## Charpter 3 Elements of Point set Topology

*Open and closed sets in R*

^{1}

*and R*

^{2}

*3.1 Prove that an open interval in R*^{1}*is an open set and that a closed interval is a*
*closed set.*

**proof: 1. Let***a, b be an open interval in R*^{1}*, and let x* * a, b. Consider*

min*x a, b x : L. Then we have Bx, 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 R*^{1}.

2. Let*a, b be a closed interval in R*^{1}*, 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*

*Bx, a x*2 *a, b 3x a*2 *, x a*2 *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 R*^{1}.

*3.2 Determine all the accumulation points of the following sets in R*^{1} *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*

*Bx,* ^{x1}_{2} * x S . So, Z has no accumulation points.*

*However, Bx,* ^{x1}_{2} * S x . So Z contains its all adherent points. It means that*
*Z is closed. Trivially, Z is not open since Bx, r is not contained in Z for all r 0.*

**Remark: 1. Definition of an adherent point: Let S be a subset of R**^{n}*, and x a point in*
*R*^{n}*, x is not necessarily in S. Then x is said to be adherent to S if every nball Bx*

*contains at least one point of S. To be roughly, Bx S .*

*2. Definition of an accumulation point: Let S be a subset of R*^{n}*, and x a point in R** ^{n}*,

*then x is called an accumulation point of S if every nball Bx contains at least one point*

*of S distinct from x. To be roughly,Bx x S . That is, x is an accumulation*

*point if, and only if, x adheres to S x. Note that in this sense,*

*Bx x S Bx 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 **nZ* *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 intervala, 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 Bx, r x a, b for any r 0. So, every*
point of*a, b is an accumulation point.*

*2. R*^{1}* 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*

*exists an nball Bx, r**x** such that Bx, r**x** x a, b . For the point a, it is easy*
to know that*Ba, r a a, b . That is, in this case, there is only one*

*accumulation point a ofa, 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 nball Bb, r is not contained in a, b for any r 0, we know that the point*
*b is not interior toa, 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 B1, 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 R*^{1}.

*So, Q is not closed. Consider x* * Q, any n ball Bx 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, . .

_{m1}* 12 5*

^{m}*, 14 5*

^{m}*, . . . , 12*

^{m}* 5*

^{n}*, . . .*

^{m}* : S*

12 1 5, 1

2 1

5^{2}, . . . , 1
2 1

5* ^{m}*, . . .

14 1 5, 1

4 1

5^{2}, . . . , 1
4 1

5* ^{m}*, . . .

. . . .

12* ^{n}* 15 , 1
2

* 1*

^{n}5^{2} , . . . , 12* ^{n}* 15

*. . . . . . .*

^{m}1 2

3

*So, we find that S*^{} _{2}^{1}*n* *: n* 1, 2, . . . _{5}^{1}*m* *: m* * 1, 2, . . . 0. So, S is not*
closed since it does not contain 0. Since ^{1}_{2} * S, and B*^{1}_{2}*, 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}

*m1*
*m*, 1

4 5^{m}

*m1*

*m*and 1

2* ^{n}* 5

^{m}*m1*

*m*, respectively.

And it means that for (1), the sequence5* ^{m}*

_{m1}*moves*

^{m}^{1}

_{2}. Similarly for others. So, it is easy to see why

^{1}

_{2}is an accumulation point of

^{1}

_{2} 5

**

^{m}

_{m1}*. And thus get the set of all accumulation points of2*

^{m}* 5*

^{n}

^{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 1*m* _{m1}

*m* 1 1*m* _{m1}

*m* : S.

*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 B2, r is not contained in S for any r 0, S is not open.*

*(g) All numbers of the form1/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*_{m1}^{m}* 1/2 1/m*_{m1}^{m}*. . . 1/n 1/m*_{m1}^{m}*. . . : 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 B1, 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 _{2k}^{1} _{k1}

*k*

1

1 _{2k1}^{1} _{k1}

*k*

: S.

