MTG 6316 HOMEWORK Spring 2017

101. (Section 26, #12) Let p : X ! Y be a closed continuous surjective map such that p�1(y)

is compact, for each y 2 Y . (Such a map is called a perfect map.) Show that if Y is

compact, then X is compact. [Hint: If U is an open set containing p�1(y), there is a

neighborhood W of y such that p�1(W ) is contained in U .]

Proof. Suppose that Y is compact and there exists a perfect map p : X ! Y .

Let {U↵}↵2I be an open cover of X. Fix an arbitrary point y 2 Y . Since p is

a perfect map, then p�1(y) is a nonempty compact subspace of X. By Lemma

26.1 (on page 164), there is a finite subcover {Uyi }

Ny

i=1 ⇢ {U↵}↵2I of p�1(y). Let

Uy=

SNy

i=1 Uyi . Then Uy

is open in X, and p�1(y) ⇢ Uy

. Thus X \ Uyis closed

in X, and p�1(y) \ (X \ Uy

) = ;, so y /2 p (X \ Uy). Since p is a closed map,

p (X \ Uy) is closed in Y . Consider

V y= Y \ [p (X \ Uy

)] ,

then V yis open in Y and y 2 V y

. Thus, {V y}y2Y forms an open cover for Y , and

since Y is compact, there exists a finite subcover {V yi}Ni=1 of Y . We want to show

that {Uyi}Ni=1 covers X. Consider any subset A ⇢ X. For any b 2 Y \ p(A), we

have p�1(b) \A = ;, so p�1

(b) ⇢ X \A. Then we have

p�1(Y \ p(A)) ⇢ X \A.

Take A = X \ Uy, we have

p�1(V y

) = p�1(Y \ p (A)) ⇢ X \A = Uy,

for any y 2 Y . Thus we have

X = p�1(Y ) ⇢ p�1

N[

i=1

V yi

!=

N[

i=1

p�1(V yi

) ⇢N[

i=1

Uyi .

Furthermore,

X ⇢N[

i=1

Uyi=

N[

i=1

0

@Nyi[

j=1

Uyij

1

A .

Therefore,

SNi=1{U

yij }Nyi

j=1 is a finite subcover of {U↵}↵2I for X.

1

Let X be a metric space and f : X ! X be a homeomorphism. Given N ⇢ X, define the

maximal invariant set in N by

Inv(N) =1\

k=�1

f

k(N) = {x 2 N : fk(x) 2 N, 8k 2 Z}.

Definition: N ⇢ X is a trapping region for f if

1. N is compact and forward invariant, i.e. f(N) ⇢ N , and

2. there exists K > 0 so that fK(N) ⇢ int(N).

(a) Show that if N is a trapping region, then

Inv(N) =1\

k=0

cl

1[

j=k

f

j(N)

!= {x 2 N : 9xn 2 N, kn > 0, kn ! 1, such thatfkn(xn) ! x as n ! 1}.

(b) Show that ifN is a trapping region and nonempty, then Inv(N) is compact and nonempty.

(c) Show that if N is a trapping region, Inv(N) ⇢ int(N).

Proof. (a) Since X is a metric space, let ⇢ denote the metric. Since N is a trapping region

and X is Hausdor↵, N is closed. Let A =T1

k=0 cl⇣S1

j=k fj(N)

⌘and B = {x 2 N : 9xn 2

N, kn > 0, kn ! 1 such thatfkn(xn) ! x as n ! 1}. We will show that

B ⇢ Inv(N) ⇢ A ⇢ B.

Let x 2 B. Then

9xn 2 N, kn > 0, kn ! 1 such thatfkn(xn) ! x as n ! 1.

Let k 2 Z. Since kn ! 1, 9M 2 N so that m � M implies km + k > 0. By forward

invariance of N , fk+km(xm) 2 N . Since f is a homeomorphism, fk is continuous, being the

composition of continuous functions. Then f

k+kn(xn) ! f

k(x) 2 N as N is closed. As k

was arbitrary, this proves B ⇢ Inv(N).

