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# Charpter 3 Elements of Point set Topology

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## Charpter 3 Elements of Point set Topology

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### and R

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3.1 Prove that an open interval in R1is an open set and that a closed interval is a closed set.

proof: 1. Leta, b be an open interval in R1, and let x  a, b. Consider

minx  a, b  x : L. Then we have Bx, L  x  L, x  L  a, b. That is, x is an interior point ofa, b. Since x is arbitrary, we have every point of a, b is interior. So,

a, b is open in R1.

2. Leta, b be a closed interval in R1, and let x be an adherent point of a, b. We want to show x  a, b. If x  a, b, then we have x  a or x  b. Consider x  a, then

Bx, a  x2   a, b   3x  a2 , x  a2   a, b   which contradicts the definition of an adherent point. Similarly for x  b.

Therefore, we have x  a, b if x is an adherent point of a, b. That is, a, b contains its all adherent points. It implies thata, b is closed in R1.

3.2 Determine all the accumulation points of the following sets in R1 and decide whether the sets are open or closed (or neither).

(a) All integers.

Solution: Denote the set of all integers by Z. Let x  Z, and consider

Bx, x12   x  S  . So, Z has no accumulation points.

However, Bx, x12   S  x  . So Z contains its all adherent points. It means that Z is closed. Trivially, Z is not open since Bx, r is not contained in Z for all r  0.

Remark: 1. Definition of an adherent point: Let S be a subset of Rn, and x a point in Rn, x is not necessarily in S. Then x is said to be adherent to S if every nball Bx

contains at least one point of S. To be roughly, Bx  S  .

2. Definition of an accumulation point: Let S be a subset of Rn, and x a point in Rn, then x is called an accumulation point of S if every nball Bx contains at least one point of S distinct from x. To be roughly,Bx  x  S  . That is, x is an accumulation point if, and only if, x adheres to S x. Note that in this sense,

Bx  x  S  Bx  S  x.

3. Definition of an isolated point: If x  S, but x is not an accumulation point of S, then x is called an isolated point.

4. Another solution for Z is closed: Since R Z  nZ n, n  1, we know that R  Z is open. So, Z is closed.

5. In logics, if there does not exist any accumulation point of a set S, then S is automatically a closed set.

(b) The intervala, b.

solution: In order to find all accumulation points ofa, b, we consider 2 cases as follows.

1.a, b : Let x  a, b, then Bx, r  x  a, b   for any r  0. So, every point ofa, b is an accumulation point.

2. R1 a, b  , a  b,  : For points in b,  and , a, it is easy to know that these points cannot be accumulation points since x  b,  or x  , a, there

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exists an nball Bx, rx such that Bx, rx  x  a, b  . For the point a, it is easy to know thatBa, r  a  a, b  . That is, in this case, there is only one

accumulation point a ofa, b.

So, from 1 and 2, we know that the set of the accumulation points ofa, b is a, b.

Since a a, b, we know that a, b cannot contain its all accumulation points. So,

a, b is not closed.

Since an nball Bb, r is not contained in a, b for any r  0, we know that the point b is not interior toa, b. So, a, b is not open.

(c) All numbers of the form 1/n, (n  1, 2, 3, . . . .

Solution: Write the set1/n : n  1, 2, . . .   1, 1/2, 1/3, . . . , 1/n, . . .  : S.

Obviously, 0 is the only one accumulation point of S. So, S is not closed since S does not contain the accumulation point 0. Since 1  S, and B1, r is not contained in S for any r  0, S is not open.

Remark: Every point of1/n : n  1, 2, 3, . . .  is isolated.

(d) All rational numbers.

Solutions: Denote all rational numbers by Q. It is trivially seen that the set of accumulation points is R1.

So, Q is not closed. Consider x  Q, any n ball Bx is not contained in Q. That is, x is not an interior point of Q. In fact, every point of Q is not an interior point of Q. So, Q is not open.

(e) All numbers of the form 2n  5m, (m, n  1, 2, . . . .

Solution: Write the set

2n  5m : m, n  1, 2, . .   m1m  12 5m, 14 5m, . . . , 12n  5m, . . . : S

  12  1 5, 1

2  1

52, . . . , 1 2  1

5m, . . . 

 14  1 5, 1

4  1

52, . . . , 1 4  1

5m, . . . 

. . . .

 12n  15 , 1 2n  1

52 , . . . , 12n  15m . . .   . . . .

1 2

3

So, we find that S  21n : n  1, 2, . . .   51m : m  1, 2, . . .   0. So, S is not closed since it does not contain 0. Since 12  S, and B12, r is not contained in S for any r  0, S is not open.

Remark: By (1)-(3), we can regard them as three sequences 12 5m

m1 m, 1

4 5m

m1

mand 1

2n  5m

m1

m, respectively.

