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### Continuity

Written by Men-Gen Tsai email: b89902089@ntu.edu.tw 1. Suppose f is a real function define on R1 which satisfies

h→0lim[f (x + h) − f (x − h)] = 0 for every x ∈ R1. Does this imply that f is continuous?

Solution: No. Take f (x) = 1, if x ∈ Z; f (x) = 0. otherwise.

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. (E denotes the closure of E.) Show, by an example, that f (E) can be a proper subset of f (E).

Proof: If f (E) is empty, the conclusion holds trivially. If f (E) is non- empty, then we take an arbitrary point y ∈ f (E). Thus, there exists p ∈ E such that y = f (p). Thus p ∈ E or p ∈ E0. Also, note that f (E) = f (E)S(f (E))0. Now we consider the following two cases:

Case 1: If p ∈ E, then y ∈ f (E) ⊂ f (E). Case 2: Suppose p ∈ E0. Since f is continuous at x = p, given  > 0, there exists δ > 0 such that

dY(f (x), f (p)) < 

whenever dX(x, p) < δ for all x ∈ X. Since p is a limit point of E, then for some δ > 0 there exists x ∈ E. Thus f (x) ∈ N(p) for some f (x) ∈ f (E). Since  is arbitrary, f (p) is a limit point of f (E) in Y . Thus f (p) ∈ f (E).

By case (1)(2), we proved that f (E) ⊂ f (E).

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Now we show that f (E) can be a proper subset of f (E). Define f (x) = 1

x, X = (0, +∞), Y = R1, E = Z+. Thus

f (E) = f (E) = {1/n : n ∈ Z+},

f (E) = {1/n : n ∈ Z+} = {0}[{1/n : n ∈ Z+}.

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: Let E = f (X) − Z(f ), that is, the set of all p ∈ X at which f (p) 6= 0. Take p[−E, and thus f (p) 6= 0. WLOG, we take f (p) > 0.

Since f is continuous at x = p, thus for every  > 0 there exists a δ > 0 such that

|f (x) − f (p)| < 

for all points x ∈ X for which dX(x, p) < δ. Especially, we take

 = f (p)/2 > 0. If x ∈ Nδ(p) for all x, then f (x) > f (p)/2 > 0, that is, Nδ(p) ⊂ E, that is, p is an interior point of E. (If f (p) < 0, we take

 = −f (p)/2 > 0). Since p is arbitrary, E is open. Thus, Z(f ) is closed.

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: First we need to show f (E) is dense in f (X), that is, every point of f (X) is a limit point of f (E), or a point of f (E) (or both).

Take any y ∈ f (X), and then there exists a point p ∈ X such that y = f (p). Since E is dense in X, thus p is a limit point of E or p ∈ E.

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If p ∈ E, then y = f (p) ∈ f (E). Thus y is a point of f (E), done. If p is a limit point of E and p /∈ E. Since f is continuous on X, for every

 > 0 there exists δ > 0 such that dY(f (x), f (p)) <  for all points x ∈ X for which dX(x, p) < δ. Since p is a limit point of E, there exists q ∈ Nδ(p) such that q 6= p and q ∈ E. Hence

f (q) ∈ N(f (p)) = N(y).

and f (q) ∈ f (E). Since p /∈ E, f (p) /∈ f (E), and f (q) 6= f (p). Hence f (p) is a limit point of f (E). Thus f (E) is dense in f (X).

Suppose p ∈ X − E. Since E is dense in X, p is a limit point of E and p /∈ E. Hence we can take a sequence {qn} → p such that qn ∈ E and qn 6= p for all n. (More precisely, since p is a limit point, every neighborhood Nr(p) of p contains a point q 6= p such that q ∈ E. Take r = rn = 1/n, and thus rn → 0 as n → ∞. At this time we can get q = qn→ p as n → ∞.) Hence

g(p) = g( lim

n→∞qn)

= lim

n→∞g(qn)

= lim

n→∞f (qn)

= f ( lim

n→∞qn)

= f (p).