Let us see that Inv(N) ⇢ A. For any k 2 Z

f

k(N) ⇢1[

j=k

f

j(N) ⇢ cl

1[

j=k

f

j(N)

!.

1

Then

Inv(N) =1\

k=�1

f

k(N) ⇢1\

k=0

f

k(N) ⇢1\

k=0

cl

1[

j=k

f

j(N)

!= A.

Now we show A ⇢ B by first showing A ⇢ N . 8k � 0,

f

k(N) ⇢ N )1[

j=k

f

j(N) ⇢ cl

1[

j=k

f

j(N)

!⇢ N,

since N is a trapping region (closed and forward invariant). Thus A ⇢ N . Let x 2 A and

n � 0. Since x 2 cl⇣S1

j=n fj(N)

⌘,

9xn 2 N, 9kn � n, so that ⇢�f

kn(xn), x�<

1

n

, (Munkres, Lemma 21.2).

Then f

kn(xn) ! x and kn ! 1 as n ! 1 showing that A ⇢ B.

(b) f is a bijection, therefore f

k(N) 6= ;, 8k 2 Z. By (a), Inv(N) = A implies Inv(N)

is closed as A is the intersection of closed sets. Since Inv(N) ⇢ N and N is compact,

Inv(N) is compact. Let Ck = cl⇣S1

j=k fj(N)

⌘for k � 0. Since N is a trapping region,

N � C0 � C1 � . . . � Ck . . . The Ck are non-empty since f is a bijection and N 6= ; imply

f

j(N) 6= ;, 8j 2 Z. The Ck are nested because a b )S1

j=b fj(N) ⇢

S1j=a f

j(N). The Ck

are compact, being closed subsets of N . By Munkres, Theorem 26.9, p.170 and the nested

sequence property, it follows that Inv(N) 6= ;.

(c) Let N be a trapping region. Then 9K > 0 so that fK(N) ⇢ int(N). Then

Inv(N) =1\

k=�1

f

k(N) ⇢ f

K(N) ⇢ int(N).

Exercise 28. Give an example of En ✓ [0, 1] so that

1X

n=1

�(En) = 1

and

�

⇣limn!1

⌘En = 0.

2

Problem 104. Recall that RK denotes R in the K-topology.(a) Show that [0, 1] is not compact as a subspace of RK .(b) Show that RK is connected [Hint: (�1, 0) and (0,1) inherit their usual topologies

as subspaces of RK .](c) Show that RK is not path connected.

Proof. The K topology on R is described in Munkres: let K denote the set of all numbers ofthe form 1/n, for n 2 Z+, and let B00 be the collection of all open intervals (a, b), along withall sets of the form (a, b)�K. The topology generated by B00 will be called the K-topologyon R.

The open sets

{( 1

n+ 1, 1 +

1

n)}1n=1 and (�1, 2)�K

form an open cover of [0, 1] with no possible finite subcover.

For part (b) the hint states that (�1, 0) and (0,1) inherit their usual topologies as asubspace of RK . Therefore both of these intervals are connected as they are in R. Let U, Vbe open sets of RK such that U [ V = RK and U \ V = ;. Then

(1) (�1, 0) = (U \ (�1, 0)) [ (V \ (�1, 0))

(2) (0,1) = (U \ (0,1)) [ (V \ (�1, 0)).

Since (1) holds, and (�1, 0) is connected, it must be true that U \ (�1, 0) = (�1, 0) orV \(�1, 0) = (�1, 0). A similar situation occurs in (2). Without loss of generality supposethat U \ (�1, 0) = (�1, 0) which implies (�1, 0) ✓ U and similarly (0,1)\V = V whichimplies (0,1) ✓ V . It must be the case that U = (�1, 0] and V = (0,1) or U = (�1, 0)and V = [0,1). In either case, V or U fails to be open. Therefore RK is connected.