And it means that for (1), the sequence5mm1m moves 12. Similarly for others. So, it is easy to see why 12 is an accumulation point of12  5mm1m. And thus get the set of all accumulation points of2n 5m : m, n  1, 2, . . .

(f) All numbers of the form1n  1/m, (m, n  1, 2, . . . .

Solution: Write the set of all numbers1n 1/m, (m, n  1, 2, . . .  as 1 1m m1

m  1  1m m1

m : S.

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And thus by the remark in (e), it is easy to know that S  1, 1. So, S is not closed since S  S. Since 2  S, and B2, r is not contained in S for any r  0, S is not open.

(g) All numbers of the form1/n  1/m, (m, n  1, 2, . . . .

Solution: Write the set of all numbers1/n  1/m, (m, n  1, 2, . . .  as

1  1/mm1m 1/2  1/mm1m. . . 1/n  1/mm1m . . . : S.

We find that S  1/n : n  N  1/m : m  N  0  1/n : n  N  0. So, S is not closed since S  S. Since 1  S, and B1, r is not contained in S for any r  0, S is not open.

(h) All numbers of the form1n/1  1/n, (n  1, 2, . . . .

Soluton: Write the set of all numbers1n/1  1/n, (n  1, 2, . . .  as 1

1 2k1 k1

k

 1

1 2k11 k1

k

: S.

We find that S  1, 1. So, S is not closed since S  S. Since 12  S, and B12 , r is not contained in S for any r  0, S is not open.

3.3 The same as Exercise 3.2 for the following sets in R2. (a) All complex z such that |z|  1.

Solution: Denotez  C : |z| 1 by S. It is easy to know that S  z  C : |z| 1.

So, S is not closed since S  S. Let z  S, then |z|  1. Consider Bz, |z|12   S, so every point of S is interior. That is, S is open.

(b) All complex z such that |z|  1.

Solution: Denotez  C : |z| 1 by S. It is easy to know that S  z  C : |z| 1.

So, S is closed since S  S. Since 1  S, and B1, r is not contained in S for any r  0, S is not open.

(c) All complex numbers of the form1/n  i/m, (m, n  1, 2, . . . .

Solution: Write the set of all complex numbers of the form1/n  i/m, (m, n  1, 2, . . .  as

1 im m1m 1 2  i

m m1

m. . .  1n  im m1m. . . : S.

We know that S  1/n : n  1, 2, . . .   i/m : m  1, 2, . . .   0. So, S is not closed since S  S. Since 1  i  S, and B1  i, r is not contained in S for any r  0, S is not open.

(d) All pointsx, y such that x2  y2  1.

Solution: Denotex, y : x2  y2  1 by S. We know that

S  x, y : x2 y2  1. So, S is not closed since S  S. Let p  x, y  S, then x2  y2  1. It is easy to find that r  0 such that Bp, r  S. So, S is open.

(e) All pointsx, y such that x  0.

Solution: Write all pointsx, y such that x  0 as x, y : x  0 : S. It is easy to know that S  x, y : x  0. So, S is not closed since S  S. Let x  S, then it is easy to find rx  0 such that Bx, rx  S. So, S is open.

(f) All pointsx, y such that x  0.

Solution: Write all pointsx, y such that x  0 as x, y : x  0 : S. It is easy to

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know that S  x, y : x  0. So, S is closed since S  S. Since 0, 0  S, and B0, 0, r is not contained in S for any r  0, S is not open.

3.4 Prove that every nonempty open set S in R1contains both rational and irratonal numbers.

proof: Given a nonempty open set S in R1. Let x  S, then there exists r  0 such that Bx, r  S since S is open. And in R1, the open ball Bx, r  x  r, x  r. Since any interval contains both rational and irrational numbers, we have S contains both rational and irrational numbers.

## 3.5

Prove that the only set in R1 which are both open and closed are the empty set and R1 itself. Is a similar statement true for R2?

Proof: Let S be the set in R1, and thus consider its complement T  R1  S. Then we have both S and T are open and closed. Suppose that S  R1and S  , we will show that it is impossible as follows.

Since S  R1, and S  , then T   and T  R1. Choose s0  S and t0  T, then we consider the new point s0t20 which is in S or T since R  S  T. If s0t20  S, we say

s0t0

2  s1,otherwise, we say s0t20  t1.

Continue these steps, we finally have two sequences namedsn  S and tm  T.

In addition, the two sequences are convergent to the same point, say p by our construction.

So, we get p  S and p  T since both S and T are closed.

However, it leads us to get a contradiction since p  S  T  . Hence S  R1 or S  .

Remark: 1. In the proof, the statement is true for Rn.

2. The construction is not strange for us since the process is called Bolzano Process.

3. 6 Prove that every closed set in R1 is the intersection of a countable collection of open sets.

proof: Given a closed set S, and consider its complement R1  S which is open. If R1  S  , there is nothing to prove. So, we can assume that R1  S  .