Thus g(p) = f (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 function g on R1 such that g(x) = f (x) for all x ∈ E. (Such functions g are called continuous extensions 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

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the complement of E (compare Exercise 29, Chap. 2). The result re- mains true if R1 is replaced by any metric space, but the proof is not so simple.

Proof: Note that the following fact:

Every open set of real numbers is the union of a countable collection of disjoint open intervals.

Thus, consider Ec = S(ai, bi), where i ∈ Z, and ai < bi < ai+1< bi+1. We extend g on (ai, bi) as following:

g(x) = f (ai) + (x − ai)f (bi) − f (ai) bi− ai

(g(x) = f (x) for x ∈ E). Thus g is well-defined on R1, and g is contin- uous on R1 clearly.

Next, consider f (x) = 1/x on a open set E = R − 0. f is continuous on E, but we cannot redefine f (0) = any real number to make new f (x) continue at x = 0.

Next, consider a vector valued function

f (x) = (f1(x), ..., fn(x)),

where fi(x) is a real valued function. Since f is continuous on E, each component of f , fi, is also continuous on E, thus we can extend fi, say gi, for each i = 1, ..., n. Thus,

g(x) = (g1(x), ..., gn(x))

is a extension of f (x) since each component of g, gi, is continuous on R1 implies g is continuous on Rn.

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Note: The above fact only holds in R1. If we change R1 into any metric spaces, we have no the previous fact.

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 plain.

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

Proof: (⇒) Let G = {(x, f (x)) : x ∈ E}. Since f is a continuous mapping of a compact set E into f (E), by Theorem 4.14 f (E) is also compact. We claim that the product of finitely many compact sets is compact. Thus G = E × f (E) is also compact.

(⇐) (Due to Shin-Yi Lee) Define

g(x) = (x, f (x))

from E to G for x ∈ E. We claim that g(x) is continuous on E.

Consider h(x, f (x)) = x from G to E. Thus h is injective, continuous on a compact set G. Hence its inverse function g(x) is injective and continuous on a compact set E.

Since g(x) is continuous on E, the component of g(x), say f (x), is continuous on a compact E.

Proof of Claim: We prove that the product of two compact spaces is compact; the claim follows by induction for any finite product.

Step 1. Suppose that we are given sets X and Y , with Y is compact.

Suppose that x0 ∈ X, and N is an open set of X × Y containing the

”slice” x0× Y of X × Y . We prove the following:

There is a neighborhood W of x0 in X such that N contains the entire set W × Y .

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The set W × Y is often called a tube about x0× Y .

First let us cover x0× Y by basis elements U × V (for the topology of X × Y ) lying in N . The set x0× Y is compact, being homeomorphic to Y . Therefore, we can cover x0× Y by finitely many such basis elements

U1× V1, ..., Un× Vn.

(We assume that each of the basis elements Ui× Vi actually intersects x0 × Y , since otherwise that basis element would be superfluous; we could discard it from the finite collection and still have a covering of x0× Y .) Define

W = U1

\...\Un.

The set W is open, and it contains x0 because each set Ui×Viintersects x0× Y .

We assert that the sets Ui × Vi, which were chosen to cover the slice x0× Y , actually cover the tube W × Y . Let x × y ∈ W × Y . Consider the point x0 × y of the slice x0 × Y having the same y-coordinate as this point. Now x0× y ∈ Ui× Vi for some i, so that y ∈ Vi. But x ∈ Uj for every j (because x ∈ W ). Therefore, we have x × y ∈ Ui × Vi, as desired.

Since all the sets Ui × Vi ⊂ N , and since they cover W × Y , the tube W × Y ⊂ N also.

Step 2. Now we prove the claim. Let X and Y be compact sets. Let A be an open covering of X ×Y . Given x0 ∈ X, the slice x0×Y is compact and may therefore be covered by finitely many elements A1, ..., Am of A. Their union N = A1S...SAm is an open set containing x0× Y ; by Step 1, the open set N contains a tube W × Y about x0 × Y , where W is open in X. Then W × Y is covered by finitely many elements A1, ..., Am of A.

