[101]
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
Ny
26.1 (on page 164), there is a finite subcover {Uiy }i=1
⇢ {U↵ }↵2I of p 1 (y). Let
S
N
y
y
U y = i=1 Ui . Then U y is open in X, and p 1 (y) ⇢ U y . Thus X \ U y is closed
in X, and p 1 (y) \ (X \ U y ) = ;, so y 2
/ p (X \ U y ). Since p is a closed map,
p (X \ U y ) is closed in Y . Consider
V y = Y \ [p (X \ U y )] ,
then V y is 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 }N
i=1 of Y . We want to show
y
N
i
that {U }i=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 \ U y , we have
p
1
(V y ) = p
1
(Y \ p (A)) ⇢ X \ A = U y ,
for any y 2 Y . Thus we have
X=p
1
(Y ) ⇢ p
1
N
[
V
yi
i=1
!
=
Furthermore,
X⇢
Therefore,
SN
y i N yi
i=1 {Uj }j=1
N
[
U yi =
i=1
N
[
i=1
N
[
p
i=1
0
@
N yi
[
j=1
1
(V yi ) ⇢
N
[
i=1
1
Ujyi A .
is a finite subcover of {U↵ }↵2I for X.
1
U yi .
[102]
Let X be a metric space and f : X ! X be a homeomorphism. Given N ⇢ X, define the
maximal invariant set in N by
1
\
Inv(N ) =
k= 1
f k (N ) = {x 2 N : f k (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 f K (N ) ⇢ int(N ).
(a) Show that if N is a trapping region, then
!
1
1
\
[
Inv(N ) =
cl
f j (N ) = {x 2 N : 9xn 2 N, kn > 0, kn ! 1, such thatf kn (xn ) ! x as n ! 1}.
k=0
j=k
(b) Show that if N 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
N is a trapping region
⇣S metric. Since
⌘
T1
1
j
and X is Hausdor↵, N is closed. Let A = k=0 cl
j=k f (N ) and B = {x 2 N : 9xn 2
N, kn > 0, kn ! 1 such thatf kn (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 thatf kn (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 , f k+km (xm ) 2 N . Since f is a homeomorphism, f k 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
k
f (N ) ⇢
1
[
j=k
j
f (N ) ⇢ cl
1
1
[
j=k
j
!
f (N ) .
[102] cont
Then
Inv(N ) =
1
\
k= 1
f k (N ) ⇢
1
\
k=0
f k (N ) ⇢
Now we show A ⇢ B by first showing A ⇢ N . 8k
f k (N ) ⇢ N
1
[
)
j=k
1
\
k=0
cl
1
[
f j (N )
j=k
0,
f j (N ) ⇢ cl
1
[
f j (N )
j=k
!
!
= A.
⇢ N,
since N is a trapping
⇣Sregion (closed
⌘ and forward invariant). Thus A ⇢ N . Let x 2 A and
1
j
n 0. Since x 2 cl
j=n f (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,
S1 j
Inv(N ) is compact. Let Ck = cl
for k
0. Since N is a trapping region,
j=k f (N )
Ck . . . The Ck are non-empty since f is a bijection and N 6= ; imply
S
S1 j
j
f j (N ) 6= ;, 8j 2 Z. The Ck are nested because a  b ) 1
f
(N
)
⇢
j=b
j=a f (N ). The Ck
N
C0
C1
...
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 f K (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
1
X
n=1
and
⇣
(En ) = 1
lim
n!1
⌘
2
En = 0.
[104]
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 of
the form 1/n, for n 2 Z+ , and let B 00 be the collection of all open intervals (a, b), along with
all sets of the form (a, b) K. The topology generated by B 00 will be called the K-topology
on R.
The open sets
1
1
{(
, 1 + )}1
and ( 1, 2) K
n+1
n n=1
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 a
subspace of RK . Therefore both of these intervals are connected as they are in R. Let U, V
be open sets of RK such that U [ V = RK and U \ V = ;. Then
(1)
(2)
( 1, 0) = (U \ ( 1, 0)) [ (V \ ( 1, 0))
(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) or
V \( 1, 0) = ( 1, 0). A similar situation occurs in (2). Without loss of generality suppose
that U \ ( 1, 0) = ( 1, 0) which implies ( 1, 0) ✓ U and similarly (0, 1) \ V = V which
implies (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 function
f : R ! Y , when Y = RK is still continuous when Y = R. Thus f satisfies the conditions for the Intermediate Value Theorem to hold. Let f be a path from 0 to 1 in RK where
f (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. By
Theorem 27.1, [0, 1] ✓ f ([0, 1]) ⇢ RK must also be compact because [0, 1] ⇢ RK is a closed
interval. But this is impossible since (a) showed [0, 1] ⇢ RK is not compact. Therefore RK
is not path-connected.