Let x  R1  S, then x is an interior point of R1  S. So, there exists an open interval

a, b such that x  a, b  R1  S. In order to show our statement, we choose a smaller intervalax, bx so that x  ax, bx and ax, bx  a, b  R1  S. Hence, we have

R1  S  xR1Sax, bx which implies that

S  R1  xR1Sax, bx

 xR1S R1  ax, bx

 n1nR1  an, bn (by Lindelof Convering Theorem).

Remark: 1. There exists another proof by Representation Theorem for Open Sets on The Real Line.

2. Note that it is true for that every closed set in R1 is the intersection of a countable collection of closed sets.

3. The proof is suitable for Rn if the statement is that every closed set in Rnis the intersection of a countable collection of open sets. All we need is to change intervals into disks.

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## 3.7

Prove that a nonempty, bounded closed set S in R1 is either a closed interval, or that S can be obtained from a closed interval by removing a countable disjoint collection of open intervals whose endpoints belong to S.

proof: If S is an interval, then it is clear that S is a closed interval. Suppose that S is not an interval. Since S  is bounded and closed, both sup S and inf S are in S. So, R1  S

 inf S, sup S  S. Denote inf S, sup S by I. Consider R1  S is open, then by Representation Theorem for Open Sets on The Real Line, we have

R1  S  m1mIm

 I  S which implies that

S  I  m1mIm.

That is, S can be obtained from a closed interval by removing a countable disjoint collection of open intervals whose endpoints belong to S.

### Open and closed sets in R

n

3.8 Prove that open nballs and n dimensional open intervals are open sets in Rn. proof: Given an open nball Bx, r. Choose y  Bx, r and thus consider

By, d  Bx, r, where d  min|x  y|, r  |x  y|. Then y is an interior point of Bx, r.

Since y is arbitrary, we have all points of Bx, r are interior. So, the open n ball Bx, r is open.

Given an ndimensional open interval a1, b1  a2, b2 . . . an, bn : I. Choose x  x1,x2, . . . , xn  I and thus consider r  mini1inri, where ri  minxi  ai, bi  xi.

Then Bx, r  I. That is, x is an interior point of I. Since x is arbitrary, we have all points of I are interior. So, the ndimensional open interval I is open.

3.9 Prove that the interior of a set in Rn is open in Rn.

Proof: Let x  intS, then there exists r  0 such that Bx, r  S. Choose any point of Bx, r, say y. Then y is an interior point of Bx, r since Bx, r is open. So, there exists d  0 such that By, d  Bx, r  S. So y is also an interior point of S. Since y is

arbitrary, we find that every point of Bx, r is interior to S. That is, Bx, r  intS. Since x is arbitrary, we have all points of intS are interior. So, intS is open.

Remark: 1 It should be noted that S is open if, and only if S  intS.

2. intintS  intS.

3. If S  T, then intS  intT.

## 3.10

If S  Rn, prove that intS is the union of all open subsets of Rn which are contained in S. This is described by saying that intS is the largest open subset of S.

proof: It suffices to show that intS  ASA, where A is open. To show the statement, we consider two steps as follows.

1. Let x  intS, then there exists r  0 such that Bx, r  S. So, x  Bx, r  AS A. That is, intS  ASA.

2. Let x  AS A, then x  A for some open set A S. Since A is open, x is an interior point of A. There exists r  0 such that Bx, r  A  S. So x is an interior point of S, i.e., x  intS. That is, ASA  intS.

From 1 and 2, we know that intS  AS A, where A is open.

Let T be an open subset of S such that intS  T. Since intS  AS A, where A is open,

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we have intS  T  ASA which implies intS  T by intS  AS A. Hence, intS is the largest open subset of S.

3.11 If S and T are subsets of Rn, prove that

intS  intT  intS  T and intS  intT  intS  T.

Proof: For the partintS  intT  intS  T, we consider two steps as follows.

1. Since intS  S and intT  T, we have intS  intT  S  T which implies that Note thatintS  intT is open.

intS  intT  intintS  intT  intS  T.

2. Since S  T  S and S  T  T, we have intS  T  intS and intS  T  intT. So,

intS  T  intS  intT.

From 1 and 2, we know thatintS  intT  intS  T.

For the partintS  intT  intS  T, we consider intS  S and intT  T. So,

intS  intT  S  T which implies that Note thatintS  intT is open.

intintS  intT  intS  intT  intS  T.

Remark: It is not necessary thatintS  intT  intS  T. For example, let S  Q, and T  Qc, then intS  , and intT  . However, intS  T  intR1  R.

3.12 Let S denote the derived set and S the closure of a set S in Rn. Prove that (a) S is closed in Rn; that isS  S.

proof: Let x be an adherent point of S. In order to show S is closed, it suffices to show that x is an accumulation point of S. Assume x is not an accumulation point of S, i.e., there exists d  0 such that

Bx, d  x  S  . *

Since x adheres to S, then Bx, d  S  . So, there exists y  Bx, d such that y is an accumulation point of S. Note that x  y, by assumption. Choose a smaller radius d so that

By, d  Bx, d  x and By, d  S  .