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Thus, for each x ∈ X, we can choose a neighborhood Wx of x such that the tube Wx× Y can be covered by finitely many element of A.

The collection of all the neighborhoods Wx is an open covering of X;

therefore by compactness of X, there exists a finite subcollection {W1, ..., Wk}

covering X. The union of the tubes

W1× Y, ..., Wk× Y

is all of X × Y ; since each may be covered by finitely many elements of A, so may X × Y be covered.

7. If 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 restriction of both f and g to every straight line in R2 are continuous!

Proof: Since x2 + y4 ≥ 2xy2, f (x, y) ≤ 2 for all (x, y) ∈ R2. That is, f is bounded (by 2). Next, select

(xn, yn) = ( 1 n3, 1

n).

(xn, yn) → (0, 0) as n → ∞, and g(xn, yn) = n/2 → ∞ as n → ∞, that is, g(x, y) is unbounded in every neighborhood of (0, 0) by choosing large enough n.

Next, select

(xn, yn) = ( 1 n2, 1

n).

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(xn, yn) → (0, 0) as n → ∞, and f (xn, yn) = 1/2 for all n. Thus,

n→∞lim f (xn, yn) = 1

2 6= 0 = f (0, 0).

for some sequence {(xn, yn)} in R2. Thus, f is not continuous at (0, 0).

Finally, we consider two cases of straight lines in R2: (1) x = c and (2) y = ax + b. (equation of straight lines).

(1) x = c: If c 6= 0, f (x, y) = cy2/(c2+y4) and g(x, y) = cy2/(c2+y6) are continuous since cy2, c2+y4, and c2+y6are continuous on R1 respect to y, and c2+ y4, c2+ y6 are nonzero. If c = 0, then f (x, y) = g(x, y) = 0, and it is continuous trivially.

(2) y = ax + b: If b 6= 0, then this line dose not pass (0, 0). Then f (x, y) = x(ax + b)2/(x2+ (ax + b)4) and g(x, y) = x(ax + b)2/(x2+ (ax + b)6). By previous method we conclude that f (x, y) and g(x, y) are continuous. If b = 0, then f (x, y) = 0 if (x, y) = (0, 0); f (x, y) = a2x/(1 + a4x2), and g(x, y) = 0 if (x, y) = (0, 0); g(x, y) = a2x/(1 + a6x4). Thus, f (x, y) → 0/1 = 0 = f (0, 0) and g(x, y) → 0/1 = 0 = f (0, 0) as x → 0. Thus, f and g are continuous.

Both of two cases implies that the restriction of both f and g to every straight line in R2 are 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: Let E is bounded by M > 0, that is, |x| ≤ M for all x ∈ E.

Since f is uniformly continuous, take  = 1 there exists δ > 0 such that

|f (x) − f (y)| < 

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whenever |x − y| < δ where x, y ∈ E. For every x ∈ E, there exists an integer n = nx such that

nδ ≤ x < (n + 1)δ.

Since E is bounded, the collection of S = {nx : x ∈ E} is finite.

Suppose x ∈ E and x is the only one element satisfying nδ ≤ x <

(n + 1)δ for some n. Let x = xn, and thus

|f (x)| ≤ |f (xn)|

for all x ∈ ET[nδ, (n + 1)δ). If there are more than two or equal to two element satisfying that condition, then take some one as xn. Since

|x − xn| < δ for all x ∈ ET[nδ, (n + 1)δ). Thus

|f (x) − f (xn)| < 1 for all x ∈ ET[nδ, (n + 1)δ), that is,

|f (x)| < 1 + |f (xn)|.

Hence

|f (x)| < max

n∈S (1 + f (xn)).

(Since S is finite, that maximum is meaningful). Thus f (x) is bounded.

Note: If boundedness of E is omitted from the hypothesis, define f (x) = x for x ∈ E = R1. Hence f is uniformly continuous on E, but f (E) = R1 is unbounded.