1
[105]
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
[106]
Let X be a compact Hausdor↵ space. Let {An } be a countable S
collection of
closed sets of X with empty interior. Then the interior of A = An = A is
empty.
Proof. Let U0 be a nonempty open subset of X. Then U0 is not contained
in 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, there
are 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 = ;. So
y2
/ cl(U1 ). Thus U1 ⇢ cl(U1 ) ⇢ U0 \ A1 ⇢ U0 .
Recursively define the descending chain of sets Un by
Un ⇢ cl(Un ) ⇢ Un
\ An ⇢ Un 1 .
T
Since T
cl(Un ) ⇢ U0 An for all n, we have cl(Un ) ⇢ U0 A. By Theorem
26.9, cl(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.
1
[108]
Problem 108. Show that [0, 1] is not limit point compact as a subspace of Rl
Proof. It is necessary and sufficient 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
n > 1. There exists an open (in the lower limit topology) interval
[1
1/n, 1
1/n for some n 2 N,
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 neighborhood [1, 2) in Rl contains x and no point of S. Now suppose x = 0. [0, 1/2) is a neighborhood
1
of 0 and contains no elements in S. Let x 2 (0, 1) S. Then 1 n1 < x < 1 n+1
(with the
possibility that n = 1). Rl is strictly finer than the standard topology on R so all the open
1
sets in the standard topology are open in the lower limit topology. U = (1 n1 , 1 n+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
[110]
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= 0
and
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 ✏ = . It
remains to show that f is surjective. Assume for a contradiction that f is not surjective, so
there exists a 2 X with a 2
/ f (X). Since X is compact, f (X) is compact and it is closed since
every metric space are Hausdor↵. So f (X)0 is open, thus there exists ✏ > 0 such that
B✏ (a) \ f (X) = ;.
We set x1 = a and for all n
(1)
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 use
the 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, since
every pair of point are at least ✏ away from each other. This is a contradiction since in metric
spaces compact spaces are also sequentially compact and the sequence {xk }k 1 is contained in
the compact space X. Therefore, f is surjective and a bijection. It also follows that f is an
homeomorphism since the inverse is also an isometry, thus continuous.
[111]
[111] cont
[111] cont
[111] cont
[111] cont
[111] cont
[112]
[118]
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 }1
n=1 at x. We define V1 = U1 and Vn = Vn 1 \ Un
for n = 2, 3, . . .. We show by induction that {Vn }1
n=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+1
n=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 }1
n=1 be any sequence in X. If A is finite, then there exists
1
a subsequence {ank }k=1 such that ank = a for some a 2 A, for all k 2 N. So
{ank }1
k=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 }1
n=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 }1
k=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 }1
n=1 is a countable base). Then we have anK 2 VK ⇢ U
and, for any k > K, ank 2 Vk ⇢ VK ⇢ U (since {Vn }1
n=1 is nested). Thus ank 2 U
for all k > K, and since U was arbitrary, we have that {ank }1
k=1 converges to
a 2 X. Therefore, X is sequentially compact.
1
[119]
a) Suppose A, B are nonempty, disjoint, closed subsets of a metric space X. Show that the function f ⇥ X
[0, 1] defined by
dist(x, A)
f (x) =
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
= 0, since dist(x, B) > 0 as A and B share no
dist(x, B)
dist(x, A)
limit points. Similarly, let x " B, so dist(x, B) = 0 and dist(x, A) > 0. Thus f (x) =
= 1. Now
dist(x, A)
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)
<
= 1.
dist(x, A) + dist(x, B) dist(x, A)
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]
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! is
connected.! 0, 2! = !. !isconnected.∎
[121]
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
[125]
Let f : X ! Y be a continuous bijection with Y hausdor↵ and X a closed subset of Rn . Show
that if for all y 2 Y there exists a neighborhood Zy such that f 1 (Zy ) is bounded, then f is a
homeomorphism. 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 ) is
contained 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 and
bounded, thus compact. It follows from assumption that for all y 2 Y there exists a neighborhood Zy such that f 1 (Zy ) is bounded and it is contained in its closure, which is compact by
the 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 to
be one to one. If a function is increasing then decreasing it reaches a relative maximum at x so
for a < x < b we have that f (a) < f (x) and f (b) < f (x) and then one can use the intermediate
value theorem to show that f will reach the value f (a) or f (b) twice. A similar argument can
be 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, v
such that f (u) = a and f (v) = b. But f is strictly increasing so f 1 ((a, b)) = (u, v) and the
result follows since (u, v) is bounded and y was arbitrary. Therefore a bijection f : R ! R is a
homeomorphism.
[129]
Problem 129. Prove Sperner’s lemma in three dimensions: Let a tetrahedron, whose vertexes 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 ABC
on 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 subtetrahedron 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
[133]
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 + I2
imples CX = I1 I2 .
1
[134]
[134] cont
[134] cont
[135]
Problem 135. Compute the winding number of V (x, y) = (y(x2
following curves.