It implies

  By, d  S  Bx, d  x  S   by (*)

which is absurb. So, x is an accumulation point of S. That is, S contains all its adherent points. Hence S is closed.

(b) If S  T, then S  T.

Proof: Let x  S, thenBx, r  x  S   for any r  0. It implies that

Bx, r  x  T   for any r  0 since S  T. Hence, x is an accumulation point of T.

That is, x  T. So, S  T. (c)S  T  S  T

Proof: For the partS  T  S  T, we show it by two steps.

1. Since S  S  T and T  S  T, we have S  S  T and T  S  T by (b).

So,

S  T  S  T

2. Let x  S  T, thenBx, r  x  S  T  . That is,

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Bx, r  x  S  Bx, r  x  T  .

So, at least one ofBx, r  x  S and Bx, r  x  T is not empty. If

Bx, r  x  S  , then x  S. And ifBx, r  x  T  , then x  T. So,

S  T  S  T. From 1 and 2, we haveS  T  S  T.

Remark: Note that sinceS  T  S  T, we have clS  T  clS  clT, where clS is the closure of S.

(d)S  S.

Proof: Since S  S  S, thenS  S  S  S  S  S sinceS  S by (a).

(e) S is closed in Rn.

Proof: SinceS  S  S by (d), then S cantains all its accumulation points. Hence, S

is closed.

Remark: There is another proof which is like (a). But it is too tedious to write.

(f) S is the intersection of all closed subsets of Rn containing S. That is, S is the smallest closed set containing S.

Proof: It suffices to show that S  ASA, where A is closed. To show the statement, we consider two steps as follows.

1. Since S is closed and S  S, then ASA  S.

2. Let x  S, then Bx, r  S   for any r  0. So, if A  S, then

Bx, r  A   for any r  0. It implies that x is an adherent point of A. Hence if A  S, and A is closed, we have x  A. That is, x  AS A. So, S  ASA.

From 1 and 2, we have S  ASA.

Let S  T  S, where T is closed. Then S  AS A  T. It leads us to get T  S.

That is, S is the smallest closed set containing S.

Remark: In the exercise, there has something to remeber. We list them below.

Remark 1. If S  T, then S  T. 2. If S  T, then S  T.

3. S  S  S.

4. S is closed if, and only if S  S.

5. S is closed.

6. S is the smallest closed set containing S.

3.13 Let S and T be subsets of Rn. Prove that clS  T  clS  clT and that S clT  clS  T if S is open, where clS is the closure of S.

Proof: Since S T  S and S  T  T, then clS  T  clS and, clS  T  clT. So, clS  T  clS  clT.

Given an open set S , and let x  S  clT, then we have

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1. x  S and S is open.

 Bx, d  S for some d  0.

Bx, r  S  Bx, r if r  d.

Bx, r  S  Bx, d if r  d.

and

2. x  clT

 Bx, r  T   for any r  0.

From 1 and 2, we know

Bx, r  S  T  Bx, r  S  T  Bx, r  T   if r  d.

Bx, r  S  T  Bx, r  S  T  Bx, d  T   if r  d.

So, it means that x is an adherent point of S T. That is, x  clS  T. Hence, S clT  clS  T.

Remark: It is not necessary that clS  T  clS  clT. For example, S  Q and T  Qc, then clS  T   and clS  clT  R1.

Note. The statements in Exercises 3.9 through 3.13 are true in any metric space.

## 3.14

A set S in Rn is called convex if, for every pair of points x and y in S and every real satisfying 0    1, we have x  1  y  S. Interpret this statement

geometrically (in R2 and R3 and prove that (a) Every nball in Rn is convex.

Proof: Given an nball Bp, r, and let x, y  Bp, r. Consider x  1  y, where 0    1.

Then

x  1  y  p  x  p  1  y  p

 x  p  1  y  p

 r  1  r

 r.

So, we havex  1  y  Bp, r for 0    1. Hence, by the definition of convex, we know that every nball in Rn is convex.

(b) Every ndimensional open interval is convex.

Proof: Given an ndimensional open interval I  a1, b1 . . . an, bn. Let x, y  I, and thus write x  x1, x2, . . . , xn and y  y1, y2, . . . yn. Consider

x  1  y  x1  1  y1,x2  1  y2, . . . ,xn  1  yn where 0    1.

Then

ai  xi  1  yi  bi, where i  1, 2, . . , n.

So, we havex  1  y  I for 0    1. Hence, by the definition of convex, we know that every ndimensional open interval is convex.

(c) The interior of a convex is convex.