RK is strictly finer than the standard topology on R. Then a continuous functionf : R ! Y , when Y = RK is still continuous when Y = R. Thus f satisfies the condi-tions for the Intermediate Value Theorem to hold. Let f be a path from 0 to 1 in RK wheref(0) = 0 and f(1) = 1. The Intermediate Value Theorem holds so it must be the case that

[0, 1] ✓ f([0, 1]) ✓ RK .

f is continuous and [0, 1] ⇢ R is compact which implies f([0, 1]) must also be compact. ByTheorem 27.1, [0, 1] ✓ f([0, 1]) ⇢ RK must also be compact because [0, 1] ⇢ RK is a closedinterval. But this is impossible since (a) showed [0, 1] ⇢ RK is not compact. Therefore RK

is not path-connected.

1

Show that a connected metric space having more than one point is uncountable.

Let X be a connected space with metric d having at least two points, say a and b. Suppose there is

some number c between 0 and d(a, b) such that no point x 2 X satisfies d(a, x) = c. Then the sets

{x 2 X| d(a, x) < c} and {x 2 X| d(a, x) > c} form a separation of X, which is a contradiction since X is

assumed to be connected. Then for every real number c between 0 and d(a, b) there is a point x 2 X such

that d(a, x) = c. Then the image of the function f : X ! R defined by f(x) = d(a, x) must contain the

entire interval [0, d(a, b)], and thus X must be uncountable.

1

Let X be a compact Hausdor↵ space. Let {An} be a countable collection ofclosed sets of X with empty interior. Then the interior of A =

SAn = A is

empty.

Proof. Let U0 be a nonempty open subset of X. Then U0 is not containedin A1 as the only open set contained in A1 is the empty set. So U0 \A1 6= ;.So there is an x0 in U0 \A1. Then x0 /2 X \ (U0 \A1). By Lemma 26.4, thereare disjoint open sets U1 and V0 containing x0 and X \ (U0 \A1) respectively.

Let y 2 X \ (U0 \ A1). Then V0 is an open nbhd of y with V0 \ U1 = ;. Soy /2 cl(U1). Thus U1 ⇢ cl(U1) ⇢ U0 \ A1 ⇢ U0.

Recursively define the descending chain of sets Un by

Un ⇢ cl(Un) ⇢ Un�1 \ An ⇢ Un�1.

Since cl(Un) ⇢ U0 � An for all n, we haveT

cl(Un) ⇢ U0 � A. By Theorem26.9,

Tcl(Un) 6= ;. So there is a point in U0 � A. Thus given a nonempty

open set U0 in X we can find a point of U0 not contained in A. Therefore A

does not contain a nonempty open subset. Hence int(A) = ;. Q.E.D.

Problem 108. Show that [0, 1] is not limit point compact as a subspace of Rl

Proof. It is necessary and su�cient to find an infinite subset of [0, 1] that does not have a

limit point under the lower limit topology. The basis of the lower limit topology on R is the

collection of all half open intervals of the form

[a, b) = {x|a x < b}.

Consider the infinite subset

S = {1� 1/n|n 2 N and n > 1}

Claim: no element of S is a limit point of S. Let s 2 S. Then s = 1� 1/n for some n 2 N,n > 1. There exists an open (in the lower limit topology) interval

[1� 1/n, 1� 1

n+ 1

)

containing s and no other element of S. Thus s is not a limit point.

Claim: S has no limit point in [0, 1]. First consider the case when x = 1. The neighbor-

hood [1, 2) in Rl contains x and no point of S. Now suppose x = 0. [0, 1/2) is a neighborhood

of 0 and contains no elements in S. Let x 2 (0, 1)� S. Then 1� 1n < x < 1� 1

n+1 (with the

possibility that n = 1). Rl is strictly finer than the standard topology on R so all the open

sets in the standard topology are open in the lower limit topology. U = (1 � 1n , 1 �

1n+1) is

also open in Rl as a result. x 2 U and U does not contain any elements of S. Therefore we

conclude that S has no limit point.