*We find that S*^{} * 1, 1. So, S is not closed since S*^{} * S. Since* ^{1}_{2} * S, and B*^{1}_{2} *, 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 R*^{2}.
*(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 Bz,* ^{|z|1}_{2} * 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 B1, r is not contained in S for any r 0, S*
is not open.

*(c) All complex numbers of the form1/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* _{m1}* ^{m}* 1
2

*i*

*m* _{m1}

*m**. . . 1n im* _{m1}^{m}*. . . : 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 B1 i, r is not contained in S for any r 0, S is not*
open.

*(d) All pointsx, y such that x*^{2} * y*^{2} 1.

**Solution: Denote***x, y : x*^{2} * y*^{2} * 1 by S. We know that*

*S*^{} * x, y : x*^{2}* y*^{2} * 1. So, S is not closed since S*^{} * S. Let p x, y S, then*
*x*^{2} * y*^{2} * 1. It is easy to find that r 0 such that Bp, r S. So, S is open.*

*(e) All pointsx, 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 r**x* * 0 such that Bx, r**x** S. So, S is open.*

*(f) All pointsx, 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 closed since S*^{} * S. Since 0, 0 S, and*
*B0, 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 R*^{1}*contains both rational and irratonal*
*numbers.*

**proof: Given a nonempty open set S in R**^{1}*. Let x* * S, then there exists r 0 such that*
*Bx, r S since S is open. And in R*^{1}*, the open ball Bx, 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 R*

^{1}

*which are both open and closed are the empty set*

*and R*

^{1}

*itself. Is a similar statement true for R*

^{2}?

**Proof: Let S be the set in R**^{1}*, and thus consider its complement T* * R*^{1} * S. Then we*
*have both S and T are open and closed. Suppose that S* * R*^{1}*and S* * , we will show that*
**it is impossible as follows.**

*Since S* * R*^{1}*, and S* * , then T and T R*^{1}*. Choose s*_{0} * S and t*0 * T, then we*
**consider the new point** ^{s}^{0}^{t}_{2}^{0} *which is in S or T since R* * S T. If* ^{s}^{0}^{t}_{2}^{0} * S, we say*

*s*0*t*0

2 * s*1,otherwise, we say ^{s}^{0}^{t}_{2}^{0} * t*1.

**Continue these steps, we finally have two sequences named***s**n** S and t**m** 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 R*^{1} or
*S* * .*

**Remark: 1. In the proof, the statement is true for R*** ^{n}*.

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

*3. 6 Prove that every closed set in R*^{1} *is the intersection of a countable collection of*
*open sets.*

**proof: Given a closed set S, and consider its complement R**^{1} * S which is open. If*
*R*^{1} * S , there is nothing to prove. So, we can assume that R*^{1} * S .*

*Let x* * R*^{1} * S, then x is an interior point of R*^{1} * S. So, there exists an open interval*

*a, b such that x a, b R*^{1} * S. In order to show our statement, we choose a smaller*
interval*a**x**, b**x** so that x a**x**, b**x** and a**x**, b**x** a, b R*^{1} * S. Hence, we have*

*R*^{1} * S *_{xR}^{1}_{S}*a**x**, b**x*
which implies that

*S* * R*^{1} _{xR}^{1}_{S}*a**x**, b**x*

_{xR}^{1}_{S}*R*^{1} * a**x**, b**x*

_{n1}^{n}*R*^{1} * a**n**, b*_{n}** (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 R*^{1} is the intersection of a countable
**collection of closed sets.**

*3. The proof is suitable for R*^{n}*if the statement is that every closed set in R** ^{n}*is the
intersection of a countable collection of open sets. All we need is to change intervals into
disks.

## 3.7

*Prove that a nonempty, bounded closed set S in R*

^{1}

*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, R*^{1} * S*

* inf S, sup S S. Denote inf S, sup S by I. Consider R*^{1} * S is open, then by*
**Representation Theorem for Open Sets on The Real Line, we have**

*R*^{1} * S *_{m1}^{m}*I*_{m}

* I S*
which implies that

*S* * I *_{m1}^{m}*I**m*.