Proof: Given a convex set S, and let x, y  intS. Then there exists r  0 such that Bx, r  S, and By, r  S. Consider x  1  y : p  S, where 0    1, since S is convex.

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Claim that Bp, r  S as follows.

Let q  Bp, r, We want to find two special points x  Bx, r, and y  By, r such that q  x  1  y.

Since the three nballs Bx, r, By, r, and Bp, r have the same radius. By parallelogram principle, we let x  q  x  p, and y  q  y  p, then

x  x  q  p  r, and y  y  q  p  r.

It implies that x  Bx, r, and y  By, r. In addition,

x  1  y

 q  x  p  1  q  y  p

 q.

Since x, y  S, and S is convex, then q  x  1  y  S. It implies that Bp, r  S since q is arbitrary. So, we have proved the claim. That is, for 0    1,

x  1  y  p  intS if x, y  intS, and S is convex. Hence, by the definition of convex, we know that the interior of a convex is convex.

(d) The closure of a convex is convex.

Proof: Given a convex set S, and let x, y  S. Consider x  1  y : p, where 0    1, and claim that p  S, i.e., we want to show that Bp, r  S  .

Suppose NOT, there exists r  0 such that

Bp, r  S  . *

Since x, y  S, then Bx, 2r  S   and By, r2  S  . And let x  Bx, r2  S and y  By, r2  S. Consider

x  1  y  p  x  1  y  x  1  y

 x  x  1  y  1  y

x  x  x  x 

1  y  1  y  1  y  1  y

 x  x  1  y  y  |  |x  y

 r2  |  |x  y

 r

if we choose a suitable number, where 0    1.

Hence, we have the pointx  1  y  Bp, r. Note that x, y  S and S is convex, we havex  1  y  S. It leads us to get a contradiction by (*). Hence, we have proved the claim. That is, for 0    1, x  1  y  p  S if x, y  S. Hence, by the

definition of convex, we know that the closure of a convex is convex.

3.15 Let F be a collection of sets in Rn, and let S  AF A and T  AF A. For each of the following statements, either give a proof or exhibit a counterexample.

(a) If x is an accumulation point of T, then x is an accumulation point of each set A in F.

Proof: Let x be an accumulation point of T, thenBx, r  x  T   for any r  0. Note that for any A  F, we have T  A. Hence Bx, r  x  A   for any r  0. That is, x is an accumulation point of A for any A  F.

The conclusion is that If x is an accumulation point of T  AF A, then x is an accumulation point of each set A in F.

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(b) If x is an accumulation point of S, then x is an accumulation point of at least one set A in F.

Proof: No! For example, Let S  Rn, and F be the collection of sets consisting of a single point x Rn. Then it is trivially seen that S  AF A. And if x is an accumulation point of S, then x is not an accumulation point of each set A in F.

## 3.16

Prove that the set S of rational numbers in the inerval0, 1 cannot be expressed as the intersection of a countable collection of open sets. Hint: Write S  x1, x2, . . ., assume that S  k1k Sk, where each Sk is open, and construct a sequenceQn of closed intervals such that Qn1  Qn  Sn and such that xn  Qn. Then use the Cantor intersection theorem to obtain a contradiction.

Proof: We prove the statement by method of contradiction. Write S  x1, x2, . . ., and assume that S  k1kSk, where each Sk is open.

Since x1  S1, there exists a bounded and open interval I1  S1 such that x1  I1. Choose a closed interval Q1  I1 such that x1  Q1. Since Q1is an interval, it contains infinite rationals, call one of these, x2. Since x2  S2, there exists an open interval I2  S2

and I2  Q1. Choose a closed interval Q2  I2such that x2  Q2. Suppose Qn has been constructed so that

1. Qn is a closed interval.

2. Qn  Qn1  Sn1. 3. xn  Qn.

Since Qn is an interval, it contains infinite rationals, call one of these, xn1. Since xn1  Sn1, there exists an open interval In1  Sn1 and In1  Qn. Choose a closed interval Qn1  In1 such that xn1  Qn1. So, Qn1satisfies our induction hypothesis, and the construction can process.

Note that

1. For all n, Qn is not empty.

2. For all n, Qn is bounded since I1 is bounded.

3. Qn1  Qn. 4. xn  Qn.

Thenn1nQn   by Cantor Intersection Theorem.

Since Qn  Sn, n1n Qn  n1nSn  S. So, we have S n1nQn  n1nQn  

which is absurb since S n1nQn   by the fact xn  Qn. Hence, we have proved that our assumption does not hold. That is, S the set of rational numbers in the inerval0, 1

cannot be expressed as the intersection of a countable collection of open sets.

Remark: 1. Often, the property is described by saying Q is not an G set.

2. It should be noted that Qc is an G set.

3. For the famous Theorem called Cantor Intersection Theorem, the reader should see another classical text book, Principles of Mathematical Analysis written by Walter Rudin, Theorem 3.10 in page 53.

