0 such that

d(p, q) > δ if p ∈ K and q ∈ F .

15

22. Let A and B be disjoint nonempty closed sets in a metric space X, and define

f(p) =ρA(p)

ρA(p) + ρB(p)(p ∈ X).

Show that f is a continuous function on X whose range lies in [0, 1], that f(p) = 0

precisely on A and f(p) = 1 precisely on B. This establishes a converse of Exercise 3:

Every closed set A ⊂ X is Z(f) for some continuous real f on X. Setting

V = f−1([

0,1

2

)), W = f−1

((1

2, 1

]),

show that V and W are open and disjoint, and that A ⊂ V,B ⊂ W . (Thus pairs

of disjoint closed sets in a metric space can be covered by pairs of disjoint open sets.

This property of metric space is called normality.)

Proof. We first show that f is well defined on X, namely, the denominator is always

positive. For p ∈ A, since p /∈ B = B, therefore ρB(p) > 0 by part (a) of Exercise 20.

For p ∈ Ac, then p /∈ A, then ρA(p) > 0. Hence f is well defined on X.

Clearly, if p ∈ A, then ρA(p) ≤ d(p, p) = 0, hence ρA(p) = 0 and f(p) = 0. If p ∈ B,

similarly ρB(p) = 0 hence f(p) = 1. Furthermore, for every p ∈ X, f(p) ≥ 0, and

f(p) ≤ (ρA(p) + ρB(p))/(ρA(p) + ρB(p)) = 1. Therefore, the range of f lies in [0, 1].

Since we have shown in part (b) of Exercise 20 that both ρA and ρB are continuous

on X, it follows that f is also continuous on X.

For every closed set A ⊂ X, if A = X, then we may trivially set f(p) = 0 for every

p ∈ X, then A = Z(f). If A $ X, then there exists nonempty closed set B ⊂ X such

that A∩B = ∅ (for instance, take B = {q}, where q /∈ A.). We can then define f to be

the function introduced in this problem so that A = Z(f), by preceding argument. In

what follows, we assume that A $ X and use the function f defined in this problem.

If we view f as a mapping of X into the metric space Y = [0, 1], then both [0, 1/2)

and (1/2, 1] are open in Y . Since f is continuous on X, by Theorem 4.8, both V and

16

W are open in X. Since f(A) = {0} and f(B) = {1}, it follows that A ⊂ V and

B ⊂W , and the proof is complete.

23. A real-valued function f defined in (a, b) is said to be convex if

f(λx+ (1− λ)y) ≤ λf(x) + (1− λ)f(y)

whenever a < x < b, a < y < b, 0 < λ < 1. Prove that every convex function

is continuous. Prove that every increasing convex function of a convex function is

convex.

If f is convex in (a, b) and if a < s < t < u < b, show that

f(t)− f(s)

t− s≤ f(u)− f(s)

u− s≤ f(u)− f(t)

u− t.

Proof. We shall prove the inequalities first. Suppose f is convex in (a, b) and a < s <

t < u < b, write

t =t− su− s

u+u− tu− s

s,

where t−su−s > 0, u−tu−s > 0 and t−s

u−s + u−tu−s = 1. Thus by convexity, we have

f(t) = f

(t− su− s

u+u− tu− s

s

)≤ t− su− s

f(u) +u− tu− s

f(s).

Rearrangements of this inequality gives

f(t)− f(s)

t− s≤ f(u)− f(s)

u− s≤ f(u)− f(t)

u− t.

The continuity of f can be shown by using these inequalities. Fix x0 ∈ (a, b) and

u > x0, since for all s < x0, we have

f(x0)− f(s)

x0 − s≤ f(u)− f(s)

u− s,

17

it follows that

f(s) ≥ f(x0)− f(u)− f(x0)

u− x0(x0 − s),

which implies lim infs↑x0f(s) ≥ f(x0). On the other hand, fix r < x0, since for all

s ∈ (r, x0), we have

f(s)− f(r)

s− r≤ f(x0)− f(r)

x0 − r,

it follows that

f(s) ≤ f(r) +f(x0)− f(r)

x0 − r(s− r),

which implies lim sups↑x0f(s) ≤ f(r) + f(x0)− f(r) = f(x0). In summary, we showed

that,

f(x0) ≤ lim infs↑x0

f(s) ≤ lim sups↑x0

f(s) ≤ f(x0).

Therefore, lims↑x0f(s) = f(x0−) exists, and f(x0−) = f(x0). Similarly, by con-

sidering the limit process from right, we can show that f(x0+) = f(x0). Hence

f(x0) = f(x0+) = f(x0−), i.e., f is continuous at x0.

Now suppose f : (a, b) → R is convex, and g is an increasing convex function defined

on the range of f , then for x ∈ (a, b), y ∈ (a, b) and λ ∈ (0, 1), we have

(g ◦ f)(λx+ (1− λ)y) = g(f(λx+ (1− λ)y)) ≤ g(λf(x) + (1− λ)f(y))

≤λg(f(x)) + (1− λ)g(f(y)) = λ(g ◦ f)(x) + (1− λ)(g ◦ f)(y).