(a) x2 + y 2
2x
2y + 1 = 0
(c) x2 + y 2 = 1
1), x(y 2
1)) on the
(b) x2 + y 2 + x + y = 1/2
(d) x2 + 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
2
2
>
<(x 1) + (y 1) = 0
y(x2 1) = 0
>
: 2
x(y
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(x2 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
1
2
2
>
<x + y + x + y = 2
y(x2 1) = 0
>
: 2
x(y
1) 0.
p
p
The two points ( 1, 32 1 ) and ( 1 2 3 , 0) satisfy this system. As one travels around the circle counterclockwise,
the x-value of V , (y(x2 1) goespfrom negative to positive at the point
p
( 1, 32 1 ) (so 1). When considering the point ( 1 2 3 , 0), (y(x2 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
2
2
>
<x + y = 1
y(x2 1) = 0
>
: 2
x(y
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
[135] cont
Solution (d): Points that satisfy the following equations represent the points on the circle
where the vector V points north:
8
2
2
>
<x + y = 4
y(x2 1) = 0
>
: 2
x(y
1) 0.
p
p
There are three points satisfying this system; namely ( 2, 0), (1,
3), (1, 3). At each
one of these points, y(x2 1) goes from positive to negative resulting in a winding number
of 3.
2
[138]
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
WL (V ) = P
C + B + L = 1.
1 about ei , 1  i  L. By the Poincaré Index Theorem
1
[141]
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 subcollection 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.∎
[142]
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 A0 be the set of limit points of A. Assume for a contradiction that A0 is countable, so that
F = A \ A0 is uncountable. Let x 2 F , since x is not a limit point of A there exists a neighborhood Ux of x that does not intersect A \ {x}. Moreover, there exists a basis element Bx such
that x 2 Bx and Bx ⇢ U . Note that for x 6= y 2 F we have that Bx and By don’t intersect A
at a di↵erent point than x and y respectively, so x 2
/ By and y 2
/ Bx and it follows that Bx 6= By .
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]
[144]
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 subset
A. For every x 2 A and n 2 N>0 there is B(x, n1 ). Since A is countable we
have An = {B(x, n1 )|x 2 A} is countable. Define B = [An . Then B is the
countable 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 that
d(x, a) < n1 . So x 2 B(a, n1 ) which is in B.
Let B(x, n1 ) and B(y, m1 ) be in B and suppose B(x, n1 ) \ B(y, m1 ) = B3 . Let
b 2 B3 . Let
1
1
1
1
< min{
d(b, x),
d(b, y)}.
k
2
n
m
Then by density there is an a 2 A so that b 2 B(a, k1 ).
Let t 2 B(a, k1 ). Then
d(t, x) < d(a, t) + d(a, x) < d(t, a) + d(a, b) + d(b, x)
1 1
2
+ + d(b, x) < + d(b, x)
k k
k
1
1
1
< min{
d(x, b),
d(y, b)} + d(x, b)  .
n
m
n
1
Hence at 2 B(x, n ). Also,
<
d(t, y) < d(a, t) + d(a, y) < d(t, a) + d(a, b) + d(b, y)
1 1
2
+ + d(b, y) < + d(b, y)
k k
k
1
1
1
< min{
d(x, b),
d(y, b)} + d(y, b)  .
n
m
m
Hence t 2 B(y, m1 ). Thus t 2 B3 . So B(a, k1 ) ⇢ B3 . Therefore B is a
countable basis for X.
Q.E.D.
<
Every metrizable Lindelöf space has a countable basis.
Proof. Let X be metrizable and Lindelöf with metric d. For every x 2 X
and n 2 N>0 there is B(x, n1 ). For a given n, X is covered by these balls.
[144] cont
Moreover,
X is Lindelöf, so there is a countable subcover, An , of X. Let
S
B = An . 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 a
ball in B.
Now let B(x, n1 ) and B(y, m1 ) be in B and suppose B(x, n1 ) \ B(y, m1 ) = B3 .
Let b 2 B3 .
Let k1 < 12 min{ n1 d(b, x) , m1
Let z 2 B(t, k1 ). Then we have
d(b, y)}. Let t 2 X such that b 2 B(t, k1 ).
d(z, x)  d(z, t) + d(t, x)  d(z, t) + d(b, t) + d(b, x)

2
1
+ d(b, x)  .
k
n
Thus B(t, k1 ) ⇢ B(x, n1 ).
Similarly, let t 2 X such that b 2 B(t, k1 ) . Let z 2 B(t, k1 ). Then we have
d(z, y)  d(z, t) + d(t, y)  d(z, t) + d(b, t) + d(b, y)

2
1
+ d(b, y)  .
k
m
So B(t, k1 ) ⇢ B(y, m1 ). Therefore B(t, k1 ) ⇢ B(x, n1 ) \ B(y, m1 ). Ergo B is a
countable basis for X.
Q.E.D.
2