*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 nballs and n dimensional open intervals are open sets in R** ^{n}*.

**proof: Given an open n**ball Bx, r. Choose y Bx, r and thus consider*By, d Bx, r, where d min|x y|, r |x y|. Then y is an interior point of Bx, r.*

*Since y is arbitrary, we have all points of Bx, r are interior. So, the open n ball Bx, r is*
open.

*Given an ndimensional open interval a*1*, b*_{1}* a*2*, b*_{2}* . . . a**n**, b**n** : I. Choose*
*x* * x*1,*x*_{2}*, . . . , x**n** I and thus consider r min*_{i1}^{in}*r**i**, where r**i* * minx**i* * a**i**, b**i* * x**i*.

*Then Bx, 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 ndimensional open interval I is open.*

*3.9 Prove that the interior of a set in R*^{n}*is open in R** ^{n}*.

**Proof: Let x*** intS, then there exists r 0 such that Bx, r S. Choose any point of*
*Bx, r, say y. Then y is an interior point of Bx, r since Bx, r is open. So, there exists*
*d* * 0 such that By, d Bx, r S. So y is also an interior point of S. Since y is*

*arbitrary, we find that every point of Bx, r is interior to S. That is, Bx, 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. intintS intS.*

*3. If S* * T, then intS intT.*

## 3.10

^{If S}* R*

^{n}*, prove that intS is the union of all open subsets of R*

^{n}*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*

*AS*

*A, 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 Bx, r S. So,*
*x* * Bx, r **AS* *A. That is, intS* *AS**A.*

2.* Let x **AS* *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 Bx, r A S. So x is an interior point*
*of S, i.e., x* * intS. That is, **AS**A* * intS.*

*From 1 and 2, we know that intS* _{AS}*A, where A is open.*

*Let T be an open subset of S such that intS* * T. Since intS **AS* *A, where A is open,*

*we have intS* * T **AS**A which implies intS* * T by intS **AS* *A. Hence, intS is the*
*largest open subset of S.*

*3.11 If S and T are subsets of R*^{n}*, prove that*

*intS intT intS T and intS intT intS T.*

**Proof: For the part***intS intT intS 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 intintS intT intS T.*

2.* Since S T S and S T T, we have intS T intS and*
*intS T intT. So,*

*intS T intS intT.*

From 1 and 2, we know that*intS intT intS T.*

For the part*intS intT intS T, we consider intS S and intT T. So,*

*intS intT S T*
which implies that Note that*intS intT is open.*

*intintS intT intS intT intS T.*

**Remark: It is not necessary that***intS intT intS T. For example, let S Q,*
*and T* * Q*^{c}*, then intS* * , and intT . However, intS T intR*^{1} * R.*

*3.12 Let S*^{} *denote the derived set and S the closure of a set S in R*^{n}*. Prove that*
*(a) S*^{} *is closed in R*^{n}*; that isS*^{}^{} * 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

*Bx, d x S .* *

*Since x adheres to S*^{}*, then Bx, d S*^{} * . So, there exists y Bx, d such that y is an*
*accumulation point of S. Note that x* * y, by assumption. Choose a smaller radius d so that*

*By, d Bx, d x and By, d S .*

It implies

* By, d S Bx, 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*Bx, r x S for any r 0. It implies that*

*Bx, 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*Bx, r x S T . That is,*

*Bx, r x S Bx, r x T .*

So, at least one of*Bx, r x S and Bx, r x T is not empty. If*

*Bx, r x S , then x S*^{}. And if*Bx, 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 clS T clS clT,*
*where clS 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 R** ^{n}*.

**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 R*^{n}*containing S. That is, S is the*
*smallest closed set containing S.*

* Proof: It suffices to show that S*

*AS*

*A, where A is closed. To show the statement,*we consider two steps as follows.