The first inequality holds because f is convex in (a, b) and g is increasing, the second

inequality holds due to the convexity of g. In summary, g ◦ f is convex in (a, b).

24. Assume that f is a continuous real function defined in (a, b) such that

f

(x+ y

2

)≤ f(x) + f(y)

2

for all x, y ∈ (a, b). Prove that f is convex.

18

Proof. For any fixed x, y ∈ (a, b), without losing of generality, assume x < y and

f(x) ≤ f(y).

First, we shall show for every n ≥ 1 and every pair (αn, βn) ∈ N × N such that

αn + βn = 2n, it holds that

f

(αnx+ βny

2n

)≤ αn

2nf(x) +

βn2nf(y). (1)

When n = 1, (1.1) repeats the condition f

(x+ y

2

)≤ f(x) + f(y)

2. Suppose the

inequality (1.1) holds when n = k. Given αk+1 + βk+1 = 2k+1, if both αk+1 and βk+1

are even, thenαk+1

2∈ N,

βk+1

2∈ N, and

αk+1

2+βk+1

2= 2k, hence by induction

assumption, we have

f

(αk+1x+ βk+1y

2k+1

)= f

(αk+1

2 x+ βk+1

2 y

2k

)≤ αk+1/2

2kf(x) +

βk+1/2

2kf(y)

=αk+1

2k+1f(x) +

βk+1

2k+1f(y).

If both αk+1 and βk+1 are odd, then αk+1 +1 is even, βk+1−1 is even (here, of course,

we assume αk+1 < 2k+1, if αk+1 = 2k+1, the inequality is trivial to verify). Thus

α′k := (αk+1 + 1)/2 ∈ N and β′k := (βk+1 − 1)/2 ∈ N and α′k + β′k = 2k, again, by

induction assumption, it follows that

f

(αk+1x+ βk+1y

2k+1

)= f

(α′kx+ β′ky

2k

)≤ α′k

2kf(x) +

β′k2kf(y)

=αk+1

2k+1f(x) +

βk+1

2k+1f(y) +

f(x)− f(y)

2k+1≤ αk+1

2k+1f(x) +

βk+1

2k+1f(y).

Therefore, the inequality (1.1) holds.

Next we show that for any λ ∈ (0, 1), f(λx + (1 − λ)y) ≤ λf(x) + (1 − λ)f(y), and

hence f is convex. For every n ≥ 1, define

kn = sup

{k ∈ N : x+

k

2n(y − x) ≤ λx+ (1− λ)y

}.

19

By definition, we have

x+kn2n

(y − x) ≤ λx+ (1− λ)y < x+kn + 1

2n(y − x),

which implies, by simple algebra, that 1−λ− 12n < kn

2n ≤ 1−λ. Therefore the sequence

{kn/2n} converges as n→∞, and limn→∞

kn2n

= 1− λ.

By (1.1), for every n ≥ 1,

f

((2n − kn)x+ kny

2n

)≤(

1− kn2n

)f(x) +

kn2nf(y).

Let n→∞ on both sides of the above inequality, using that f is continuous in (a, b)

and limn→∞

kn2n

= 1− λ, it follows that

f(λx+ (1− λ)y) ≤ λf(x) + (1− λ)f(y).

That is, f is convex in (a, b).

25. If A ⊂ Rk and B ⊂ Rk, define A+B be the set of all sums x + y with x ∈ A,y ∈ B.

(a) If K is compact and C is closed in Rk, prove that K + C is closed. Hint: Take

z /∈ K + C, put F = z − C, the set of all z − y with y ∈ C. Then K and F are

disjoint. Choose δ as in Exercise 21. Show that the open ball with center z and

radius δ does not intersect K + C.

(b) Let α be an irrational real number. Let C1 be the set of all integers, let C2 be

the set of all nα with n ∈ C1. Show that C1 and C2 are closed subsets of R1

whose sum C1 + C2 is a countable dense subset of R1.

Proof.

(a) To show K + C is closed, we shall show its complement (K + C)c is open. Take

z /∈ K + C, put F = z − C, the set of all z − y with y ∈ C. Then K ∩ F = ∅.

It is easily seen F , as a translation of the closed set C, is also closed. (For a

20

rigorous proof, let p ∈ F ′, then for any ε > 0, there exists q = z − y ∈ F such

that |p − q| < ε, which implies z − p ∈ C ′, since C is closed, z − p ∈ C, thus

p ∈ F .) Since K is compact, F is closed, and K ∩ F = ∅, by Exercise 21, there

exists δ > 0 such that |p− q| > δ if p ∈ K and q ∈ F . For every r ∈ K+C, there

exists p ∈ K and q ∈ C such that r = p+ q, then

|z − r| = |z − p− q| = |(z − q)− p| > δ,

since z − q ∈ F and p ∈ K. The inequality implies r /∈ Nδ(z), i.e., Nδ(z) ∩ (K +

C) = ∅, or Nδ(z) ⊂ (K + C)c. Therefore, (K + C)c is open, K + C is closed.