1

Let (X, d) be a metric space and f : X ! X that satisfies

d(x, y) = d(f(x), f(y))

for all x, y 2 X. We say f is an isometry. Show that if X is compact, then f is bijective.It follows from the fact that f is an isometry that f is injective since for x 6= y, then d(x, y) 6= 0and

d(f(x), f(y)) = d(x, y) 6= 0

so f(x) 6= f(y). It also follows that f is continuous, using the ✏ � � criterion with ✏ = �. Itremains to show that f is surjective. Assume for a contradiction that f is not surjective, sothere exists a 2 X with a /2 f(X). Since X is compact, f(X) is compact and it is closed sinceevery metric space are Hausdor↵. So f(X)0 is open, thus there exists ✏ > 0 such that

B✏(a) \ f(X) = ;. (1)

We set x1 = a and for all n � 1 we set xn+1 = f(xn). So that

d(xk, x1) > ✏,

for all k � 2 since xk 2 f(X) and we have (1). More generally, for n 6= m with n > m, we usethe inequatity with k = n�m+ 1 and apply f m� 1 times, thus

d(xn, xm) > ✏,

since f is an isometry. Thus, this sequence does not have a converging subsequence, sinceevery pair of point are at least ✏ away from each other. This is a contradiction since in metricspaces compact spaces are also sequentially compact and the sequence {xk}k�1 is contained inthe compact space X. Therefore, f is surjective and a bijection. It also follows that f is anhomeomorphism since the inverse is also an isometry, thus continuous.

[111]

[111] cont

[111] cont

[111] cont

[111] cont

[111] cont

[112]

MTG 6316 HOMEWORK Spring 2017

118. Prove the following lemma. Use the lemma to prove the proposition.

Lemma 1. A space X is first-countable if and only if for each x 2 X there exist nested

open neighborhoods Vn of x for n 2 N (i.e., V1 � V2 � V3 � · · · ) such that every

neighborhood of x contains Vn for at least one n 2 N.

Proof. The (= direction is trivial by the definition of first-countable. Now we

show the =) direction. Suppose X is first-countable, then for any fixed x 2 X,

there is a countable base {Un}1n=1 at x. We define V1 = U1 and Vn = Vn�1 \ Un

for n = 2, 3, . . .. We show by induction that {Vn}1n=1 is a countable collection of

nested open neighborhoods of x. Note that V1 = U1 is an open neighborhood of x,

V2 = V1 \ U2 is open, x 2 V1, and x 2 U2, so V2 ⇢ V1 is an open neighborhood of

x. Now assume that {Vn}kn=1 is a collection of nested open neighborhoods of x for

some k 2 N. Similarly, Vk+1 = Vk \Uk+1 ⇢ Vk is open and contains x, so {Vn}k+1n=1

is a collection of nested open neighborhoods of x. Thus the induction is complete.

Now for any open neighborhood U of x, there is some N 2 N so that UN ⇢ U . By

construction, we have VN ⇢ UN ⇢ U .

Proposition 2. If X is limit point compact, Hausdor↵, and first countable, then X is

sequentially compact.

Proof. Let A = {an}1n=1 be any sequence in X. If A is finite, then there exists

a subsequence {ank}1k=1 such that ank = a for some a 2 A, for all k 2 N. So

{ank}1k=1 converges to this a 2 X. Thus we may assume that A is infinite. Since

X is limit point compact, there exists a limit point a 2 X of A. Note that X is first

countable, then by Lemma 1, there exists a nested countable base {Vn}1n=1 at a.

Also note that X is Hausdor↵, so X satisfies the T1 axiom. Then by Theorem 17.9

(page 99), every neighborhood of a contains infinitely many point of A. Then we

can construct a subsequence {ank}1k=1 as follows: Pick an1 2 V1 \A with an1 6= a.