1.* Since S is closed and S S, then **AS**A* * S.*

2.* Let x S, then Bx, r S for any r 0. So, if A S, then*

*Bx, 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 **AS* *A. So, S* *AS**A.*

*From 1 and 2, we have S* *AS**A.*

*Let S* * T S, where T is closed. Then S **AS* *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 R*^{n}*. Prove that clS T clS clT and that*
*S clT clS T if S is open, where clS is the closure of S.*

**Proof: Since S** T S and S T T, then clS T clS and,*clS T clT. So, clS T clS clT.*

*Given an open set S , and let x* * S clT, then we have*

*1. x* * S and S is open.*

* Bx, d S for some d 0.*

*Bx, r S Bx, r if r d.*

*Bx, r S Bx, d if r d.*

and

*2. x* * clT*

* Bx, r T for any r 0.*

From 1 and 2, we know

*Bx, r S T Bx, r S T Bx, r T if r d.*

*Bx, r S T Bx, r S T Bx, d T if r d.*

*So, it means that x is an adherent point of S T. That is, x clS T. Hence,*
*S clT clS T.*

**Remark: It is not necessary that clS T clS clT. For example, S Q and***T* * Q*^{c}*, then clS T and clS clT R*^{1}.

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

## 3.14

*A set S in R*

^{n}

**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 R*^{2} *and R*^{3}* and prove that*
*(a) Every nball in R*^{n}*is convex.*

* Proof: Given an nball Bp, r, and let x, y Bp, 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 Bp, r for 0 1. Hence, by the definition of convex,*
*we know that every nball in R*^{n}*is convex.*

*(b) Every ndimensional open interval is convex.*

* Proof: Given an ndimensional open interval I a*1

*, b*

_{1}

* . . . a*

*n*

*, b*

_{n}*. Let x, y I,*

*and thus write x*

* x*1

*, x*

_{2}

*, . . . , x*

*n*

* and y y*1

*, y*

_{2}

*, . . . y*

*n*. Consider

*x 1 y x*1 * 1 y*1,*x*2 * 1 y*2, . . . ,*x**n* * 1 y**n** where 0 1.*

Then

*a*_{i}* x**i* * 1 y**i* * b**i**, 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*
*Bx, r S, and By, r S. Consider x 1 y : p S, where 0 1, since S*
is convex.

*Claim that Bp, r S as follows.*

*Let q* * Bp, r, We want to find two special points x Bx, r, and y By, r such*
*that q* * x 1 y.*

*Since the three nballs Bx, r, By, r, and Bp, 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 Bx, r, and y By, 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 Bp, 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 Bp, r S .*

* Suppose NOT, there exists r* 0 such that

*Bp, r S .* *

*Since x, y* * S, then Bx,* _{2}^{r}* S and By,* ^{r}_{2}* S . And let x Bx,* ^{r}_{2}* S and*
*y By,* ^{r}_{2}* 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 Bp, 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 R*^{n}*, and let S* _{AF}*A and T* _{AF}*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**Bx, r x T for any*r* * 0. Note that for any A F, we have T A. Hence Bx, 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*

*AF*

*A, then x is an*

*accumulation point of each set A in F.*

*(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*** R*^{n}*, and F be the collection of sets consisting of a*
*single point x R*^{n}*. Then it is trivially seen that S **AF* *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 inerval0, 1 cannot be*

*expressed as the intersection of a countable collection of open sets. Hint: Write*

*S*

* x*1

*, x*

_{2}, . . .

*, assume that S *

_{k1}

^{k}*S*

_{k}*, where each S*

_{k}*is open, and construct a*

*sequenceQ*

*n*

* of closed intervals such that Q*

*n1*

* Q*

*n*

* S*

*n*

*and such that x*

_{n}* Q*

*n*.

*Then use the Cantor intersection theorem to obtain a contradiction.*

**Proof: We prove the statement by method of contradiction. Write S*** x*1*, x*_{2}, . . ., and
*assume that S* _{k1}^{k}*S*_{k}*, where each S** _{k}* is open.