9. Show that the requirement in the definition of uniform continuity can be rephrased as follows, in terms of diameters of sets: To every  > 0 there exists a δ > 0 such that diamf (E) <  for all E ⊂ X with diamE < δ.

Proof: Recall the original definition of uniformly continuity:

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for every  > 0 there exists δ > 0 such that dY(f (p), f (q)) <  for all p and q in X for which dX(p, q) < δ.

(⇐) Given  > 0. ∀p, q ∈ X for which dX(p, q) < δ. Take E = {p, q},

and thus diamE = supp,q∈Ed(p, q) = d(p, q) < δ. Hence diamf (E) < .

Note that diamf (E) ≥ d(f (p), f (q)) since p, q ∈ E. Hence d(f (p), f (q)) <

. Thus for every  > 0 there exists δ > 0 such that dY(f (p), f (q)) <  for all p and q in X for which dX(p, q) < δ.

(⇒) ∀E ⊂ X with diamE < δ. ∀p, q ∈ R, d(p, q) ≤ diamE < δ. Thus we have

d(f (p), f (q)) <  2

for all p, q ∈ E. Hence diamf (E) ≤ /2 < . Thus to every  > 0 there exists a δ > 0 such that diamf (E) <  for all E ⊂ X with diamE < δ.

10. Complete the details of the following alternative proof of Theorem 4.19:

If f is not uniformly continuous, then for some  > 0 there are sequences {pn}, {qn} in X such that dX(pn, qn) → 0 but dY(f (pn), f (qn)) > . Use Theorem 2.37 to obtain a contradiction.

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. ∀0 > 0, ∃N such that dX(xn, xm) < 0 whenever m, n ≥ N . Since f is a uniformly continuous,

∀ > 0, ∃δ > 0 such that dY(f (x), f (y)) <  whenever d(x, y) < δ. Take

0 = δ. Thus

dY(f (xn), f (xm)) < 

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whenever m, n ≥ N for some N ; that is, {f (xn)} is a Cauchy sequence.

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

State this more precisely and prove it.

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 define on X is the desired extension of f . 14. Let I = [0, 1] be the closed unit interval. Suppose f is continuous

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

Proof: Let g(x) = f (x) − x. If g(1) = 0 or g(0) = 0, then the conclusion holds trivially. Now suppose g(1) 6= 0 and g(0) 6= 0. Since f is from I to I, 0 6= f (x) 6= 1. Thus,

g(1) = f (1) − 1 < 0, g(0) = f (0) − 0 > 0.

Since g is continuous on [0, 1], by Intermediate Value Theorem (Theo- rem 4.23)

g(c) = 0

for some c ∈ (0, 1). Hence f (c) = c for some c ∈ (0, 1).

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

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

Proof: Suppose not, there exist three points x1 < x2 < x3 ∈ R1 such that

f (x2) > f (x1), f (x2) > f (x3) or

f (x2) < f (x1), f (x2) < f (x3).

WLOG, we only consider the case that f (x2) > f (x1), f (x2) > f (x3) for some x1 < x2 < x3. Since f is continuous on R1, for

 = f (x2) − f (x1) 2 > 0 there exists δ1 > 0 such that

|f (x) − f (x1)| <  whenever |x − x1| < δ1. That is,

f (x) < f (x1) + f (x2)

2 < f (x2)

whenever x < x1 + δ1. Note that δ1 < x2 − x1. Hence we can take y1 ∈ (x1, x1+ δ1). Similarly, for

 = f (x2) − f (x3) 2 > 0 there exists δ2 > 0 such that

|f (x) − f (x3)| <  whenever |x − x3| < δ2. That is,

f (x) < f (x2) + f (x3)

2 < f (x2)