4. For the method of proof, the reader should see another classical text book, Principles of Mathematical Analysis written by Walter Rudin, Theorem 2.43, in page 41.

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### Covering theorems in R

n

3.17 If S  Rn, prove that the collection of isolated points of S is countable.

Proof: Denote the collection of isolated points of S by F. Let x  F, there exists an nball Bx, rx  x  S  . Write Qn  x1, x2, . . ., then there are many numbers in Qn lying on Bx, rx  x. We choose the smallest index, say m  mx, and denote x by xm.

So, F  xm : m  P, where P N, a subset of positive integers. Hence, F is countable.

3.18 Prove that the set of open disks in the xyplane with center x, x and radius x  0, x rational, is a countable covering of the set x, y : x  0, y  0.

Proof: Denote the set of open disks in the xyplane with center x, x and radius x  0 by S. Choose any pointa, b, where a  0, and b  0. We want to find an 2 ball

Bx, x, x  S which contains a, b. It suffices to find x  Q such that

x, x  a, b  x. Since

x, x  a, b  x  x, x  a, b2  x2  x2 2a  bx  a2  b2  0.

Since x2  2a  bx  a2  b2  x  a  b2  2ab, we can choose a suitable rational number x such that x2 2a  bx  a2  b2  0 since a  0, and b  0. Hence, for any pointa, b, where a  0, and b  0, we can find an 2 ball Bx, x, x  S which containsa, b.

That is, S is a countable covering of the setx, y : x  0, y  0.

Remark: The reader should give a geometric appearance or draw a graph.

3.19 The collection Fof open intervals of the form1/n, 2/n, where n  2, 3, . . . , is an open covering of the open interval0, 1. Prove (without using Theorem 3.31) that no finite subcollection of F covers0, 1.

Proof: Write F as12, 1, 13, 23 , . . . , 1n, 2n, . . . . Obviously, F is an open covering of0, 1. Assume that there exists a finite subcollection of F covers 0, 1, and thus write them as F  n11, m11 , . . . . , n1k, m1k . Choose p  0, 1 so that p  min1ikn1i. Then p  n1i, m1i , where 1  i  k. It contracdicts the fact F covers0, 1.

Remark: The reader should be noted that if we use Theorem 3.31, then we cannot get the correct proof. In other words, the author T. M. Apostol mistakes the statement.

3.20 Give an example of a set S which is closed but not bounded and exhibit a coubtable open covering F such that no finite subset of F covers S.

Solution: Let S  R1, then R1 is closed but not bounded. And let

F  n, n  2 : n  Z, then F is a countable open covering of S. In additon, it is trivially seen that no finite subset of F covers S.

3.21 Given a set S in Rn with the property that for every x in S there is an nball Bx

such that Bx  S is coubtable. Prove that S is countable.

Proof: Note that F  Bx : x  S forms an open covering of S. Since S  Rn, then there exists a countable subcover F F of S by Lindelof Covering Theorem. Write F  Bxn : n  N. Since

S  S  nN Bxn  nN S  Bxn, and

S Bxn is countable by hypothesis.

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Then S is countable.

Remark: The reader should be noted that exercise 3.21 is equivalent to exercise 3.23.

3.22 Prove that a collection of disjoint open sets in Rn is necessarily countable. Give an example of a collection of disjoint closed sets which is not countable.

Proof: Let F be a collection of disjoint open sets in Rn, and write Qn  x1, x2, . . ..

Choose an open set S  in F, then there exists an n ball By, r  S. In this ball, there are infinite numbers in Qn. We choose the smallest index, say m  my. Then we have F  Sm : m  P  N which is countable.

For the example that a collection of disjoint closed sets which is not countable, we give it as follows. Let G  x : x  Rn, then we complete it.

3.23 Assume that S  Rn. A point x in Rn is said to be condensation point of S if every nball Bx has the property that Bx  S is not countable. Prove that if S is not

countable, then there exists a point x in S such that x is a condensation point of S.

Proof: It is equivalent to exercise 3.21.

Remark: Compare with two definitions on a condensation point and an accumulation point, it is easy to know that a condensation point is an accumulation point. However, am accumulation point is not a condensation point, for example, S  1/n : n  N. We have 0 is an accumulation point of S, but not a condensation point of S.

## 3.24

Assume that S  Rn and assume that S is not countable. Let T denote the set of condensation points of S. Prove that

(a) S T is countable.

Proof: If S T is uncountable, then there exists a point x in S  T such that x is a condensation point of S T by exercise 3.23. Obviously, x S is also a condensation point of S. It implies x  T. So, we have x  S  T which is absurb since x  S  T.

Remark: The reader should regard T as a special part of S, and the advantage of T helps us realize the uncountable set S Rn. Compare with Cantor-Bendixon Theorem in exercise 3.25.

(b) S T is not countable.