*Since x*_{1} * S*1*, there exists a bounded and open interval I*_{1} * S*1 *such that x*_{1} * I*1.
*Choose a closed interval Q*_{1} * I*1 *such that x*_{1} * Q*1*. Since Q*_{1}is an interval, it contains
*infinite rationals, call one of these, x*_{2}*. Since x*_{2} * S*2*, there exists an open interval I*_{2} * S*2

*and I*_{2} * Q*1*. Choose a closed interval Q*_{2} * I*2*such that x*_{2} * Q*2*. Suppose Q**n* has been
constructed so that

*1. Q**n* is a closed interval.

*2. Q**n* * Q**n1* * S**n1*.
*3. x**n* * Q**n*.

*Since Q**n* *is an interval, it contains infinite rationals, call one of these, x** _{n1}*. Since

*x*

_{n1}* S*

*n1*

*, there exists an open interval I*

_{n1}* S*

*n1*

*and I*

_{n1}* Q*

*n*. Choose a closed

*interval Q*

_{n1}* I*

*n1*

*such that x*

_{n1}* Q*

*n1*

*. So, Q*

*satisfies our induction hypothesis, and the construction can process.*

_{n1}Note that

*1. For all n, Q**n* is not empty.

*2. For all n, Q**n* *is bounded since I*_{1} is bounded.

*3. Q*_{n1}* Q**n*.
*4. x**n* * Q**n*.

Then_{n1}^{n}*Q**n* ** by Cantor Intersection Theorem.**

*Since Q**n* * S**n*, _{n1}^{n}*Q**n* _{n1}^{n}*S**n* * S. So, we have*
*S* _{n1}^{n}*Q**n* _{n1}^{n}*Q**n* * *

*which is absurb since S* _{n1}^{n}*Q**n** by the fact x**n* * Q**n*. 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 Q*^{c}*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.

*Covering theorems in R*

^{n}*3.17 If S* * R*^{n}*, 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*
*nball Bx, r**x** x S . Write Q*^{n}* x*1*, x*_{2}, . . ., then there are many numbers in
*Q*^{n}*lying on Bx, r**x** x. We choose the smallest index, say m mx, and denote x by*
*x** _{m}*.

*So, F* * x**m* *: 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 xyplane 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 pointa, b, where a 0, and b 0. We want to find an 2 ball*

*Bx, 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} * x*^{2} * x*^{2}* 2a bx a*^{2} * b*^{2} 0.

*Since x*^{2} * 2a bx a*^{2} * b*^{2}* x a b*^{2} * 2ab, we can choose a suitable rational*
*number x such that x*^{2}* 2a bx a*^{2} * b*^{2}* 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 Bx, x, x S which*
contains*a, b.*

*That is, S is a countable covering of the setx, 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 form1/n, 2/n, where n 2, 3, . . . , is an*
*open covering of the open interval0, 1. Prove (without using Theorem 3.31) that no*
*finite subcollection of F covers*0, 1.

* Proof: Write F as*

^{1}

_{2}, 1,

^{1}

_{3},

^{2}

_{3}, . . . ,

^{1}

*n*,

^{2}

_{n}*, . . . . 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*

^{}

_{n}^{1}

_{1},

_{m}^{1}

_{1}, . . . . ,

*n*

^{1}

*k*,

_{m}^{1}

_{k}*. Choose p 0, 1 so that p min*1ik

*n*

^{1}

*i*. Then

*p*

_{n}^{1}

*,*

_{i}

_{m}^{1}

_{i}*, where 1 i k. It contracdicts the fact F*

^{}covers0, 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*** R*^{1}*, then R*^{1} 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 R*^{n}*with the property that for every x in S there is an nball Bx*

*such that Bx S is coubtable. Prove that S is countable.*

**Proof: Note that F*** Bx : x S forms an open covering of S. Since S R** ^{n}*, then

*there exists a countable subcover F*

^{}

** F of S by Lindelof Covering Theorem. Write***F*

^{}

* Bx*

*n*

* : n N. Since*

*S* * S **nN* *Bx**n* *nN* *S Bx**n*,
and

*S Bx**n* is countable by hypothesis.