For k = 2, 3, . . ., pick {ank} 2 Vk \ A such that ank 6= a and nk > nk�1. Now we

show ank ! a as k ! 1. For any neighborhood U of a, there is some K 2 N so

that VK ⇢ U (since {Vn}1n=1 is a countable base). Then we have anK 2 VK ⇢ U

and, for any k > K, ank 2 Vk ⇢ VK ⇢ U (since {Vn}1n=1 is nested). Thus ank 2 U

for all k > K, and since U was arbitrary, we have that {ank}1k=1 converges to

a 2 X. Therefore, X is sequentially compact.

1

a) Suppose A,B are nonempty, disjoint, closed subsets of a metric space X. Show that the function f ⇥ X �[0, 1] defined by

f(x) = dist(x,A)

dist(x,A) + dist(x,B)

is continuous with f(x) = 0 for all x " A, f(x) = 1 for all x " B, and 0 < f(x) < 1 for all x " X \ (A<B).

b) Show that if X is a connected metric space with at least two distinct points, then X is uncountable.

a) First note that the distance funtion is continuous on X. Thus f(x) is continuous on X except for

the points x " X in which dist(x,A) + dist(x,B) = 0. Since dist(x,C) = infy"C{dist(x, y)} ' 0 for all

C L X, then dist(x,A)+ dist(x,B) = 0 only when dist(x,A) = 0 and dist(x,B) = 0. This would imply that

x " cl(A)=B = o or x " A=cl(B) = o since A and B are disjoint and closed. Thus dist(x,A)+dist(x,B) > 0

for all x " X.

Let x " A. Then dist(x,A) = 0 implies f(x) = 0

dist(x,B)

= 0, since dist(x,B) > 0 as A and B share no

limit points. Similarly, let x " B, so dist(x,B) = 0 and dist(x,A) > 0. Thus f(x) = dist(x,A)

dist(x,A)

= 1. Now

suppose x " X \(A<B). Since X being a metric space implies X is regular, then x and A may be separated

by disjoint open sets (similarly with x and B). Thus dist(x,A) > 0 and dist(x,B) > 0. So f(x) > 0 since

both the numerator and denominator are positive. Finally,

f(x) = dist(x,A)

dist(x,A) + dist(x,B)

< dist(x,A)

dist(x,A)

= 1.

b) Let X contain the distinct points x1 and x2. Letting A = {x1} and B = {x2}, then using f(x) as in part

a, f is a continuous function from X � [0, 1] where f(A) = f(x1) = 0 and f(B) = f(x2) = 1. Since f

is continuous and X is connected, then f(X) is also connected in [0, 1]. But the only connected subset of

[0, 1] which also contains 0 and 1 is all of [0, 1]. Since the image of X under f is uncountable, then so is X.

1

120.Showthattheset! = !,! ∈ ℝ! !! + !! = 1 isacompact,connectedspaceofℝ!. P:{1}∈ℝisasingletonset,thusclosed.

Let!: ℝ! → ℝbedefinedby !,! → !! + !!.!iscontinuous.! = !!!( 1 )isclosedinℝ!. Let!: 0, 2! → ℝ!bedefinedby! → (cos !, !"#$).!iscontinuous. 0, 2! isconnected.! 0, 2! = !.!isconnected.∎

Definition: A subset A of a metric space (X, d) is precompact if its closure cl(A) is compact.

Show that if A is precompact, then for every ✏ > 0 there exists a finite covering of A

by open balls of radius ✏ with centers in A.

Proof. Let ✏ > 0 and let O = {B(a, ✏) : a 2 A}. Then O is an open cover of A. Suppose

x is a limit point of A. Then 9a 2 A so that a 2 B(x, ✏2) \ {x}. Then d(x, a) < ✏

2 implies

x 2 B(a, ✏), so O is also an open cover of cl(A). Since A is precompact, there is a finite

subcover of O that covers cl(A). Since A ⇢ cl(A), this subcover is also a finite covering of

A by open balls of radius ✏ with centers in A.