(b)2 For each real number r, let brc denote the largest integer ≤ r and {r} denote

the fractional part of r.

Notice that {r} = r − brc.

Let α be an irrational number. Then for distinct i, j ∈ Z, we must have {iα} 6=

{jα}. If this were not true, then

iα− biαc = {iα} = {jα} = jα− bjαc,

which yields the false statement α =biαc − bjαc

i− j∈ Q. Hence,

S := {{iα} : i ∈ Z}

is an infinite subset of [0, 1]. By the Bolzano-Weierstrass Theorem, S has a limit

point in [0, 1]. One can thus find pairs of elements of S that are arbitrarily close.

Now, fix an n ∈ N. By the previous paragraph, there exist distinct i, j ∈ Z such

that

0 < |{iα} − {jα}| < 1

n.

Without losing of generality, it may be assumed that 0 < {iα} − {jα} < 1

n. Let

2The proof of that {{nα} : n ∈ Z} is dense in [0, 1] is due to the first answer of the following link:http://math.stackexchange.com/questions/272545/.

21

M be the largest positive integer such that M({iα}−{jα}) ≤ 1. The irrationality

of α then yields

(♠) M({iα} − {jα}) < 1.

Next, observe that for any m ∈ {0, . . . , n−1}, we can find a k ∈ {1, . . . ,M} such

that

k({iα} − {jα}) ∈[m

n,m+ 1

n

].

This is because

• the length of the interval

[m

n,m+ 1

n

]equals

1

n, while

• the distance between l({iα} − {jα}) and (l + 1)({iα} − {jα}) is <1

nfor all

l ∈ N.

On the other hand, there is another expression for k({iα} − {jα}):

k({iα} − {jα}) = {k({iα} − {jα})} (As 0 < k({iα} − {jα}) < 1; see (♠).)

= {k[(iα− biαc)− (jα− bjαc)]}

= {k(i− j)α+ k(bjαc − biαc)}

= {k(i− j)α}. (The {·} function discards any integer part.)

Hence,

{k(i− j)α} ∈[m

n,m+ 1

n

]∩ S.

As n is arbitrary, every non-degenerate sub-interval of [0, 1], no matter how small,

must contain an element of S. Therefore S is dense in [0, 1].

Given ε > 0, and let x0 be any real number. Since {x0} ∈ [0, 1] and S is dense in

[0, 1], there exists i ∈ Z such that |{x0} − {iα}| < ε, that is,

|(x0 − bx0c)− (iα− biαc)| = |x0 − [(bx0c − biαc) + iα]| < ε,

where bx0c− biαc ∈ C1, iα ∈ C2. Therefore, C1 +C2 is dense in R1. Clearly, C1 +C2

22

is countable since there exists an obvious injection of C1 +C2 into Z ×Z and C1 +C2

is infinite. Therefore, C1 + C2 is not closed, otherwise R1 = C1 + C2 = C1 + C2 is

countable, which is absurd.

26. Suppose X,Y, Z are metric spaces, and Y is compact. Let f map X into Y , let g be

a continuous one-to-one mapping of Y into Z, and put h(x) = g(f(x)) for x ∈ X.

Prove that f is uniformly continuous if h is uniformly continuous. Hint: g−1 has

compact domain g(Y ), and f(x) = g−1(h(x)).

Prove also that f is continuous if h is continuous.

Show (by modifying Example 4.21, or by finding a different example) that the com-

pactness of Y cannot be omitted from the hypothesis, even when X and Z are compact.

Proof. Since g is a continuous mapping of Y into Z and Y is compact, by Theorem 4.14,

g(Y ) is compact. By Theorem 4.17, g−1 is continuous mapping of g(Y ) onto Y . Since

g(Y ) is compact, by Theorem 4.19, g−1 is uniformly continuous on g(Y ). Therefore

if h is uniformly continuous on X, by the result of Exercise 12, f(x) = g−1(h(x)) is

uniformly continuous on X.

If h is continuous on X, since g is always (uniformly) continuous on g(Y ), it follows

by Theorem 4.7 that f(x) = g−1(h(x)) is continuous on X.

(Modifying Example 4.21) Let X = Z = {(cos t, sin t) : 0 ≤ t < 2π}, Y = [0, 2π). Let

f be the mapping of X onto Y that sends (cos t, sin t) to t, and let g be the mapping

of Y onto Z that sends t to (cos t, sin t). Clearly h = g ◦ f is the identical mapping of

X onto Z, hence it is (uniformly) continuous on X. However, f is not continuous at

(1, 0). Note X(= Z) is a compact subset of R2, as it is closed and bounded.

23