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whenever x > x3 − δ2. Note that δ2 < x3 − x2. Hence we can take y2 ∈ (x3− δ2, x3). Note that y2 > y1. Since f is continuous on a closed set [y1, y2], f take a maximum value at p ∈ [y1, y2]. Note that

sup

x∈(x1,x3)

f (x) ≤ sup

x∈[y1,y2]

f (x)

by previous inequations. Also, sup

x∈(x1,x3)

f (x) ≥ sup

x∈[y1,y2]

f (x)

Hence supx∈(x1,x3)f (x) = supx∈[y1,y2]f (x). Since (x1, x3) is an open set, f ((x1, x3)) is also open. Note that f (p) ∈ f ((x1, x3)) but f (p) is not an interior point of f ((x1, x3)). (otherwise f (p) +  ∈ f ((x1, x3)) for some

 > 0. That is, f (p) +  > f (p), a contradiction with the maximum of f (p)).

16. Let [x] denote the largest integer contained in x, this is, [x] is a integer such that x − 1 < [x] ≤ x; and let (x) = x − [x] denote the fractional part of x. What discontinuities do the function [x] and (x) have?

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 as- sociated with at most one point of E. Deal similarly with the other possible types of simple discontinuities.

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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),

1

n (x = mn).

Prove that f is continuous at every irrational point, and that f has a simple discontinity at every rational point.

19. Suppose f is a real function with domain R1 which has the interme- diate value property: 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 contra- diction. (N. J. Fine, Amer. Math. Monthly, vol. 73, 1966, p. 782.)

Proof: Let S = {x : f (x) = r}. If xn → x0, but f (xn) > r > f (x0) for some r and all n since Q is dense in R1, then f (tn) = r for some tn between x0 and xn; thus tn→ x0. Hence x0 is a limit point of S. Since S is closed, f (x0) = r, a contradiction. Hence, lim sup f (xn) ≤ f (x0).

Similarly, lim inf f (xn) ≥ f (x0). Hence, lim f (xn) = f (x0), and f is continuous at x0.

Note: Original problem is stated as follows:

Let f be a function from the reals to the reals, differentiable at every point. Suppose that, for every r, the set of points

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x, where f0(x) = r, is closed. Prove that f0 is continuous.

If we replace Q into any dense subsets of R1, the conclusion also holds.

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

ρE(x) = inf

z∈Ed(x, z).

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

(b) Prove that ρE(x) is a uniformly continuous function on X, by show- ing that

E(x) − ρE(y)| ≤ d(x, y)

for all x ∈ X, y ∈ X. Hint: ρE(x) ≤ d(x, z) ≤ d(x, y) + d(y, z), so that ρE(x) ≤ d(x, y) + ρE(y).

Proof of (a): (⇐) If x ∈ E ⊂ E, then

z∈Einf d(x, z) ≤ d(x, x) = 0

since we take z = x ∈ E. Hence ρE(x) = 0 if x ∈ E. Suppose x ∈ E −E, that is, x is a limit point of E. Thus for every neighborhood of x contains a point y 6= x such that q ∈ E. It implies that d(x, y) → 0 for some y ∈ E, that is, ρE(x) = infz∈Ed(x, z) = 0 exactly.

(⇒) Suppose ρE(x) = infz∈Ed(x, z) = 0. Fixed some x ∈ X. If d(x, z) = 0 for some z ∈ E, then x = z, that is x ∈ E ⊂ E. If d(x, z) > 0 for all z ∈ E, then by infz∈Ed(x, z) = 0, for any  > 0 there exists z ∈ E such that

d(x, z) < , that is,

z ∈ N(x).

Since  is arbitrary and z ∈ E, x is a limit point of E. Thus x ∈ E0 ⊂ E.

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Proof of (b): For all x ∈ X, y ∈ X, z ∈ E,

ρE(x) ≤ d(x, z) ≤ d(x, y) + d(y, z).

Take infimum on both sides, and we get that ρE(x) ≤ d(x, y) + ρE(y).

Similarly, we also have

ρE(y) ≤ d(x, y) + ρE(x).

Hence

E(x) − ρE(y)| ≤ d(x, y)

for all x ∈ X, y ∈ X. Thus ρE is a uniformly continuous function on X.