1

Let f : X ! Y be a continuous bijection with Y hausdor↵ and X a closed subset of Rn. Showthat if for all y 2 Y there exists a neighborhood Zy such that f�1(Zy) is bounded, then f is ahomeomorphism. Prove that a continuous bijection f : R ! R is a homeomorphism.

To show that f : X ! Y is a homeomorphism we have to show that for all y 2 Y , f�1(Zy) iscontained in a compact subset and the result will follow from exercise 124. Note that since X

is a closed subset of Rn, we only need that f

�1(Zy) is bounded, since its closure is closed andbounded, thus compact. It follows from assumption that for all y 2 Y there exists a neighbor-hood Zy such that f

�1(Zy) is bounded and it is contained in its closure, which is compact bythe previous argument. Therefore f is a homeomorphism by exercise 124.

To show that f : R ! R is a homeomorphism we can use this result since R is Hausdor↵. First,note that a continuous bijection from R to R need to be strictly monotone, otherwise it fails tobe one to one. If a function is increasing then decreasing it reaches a relative maximum at x sofor a < x < b we have that f(a) < f(x) and f(b) < f(x) and then one can use the intermediatevalue theorem to show that f will reach the value f(a) or f(b) twice. A similar argument canbe made for the case decreasing then increasing with a relative minimum.

Assume without loss of generality that f is strictly increasing. Otherwise one could argue with�f but f is a homeomorphism if �f is. Let y 2 R, we need to find a neighborhood of y

with bounded inverse image. Let a, b such that y 2 (a, b), since f is surjective there exists u, vsuch that f(u) = a and f(v) = b. But f is strictly increasing so f

�1((a, b)) = (u, v) and theresult follows since (u, v) is bounded and y was arbitrary. Therefore a bijection f : R ! R is ahomeomorphism.

Problem 129. Prove Sperner’s lemma in three dimensions: Let a tetrahedron, whose ver-

texes have been labeled A,B,C, and D be divided into subtetrahedra with vertexes labeled

so that only three labels appear on each face of the original tetrahedron. Then at least one

subtetrahedron has all four labels.

Proof. First consider the number of sides labeled ABC in the division of the tetrahedron

into subtetrahedron. The sides labeled ABC on the inside of the tetrahedron belong to

exactly two subtetrahedron. Thus the number of sides labeled ABC on the inside of the

tetrahedron must be even. The labeling of ABC on the outside of the main tetrahedron

only occurs at one side of the main tetrahedron (namely the side with vertexes ABC). From

Sperner’s Lemma for the two dimensional case, we know the number of sides labeled ABCon the outside must be odd. Therefore the total number of sides labeled ABC must be odd.

It is analogous to find the amount of sides labeled ABD, BCD, and ACD are all odd, as well.

Let d be the number of subtetrahedron with vertexes ABCD and a be the number of sub-

tetrahedron with sides ABC that do not have vertexes ABCD. The other subtetrahedra

with sides ABC are those with vertexes ABCA, ABCB, and ABCC. Each subtetrahedra

of this form must have two sides with vertexes ABC as opposed to ABCD only having one

side with vertexes ABC. Thus the total amount of sides with vertexes ABC is 2a+ d which

is odd by the results of the previous paragraph. Therefore d is odd. This is a stronger result

compared to proving there is a single complete subtetrahedron.

1

Consider an annulus triangulated with vertices labeled A,B, or C. The content may be defined as for cells,

while there are now two indexes: one for the outside boundary, I1, and one for the inside boundary, I2.Prove that C = I1 � I2.

Let X denote the triangulated annulus and Y denote the inner complement

complement of X (see figure to right). Note that the boundary of X1 is the

inner boundary of X.

Using the labeling of the inner boundary of X, we can create a triangulation of

X1, in any manner we choose. So the triangulation of X1 combined with the

triangulation of X gives a triangulation of X <X1. Thus by the Index Lemma,

CX<X1= IX<X1

= I1. But, as the content is defined as for cells, then CX<X1=

CX + CX1= CX + I2, again by the Index Lemma. So I1 = CX<X1

= CX + I2imples CX = I1 � I2.