*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 R*^{n}*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 R**^{n}*, and write Q*^{n}* x*1*, x*_{2}, . . ..

*Choose an open set S in F, then there exists an n ball By, r S. In this ball, there*
*are infinite numbers in Q*^{n}*. We choose the smallest index, say m* * my. Then we have*
*F* * S**m* *: 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 R** ^{n}*, then we complete it.

*3.23 Assume that S* * R*^{n}*. A point x in R*^{n}**is said to be condensation point of S if every***nball Bx has the property that Bx 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*

* R*

^{n}*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 R*^{n}**. 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 Bx, r S is*
*uncountable for any r* 0.

**Suppose NOT, then there exists an n**ball Bx, d S which is countable. Since x is an*adherent point of T, then Bx, d T . Choose y Bx, d T so that By, Bx, d*

*and By, S is uncountable. However, we get a contradiction since*
*By, S is uncountable Bx, d S is countable .*

*Hence, Bx, 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 Bx, d*

*such that Bx, d T x. On the other hand, x T means that Bx, d x S is*

*uncountable. Hence, by exercise 3.23, we know that there exists y* * Bx, d x S*
*such that y is a condensation point ofBx, 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* * Bx, d.*

*3. Bx, 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 R}

^{n}

**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 R*

^{n}*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 R*** ^{n}*. 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 R**^{k}*. 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 R*^{1}*. Prove that there is a nonempty perfect set P in R*^{1} such
*that P S .*

**Proof: Since S has measure zero, there exists a collection of open intervals**I*k* such
that

*S* * I**k* and

##

^{|I}

^{k}^{|}

^{ 1.}

Consider its complement*I**k** ^{c}* 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

*I**k*^{c}* 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 R**^{1} 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 R** ^{k}*, it still holds.

### Metric Spaces

*3.26 In any metric spaceM, 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, B**M**x, r M. That is, x is an interior*

*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 R** ^{n}* :

*d*

_{1}

*x, y max*1in

*|x*

*i*

* y*

*i*

*|, d*2

*x, y *

###

_{i1}

^{in}

^{|x}*i*

* y*

*i*|.

*In each of the following metric spaces prove that the ball Ba; r has the geometric*
*appearance indicated:*

*(a) InR*^{2}*, d*_{1}*, a square with sides parallel to the coordinate axes.*

* Solution: It suffices to consider the case B0, 0, 1. Let x x*1

*, x*

_{2}

* B0, 0, 1,*then we have

*|x*1| * 1, and |x*2| 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 Ba; r is a square with sides parallel to the coordinate axes.*

*(b) InR*^{2}*, d*_{2}*, a square with diagonals parallel to the axes.*

* Solution: It suffices to consider the case B0, 0, 1. Let x x*1

*, x*

_{2}

* B0, 0, 1,*then we have

*|x*1 * x*2| 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 Ba; r is a square with diagonals parallel to the coordinate*
axes.

*(c) A cube inR*^{3}*, d*_{1}.

* Solution:It suffices to consider the case B*0, 0, 0, 1. Let

*x*

* x*1

*, x*

_{2}

*, x*

_{3}

* B0, 0, 0, 1, then we have*

*|x*1| * 1, |x*2| * 1, and |x*3| 1.

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

*(d) An octahedron inR*^{3}*, d*_{2}.

* Solution: It suffices to consider the case B*0, 0, 0, 1. Let

*x*

* x*1

*, x*

_{2}

*, x*

_{3}

* B0, 0, 0, 1, then we have*

*|x*_{1} * x*2* x*3| 1.

*It means that the ball B0, 0, 0, 1 is an octahedron. Hence, Ba; r is an octahedron.*

* Remark: The exercise tells us one thing that Ba; r may not be an n ball if we*
consider some different matrices.