Exercise 1: (Due to Shin-Yi Lee) In a metric space (S, d), let A be a nonempty subset of S. Define a function fA(x) : S → R by the equation

fA(x) = inf{d(x, y) : y ∈ A}

for every x ∈ S. The value fA(x) is called the distance from x to A.

(a) Prove that fA is uniformly continuous on S.

(b) Prove A = {x ∈ S : fA(x) = 0}.

Exercise 2: (Due to Shin-Yi Lee) In a metric space (S, d). Let A and B be two disjoint closed subsets of S. Prove that there are two open subset of S, say U and V such that A ⊂ U and B ⊂ V with UTV = φ.

It will be shown in Exercise 4.22.

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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: Let

ρF(x) = inf

z∈Fd(x, z) for all x ∈ K. By Exercise 4.20(a), we know that

ρF(x) = 0 ⇔ x ∈ F = F

(since F is closed). That is, ρF(x) = 0 if and only if x ∈ F . Since K and F are disjoint, ρF(x) is a positive function. Also, by Exercise 4.20(b) ρF(x) is continuous. Thus ρF(x) is a continuous positive function.

Since K is compact, ρF(x) takes minimum m > 0 for some x0 ∈ K.

Take δ = m/2 > 0 as desired.

Next, let X = R1,

A = Z+− {2},

B = {n + 1/n : n ∈ Z+}.

Hence A and B are disjoint, and they are not compact. Suppose there exists such δ > 0. Take

x = [1

δ] + 1 ∈ A, y = x + 1 x ∈ B.

However,

d(x, y) = 1 x < 1

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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 contained in 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 is contained in V , B is contained in W . (Thus pairs of disjoint closed set in a metric space can be covered by pairs of disjoint open sets. This property of metric spaces is called normality.)

Proof: Note that ρA(p) and ρB(p) are (uniformly) continuous on X, and ρA(p) + ρB(p) > 0. (Clearly, ρA(p) + ρB(p) ≥ 0 by the definition.

If ρA(p) + ρB(p) = 0, then p ∈ ATB by Exercise 20, a contradiction).

Thus f (p) = ρA(p)/(ρA(p)+ρB(p)) is continuous on X. Next, f (p) ≥ 0, and f (p) ≤ 1 since ρA(p) ≤ ρA(p) + ρB(p). Thus f (X) lies in [0, 1].

Next, f (p) = 0 ⇔ ρA(p) = 0 ⇔ p ∈ A precisely, and f (p) = 1 ⇔ ρB(p) = 0 ⇔ p ∈ B precisely by Exercise 20.

Now we prove a converse of Exercise 3: Every closed set A ⊂ X is Z(f ) for some continuous real f on X. If Z(f ) = φ, then f (x) = 1 for all x ∈ X satisfies our requirement. If Z(f ) 6= φ, we consider two possible cases: (1) Z(f ) = X; (2) Z(f ) 6= X. If Z(f ) = X, then f (x) = 0 for all x ∈ X. If Z(f ) 6= X, we can choose p ∈ X such that f (p) 6= 0.

Note that Z(f ) and {p} are one pair of disjoint closed sets. Hence we

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let

f (x) = ρZ(f )(x) ρZ(f )(x) + ρ{p}(x).

By the previous result, we know that f (x) satisfies our requirement.

Hence we complete the whole proof.

Note that [0, 1/2) and (1/2, 1] are two open sets of f (X). Since f is continuous, V = f−1([0, 1/2)) and W = f−1((1/2, 1]) are two open sets.

f−1({0}) ⊂ f−1([0, 1/2)), and f−1({1}) ⊂ f−1((1/2, 1]). Thus, A ⊂ V and B ⊂ W . Thus a metric space X is normal.

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. (For example, if f is convex, so is ef.) 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 .

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.

25. If A ⊂ Rk and B ⊂ Rk, define A + B to 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.

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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 not closed, by showing that C1+ C2 is a countable dense subset of R1.

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

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