1

[134]

[134] cont

[134] cont

Problem 135. Compute the winding number of V (x, y) = (y(x

2 � 1), x(y

2 � 1)) on the

following curves.

(a) x

2+ y

2 � 2x� 2y + 1 = 0 (b) x

2+ y

2+ x+ y = 1/2

(c) x

2+ y

2= 1 (d) x

2+ y

2= 4

Solution (a). Points that satisfy the following equations represent the points on the circle

(of radius 1) where the vector V points north:

8><

>:

(x� 1)

2+ (y � 1)

2= 0

y(x

2 � 1) = 0

x(y

2 � 1) � 0.

The only point that satisfies these equations is the point (1, 2) on the circle. As one travels

around the circle counterclockwise, the x-value of V , (y(x

2�1) goes from positive to negative

which means V moves from quadrant one to quadrant two (so +1). Since this is the only

solution, the winding number is equal to 1.

Solution (b): Points that satisfy the following equations represent the points on the circle

where the vector V points north:

8><

>:

x

2+ y

2+ x+ y =

12

y(x

2 � 1) = 0

x(y

2 � 1) � 0.

The two points (�1,

p3�12 ) and (

�1�p3

2 , 0) satisfy this system. As one travels around the cir-

cle counterclockwise, the x-value of V , (y(x

2� 1) goes from negative to positive at the point

(�1,

p3�12 ) (so �1). When considering the point (

�1�p3

2 , 0), (y(x

2 � 1) goes from positive to

negative (so +1). This results in a winding number of 0.

Solution (c): Points that satisfy the following equations represent the points on the circle

where the vector V points north:

8><

>:

x

2+ y

2= 1

y(x

2 � 1) = 0

x(y

2 � 1) � 0.

The only point that satisfies these conditions is (�1, 0). When traveling around the unit

circle counterclockwise, y(x

�1) changes from negative to positive at this point resulting in a

winding number of �1.

1

Solution (d): Points that satisfy the following equations represent the points on the circle

where the vector V points north:

8><

>:

x

2+ y

2= 4

y(x

2 � 1) = 0

x(y

2 � 1) � 0.

There are three points satisfying this system; namely (�2, 0), (1,�p3), (1,

p3). At each

one of these points, y(x

2 � 1) goes from positive to negative resulting in a winding number

of 3.

2

Let � be the height function on a desert island. Let V be the corresponding vector field. Let

P be the number of peaks, C the number of cols, and B the number of bottoms, and L the

number of lakes denote the only critical points of the island. Prove that P �C +B = 1�L.

Verify this by carefully identifying all the critical points on the island in Figure 11.4 (Henle,

page 72).

Fact The winding number of V along the shoreline is 1, and P � C + B = 1 for the same

island with no lakes (from a previous problem).

Proof. Around each lake, the height of the shoreline must be higher than the lake. Then

V is non-zero on the shoreline. Let e0 denote the counterclockwise curve around the island

and e1, e2, . . . , eL be the clockwise curves around each of the L lakes. Then the winding

number is 1 around e0 and �1 about ei, 1 i L. By the Poincare Index Theorem

WL(V ) = P � C +B + L = 1.

1

141. Show that if X has a countable basis then every basis ! for X contains a countable basis for X. [Hint. For every pair of indices n, m for which it is possible, choose Cn,m ∈ ! such that

Bn ⊂ Cn, m ⊂ Bm. ]

P: Let B = !! !!!! be a countable basis. Pick Cn, m (when possible). It is a countable sub-

collection of ! which is a (countable) basis; ∀ open U ∈ X and x ∈ U ∃

• an open set Bm ⊂ U containing x • an open set C ⊂ Bm containing x • open set Bn ⊂ C containing x

⇒ x ∈ Cn, m. ∃ Cn, m ⊂ U. {Cn, m } is a countable basis by Lemma 13.2.∎

Let X have a countable basis and A an uncountable subset of X. Show that uncountably many

points of A are limit points of A.