Proof: Suppose S T is countable, then S  S  T  S  T is countable by (a) which is absurb. So, S T is not countable.

(c) T is a closed set.

Proof: Let x be an adherent point of T, then Bx, r  T   for any r  0. We want to show x  T. That is to show x is a condensation point of S. Claim that Bx, r  S is uncountable for any r  0.

Suppose NOT, then there exists an nball Bx, d  S which is countable. Since x is an adherent point of T, then Bx, d  T  . Choose y  Bx, d  T so that By,   Bx, d

and By,   S is uncountable. However, we get a contradiction since By,   S is uncountable  Bx, d  S is countable .

Hence, Bx, r  S is uncountable for any r  0. That is, x  T. Since T contains its all adherent points, T is closed.

(d) T contains no isolated points.

Proof: Let x  T, and if x is an isolated point of T, then there exists an n ball Bx, d

such that Bx, d  T  x. On the other hand, x  T means that Bx, d  x  S is

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uncountable. Hence, by exercise 3.23, we know that there exists y  Bx, d  x  S such that y is a condensation point ofBx, d  x  S. So, y is a condensation point of S. It implies y  T. It is impossible since

1. y x  T.

2. y  Bx, d.

3. Bx, d  T  x.

Hence, x is not an isolated point of T, if x  T. That is, T contains no isolatd points.

Remark: Use exercise 3.25, by (c) and (d) we know that T is perfect.

Note that Exercise 3.23 is a special case of (b).

## 3.25

A set in Rn is called perfect if S  S, that is, if S is a closed set which contains no isolated points. Prove that every uncountable closed set F in Rn can be expressed in the form F  A  B, where A is perfect and B is countable (Cantor-Bendixon theorem).

Hint. Use Exercise 3.24.

Proof: Let F be a uncountable closed set in Rn. Then by exercise 3.24,

F  F  T  F  T, where T is the set of condensation points of F. Note that since F is closed, T  F by the fact, a condensation point is an accumulation point. Define

F T  A and F  T  B, then B is countable and A T is perfect.

Remark: 1. The reader should see another classical text book, Principles of Mathematical Analysis written by Walter Rudin, Theorem 2.43, in page 41. Since the theorem is famous, we list it below.

Theorem 2.43 Let P be a nonempty perfect set in Rk. Then P is uncountable.

Theorem Modefied 2.43 Let P be a nonempty perfect set in a complete separable metric space. Then P is uncountable.

2. Let S has measure zero in R1. Prove that there is a nonempty perfect set P in R1 such that P S  .

Proof: Since S has measure zero, there exists a collection of open intervalsIk such that

S  Ik and

## 

|Ik|  1.

Consider its complementIkc which is closed with positive measure. Since the complement has a positive measure, we know that it is uncountable. Hence, by Cantor-Bendixon Theorem, we know that

Ikc  A  B, where A is perfect and B is countable.

So, let A  P, we have prove it.

Note: From the similar method, we can show that given any set S in R1 with measure 0  d  , there is a non-empty perfect set P such that P  S  . In particular, S  Q, S the set of algebraic numbers, and so on. In addition, even for cases in Rk, it still holds.

### Metric Spaces

3.26 In any metric spaceM, d prove that the empty set  and the whole set M are both open and closed.

proof: In order to show the statement, it suffices to show that M is open and closed since M M  . Let x  M, then for any r  0, BMx, r  M. That is, x is an interior

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point of M. Sinc x is arbitrary, we know that every point of M is interior. So, M is open.

Let x be an adherent point of M, it is clearly x  M since we consider all points lie in M. Hence, M contains its all adherent points. It implies that M is closed.

Remark: The reader should regard the statement as a common sense.

3.27 Consider the following two metrics in Rn : d1x, y  max1in|xi  yi|, d2x, y 

### 

i1in|xi  yi|.

In each of the following metric spaces prove that the ball Ba; r has the geometric appearance indicated:

(a) InR2, d1, a square with sides parallel to the coordinate axes.

Solution: It suffices to consider the case B0, 0, 1. Let x  x1, x2  B0, 0, 1, then we have

|x1|  1, and |x2|  1.

So, it means that the ball B0, 0, 1 is a square with sides lying on the coordinate axes.

Hence, we know that Ba; r is a square with sides parallel to the coordinate axes.

(b) InR2, d2, a square with diagonals parallel to the axes.

Solution: It suffices to consider the case B0, 0, 1. Let x  x1, x2  B0, 0, 1, then we have

|x1  x2|  1.

So, it means that the ball B0, 0, 1 is a square with diagonals lying on the coordinate axes. Hence, we know that Ba; r is a square with diagonals parallel to the coordinate axes.

(c) A cube inR3, d1.

Solution:It suffices to consider the case B0, 0, 0, 1. Let x  x1, x2, x3  B0, 0, 0, 1, then we have

|x1|  1, |x2|  1, and |x3|  1.