*3.28 Let d*_{1} *and d*_{2} *be the metrics of Exercise 3.27 and letx y denote the usual*
*Euclidean metric. Prove that the following inequalities for all x and y in R** ^{n}* :

*d*_{1}*x, y x y d*2*x, y and d*2*x, y n x y nd*1*x, y.*

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

*1. d*_{1}*x, y max*1in*|x*_{i}* y**i*|.

2.*x y *

###

_{i1}

^{in}*x*

*i*

* y*

*i*

^{2}

^{1/2}.

*3. d*

_{2}

*x, y *

###

_{i1}

^{in}

^{|x}*i*

* y*

*i*|.

Then we have (a)

*d*_{1}*x, y max*_{1in}*|x**i* * y**i*| max

1in*|x**i* * y**i*|^{2} ^{1/2}

##

*i1*
*in*

*x**i* * y**i*^{2}

1/2

* x y.*

(b)

*x y *

##

*i1*
*in*

*x**i* * y**i*^{2}

1/2

##

*i1*
*in*

*|x**i* * y**i*|

2 1/2

##

*i1*
*in*

*|x**i* * y**i*| * d*2*x, y.*

(c)

*nx y n*

##

*i1*
*in*

*x**i* * y**i*^{2}

1/2

*n*

##

*i1*
*in*

*x**i* * y**i*^{2}

1/2

* n n max*

1in*|x**i* * y**i*| ^{2}

1/2

* n max*

1in*|x**i* * y**i*|

* d*1*x, y.*

(d)

*d*2*x, y*^{2}

##

*i1*
*in*

*|x**i* * y**i*|

2

##

*i1*
*in*

*x**i* * y**i*^{2}

##

1ijn

*2|x**i* * y**i**||x**j* * y**j*|

##

*i1*
*in*

*x**i* * y**i*^{2}* n 1*

##

*i1*
*in*

*x**i* * y**i*^{2} *by A. P. G. P.*

* n*

##

*i1*
*in*

*x**i* * y**i*^{2}

* nx y*^{2}.
So,

*d*_{2}*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*

*dx, y dx, y dx, y.*

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

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

*R*^{k}*, d*_{1}*, R*^{k}*, d*_{2}*, and R** ^{k}*,

**. 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 IfM, d is a metric space, define d*^{}*x, y * _{1dx,y}^{dx,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 dx, x 0.*

*(2) For x* * y, d*^{}*x, y * _{1dx,y}^{dx,y}* 0 since dx, y 0.*

*(3) For x, y* * M, d*^{}*x, y * _{1dx,y}* ^{dx,y}*

_{1dy,x}

^{dy,x}* d*

^{}

*y, x*

*(4) For x, y, z* * M,*

*d*^{}*x, y * *dx, y*

1* dx, y* 1 1
1* dx, y*

1 1

1* dx, z dz, y* *since dx, y dx, z dz, y*

*dx, z dz, y*

1* dx, z dz, y*

*dx, z*

1* dx, z dz, y* *dz, y*

1* dx, z dz, y*

*dx, z*

1* dx, z* *dz, y*

1* dz, 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*** x**i* *: i* * 1, 2, . . . , n of a metric*
space*M, d. Then for any r 0, Bx, r S . If x S, then B**M**x, S where*

* min*_{1ijn}*dx**i**, x** _{j}*.

* It is impossible. Hence, x S. That is, S contains its all adherent*

*points. So, S is closed.*

*3.31 In a metric spaceM, d the closed ball of radius r 0 about a point a in M is the*
*set Ba; r x : dx, a r.*

*(a) Prove that Ba; r is a closed set.*

**Proof: Let x*** M Ba; r, then dx, a r. Consider Bx, , where * ^{dx,ar}_{2} ,
*then if y* * Bx, , we have dy, a dx, a dx, y dx, a * ^{dx,ar}_{2} * r. Hence,*
*Bx, M Ba; r. That is, every point of M Ba; r is interior. So, M Ba; r is*
*open, or equivalently, Ba; r is a closed set.*