Let A

0be the set of limit points of A. Assume for a contradiction that A

0is countable, so that

F = A \A0is uncountable. Let x 2 F , since x is not a limit point of A there exists a neighbor-

hood U

x

of x that does not intersect A \ {x}. Moreover, there exists a basis element B

x

such

that x 2 B

x

and B

x

⇢ U . Note that for x 6= y 2 F we have that B

x

and B

y

don’t intersect A

at a di↵erent point than x and y respectively, so x /2 B

y

and y /2 B

x

and it follows that B

x

6= B

y

.

So each element of F is contained in a distinct basis element. This is a contradiction since F is

uncountable and X have a countable basis. Therefore A as uncountably many limit points.

[143]

Every metrizable space with a countable dense subset has a countable basis.

Proof. Let X be metrizable with metric d, and have a countable dense subsetA. For every x 2 A and n 2 N>0 there is B(x, 1

n). Since A is countable wehave An = {B(x, 1

n)|x 2 A} is countable. Define B = [An. Then B is thecountable union of countable sets. Hence B is countable.

Let x 2 X and n > 0. By density of A in X, there is an a in A so thatd(x, a) < 1

n . So x 2 B(a, 1n) which is in B.

Let B(x, 1n) and B(y, 1

m) be in B and suppose B(x, 1n) \ B(y, 1

m) = B3. Letb 2 B3. Let

1

k

<

1

2min{ 1

n

� d(b, x),1

m

� d(b, y)}.

Then by density there is an a 2 A so that b 2 B(a, 1k ).

Let t 2 B(a, 1k ). Then

d(t, x) < d(a, t) + d(a, x) < d(t, a) + d(a, b) + d(b, x)

<

1

k

+1

k

+ d(b, x) <2

k

+ d(b, x)

< min{ 1n

� d(x, b),1

m

� d(y, b)}+ d(x, b) 1

n

.

Hence a 2 B(x, 1n). Also,

d(t, y) < d(a, t) + d(a, y) < d(t, a) + d(a, b) + d(b, y)

<

1

k

+1

k

+ d(b, y) <2

k

+ d(b, y)

< min{ 1n

� d(x, b),1

m

� d(y, b)}+ d(y, b) 1

m

.

Hence t 2 B(y, 1m). Thus t 2 B3. So B(a, 1

k ) ⇢ B3. Therefore B is acountable basis for X. Q.E.D.

Every metrizable Lindelof space has a countable basis.

Proof. Let X be metrizable and Lindelof with metric d. For every x 2 X

and n 2 N>0 there is B(x, 1n). For a given n, X is covered by these balls.

Moreover, X is Lindelof, so there is a countable subcover, An, of X. LetB =

SAn. Then B is a countable union of countable sets. So B is countable.

Let x 2 X. For every n, An covers X. So x is in a ball in An. Thus x is in aball in B.

Now let B(x, 1n) and B(y, 1

m) be in B and suppose B(x, 1n) \ B(y, 1

m) = B3.Let b 2 B3.

Let 1k <

12 min{ 1

n � d(b, x) , 1m � d(b, y)}. Let t 2 X such that b 2 B(t, 1

k ).Let z 2 B(t, 1

k ). Then we have

d(z, x) d(z, t) + d(t, x) d(z, t) + d(b, t) + d(b, x)

2

k

+ d(b, x) 1

n

.

Thus B(t, 1k ) ⇢ B(x, 1

n).

Similarly, let t 2 X such that b 2 B(t, 1k ) . Let z 2 B(t, 1

k ). Then we have

d(z, y) d(z, t) + d(t, y) d(z, t) + d(b, t) + d(b, y)

2

k

+ d(b, y) 1

m

.

So B(t, 1k ) ⇢ B(y, 1

m). Therefore B(t, 1k ) ⇢ B(x, 1

n) \ B(y, 1m). Ergo B is a

countable basis for X.

Q.E.D.

2