So, it means that the ball B0, 0, 0, 1 is a cube with length 2. Hence, we know that Ba; r is a cube with length 2a.

(d) An octahedron inR3, d2.

Solution: It suffices to consider the case B0, 0, 0, 1. Let x  x1, x2, x3  B0, 0, 0, 1, then we have

|x1  x2 x3|  1.

It means that the ball B0, 0, 0, 1 is an octahedron. Hence, Ba; r is an octahedron.

Remark: The exercise tells us one thing that Ba; r may not be an n ball if we consider some different matrices.

3.28 Let d1 and d2 be the metrics of Exercise 3.27 and letx  y denote the usual Euclidean metric. Prove that the following inequalities for all x and y in Rn :

d1x, y  x  y  d2x, y and d2x, y  n x  y  nd1x, y.

Proof: List the definitions of the three metrics, and compare with them as follows.

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1. d1x, y  max1in|xi  yi|.

2.x  y 

### 

i1inxi  yi2 1/2. 3. d2x, y 

### 

i1in|xi  yi|.

Then we have (a)

d1x, y  max1in|xi  yi|  max

1in|xi  yi|2 1/2

i1 in

xi  yi2

1/2

 x  y.

(b)

x  y 

i1 in

xi  yi2

1/2

i1 in

|xi  yi|

2 1/2

## 

i1 in

|xi  yi|  d2x, y.

(c)

nx  y  n

i1 in

xi  yi2

1/2

n

i1 in

xi  yi2

1/2

 n n max

1in|xi  yi| 2

1/2

 n max

1in|xi  yi|

 d1x, y.

(d)

d2x, y2

i1 in

|xi  yi|

2

i1 in

xi  yi2

## 

1ijn

2|xi  yi||xj  yj|

## 

i1 in

xi  yi2 n  1

## 

i1 in

xi  yi2 by A. P. G. P.

 n

## 

i1 in

xi  yi2

 nx  y2. So,

d2x, y  n x  y.

From (a)-(d), we have proved these inequalities.

Remark: 1. Let M be a given set and suppose thatM, d and M, d are metric spaces.

We define the metrics d and d are equivalent if, and only if, there exist positive constants

,  such that

dx, y  dx, y  dx, y.

The concept is much important for us to consider the same set with different metrics. For

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example, in this exercise, Since three metrics are equivalent, it is easy to know that

Rk, d1, Rk, d2, and Rk,.  are complete. (For definition of complete metric space, the reader can see this text book, page 74.)

2. It should be noted that on a finite dimensional vector space X, any two norms are equivalent.

3.29 IfM, d is a metric space, define dx, y  1dx,ydx,y . Prove that d is also a metric for M. Note that 0  dx, y  1 for all x, y in M.

Proof: In order to show that d is a metric for M, we consider the following four steps.

(1) For x  M, dx, x  0 since dx, x  0.

(2) For x  y, dx, y  1dx,ydx,y  0 since dx, y  0.

(3) For x, y  M, dx, y  1dx,ydx,y1dy,xdy,x  dy, x

(4) For x, y, z  M,

dx, y  dx, y

1 dx, y  1  1 1 dx, y

 1  1

1 dx, z  dz, y since dx, y  dx, z  dz, y

dx, z  dz, y

1 dx, z  dz, y

dx, z

1 dx, z  dz, ydz, y

1 dx, z  dz, y

dx, z

1 dx, zdz, y

1 dz, y

 dx, z  dz, y

Hence, from (1)-(4), we know that d is also a metric for M. Obviously, 0  dx, y  1 for all x, y in M.

Remark: 1. The exercise tells us how to form a new metric from an old metric. Also, the reader should compare with exercise 3.37. This is another construction.

2. Recall Discrete metric d, we find that given any set nonempty S, S, d is a metric space, and thus use the exercise, we get another metric spaceS, d, and so on. Hence, here is a common sense that given any nonempty set, we can use discrete metric to form many and many metric spaces.

3.30 Prove that every finite subset of a metric space is closed.

Proof: Let x be an adherent point of a finite subet S  xi : i  1, 2, . . . , n of a metric spaceM, d. Then for any r  0, Bx, r  S  . If x  S, then BMx,   S   where

  min1ijndxi, xj. It is impossible. Hence, x  S. That is, S contains its all adherent points. So, S is closed.

3.31 In a metric spaceM, d the closed ball of radius r  0 about a point a in M is the set Ba; r x : dx, a  r.

(a) Prove that Ba; r is a closed set.

Proof: Let x  M  Ba; r, then dx, a  r. Consider Bx, , where   dx,ar2 , then if y  Bx, , we have dy, a  dx, a  dx, y  dx, a    dx,ar2  r. Hence, Bx,   M  Ba; r. That is, every point of M  Ba; r is interior. So, M  Ba; r is open, or equivalently, Ba; r is a closed set.

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