Math 516
Professor Lieberman
February 6, 2009
HOMEWORK #2 SOLUTIONS
Chapter 8
13. Let T be the weakest topology containing C . Then B ⊂ T . In addition, each x ∈ X
is in some B ∈ B. (Specifically, every x is in X ∈ B.) If B1 and B2 are two elements
of B, then we can write
n
m
B1 = ⋂ C1i ,
B2 = ⋂ C2j ,
i=1
j=1
where n and m are positive integers and C1i and C2j are elements of C. It follows
that
B1 ∩ B2 = C11 ∩ ⋅ ⋅ ⋅ ∩ C1n ∩ C21 ∩ ⋅ ⋅ ⋅ ∩ C2m ,
so B1 ∩ B2 ∈ B. Therefore B is the base for some topology, which must be T .
14. First, X is in T because X = X ∖ ∅ and ∅ ∈ T by hypothesis. Second, if (Oα )α∈A is
a family of sets in T , then
X ∖ ∪Oα = ∩(X ∖ Oα ).
To give a complete proof, you have to consider two possibilities. Either X ∖ Oα is
infinite for all α, in which case Oα is empty for all α, so the union is the empty set,
which is in T or X ∖ Oα is finite for α, in which case the intersection is a subset of
a finite set and hence finite, so O ∖ ∪Oα ∈ T . Finally, if O1 , . . . , On are in T , then
X ∖ ∩Oj = ∪(X ∖ Oj ).
This union is a union of finitely many finite sets, so it’s in T . To see that T is not
first countable, fix x ∈ X and let Y = X ∖ {x}. Because X is uncountable, so is Y .
Now let (Uj ) be a countable collection of neighborhoods of x. It follows that X ∖ ∪Uj
is still uncountable, so there is a point y ∈ Y such that y ∉ Uj for all positive integers
j. Hence, for any positive integer j, Uj is not a subset of the open set O = X ∖ {y}.
Since x ∈ O, it follows that (Uj ) is not a base for the topology at x, so T is not first
countable.
15. If x ∈ R, then x ∈ [x, x + 1). If x ∈ [a1 , b1 ) ∩ [a2 , b2 ) for some real numbers a1 < b1 and
a2 < b2 , then b1 > x and b2 > x, so x ∈ [x, min{b1 , b2 }). Hence, by Proposition 5, B is
the base for a topology.
For each x ∈ R, we can take the collection of all intervals of the form [x, x + n1 )
for some positive integer n as a base at x. Hence the half-open interval topology is
first countable. On the other hand, to examine the second countability axiom, we
let x ∈ R and let O be an open set, and we introduce the open set V = [x, x + 1). If
x < inf O, then x ∉ O, and if x > inf O, then there is a number y ∈ (x, inf O). Hence
if x ≠ inf O, we cannot have x ∈ O ⊂ V . Hence a base for the half-open topology
1
2
must contain at least one open set Ox for each x ∈ R. (This is a set Ox such that
x ∈ Ox ⊂ [x, x + 1).) Since R is uncountable, it follows that any base for this topology
is uncountable. Finally, if x ∈ R and if O is an open set containing x, then there is a
positive integer n such that [x, x + n1 ) ⊂ O. Since there is a rational number r in the
interval (x, x + x1 ), it follows that r ∈ O. Therefore Q is dense.
it follows from Proposition 7.6 that this topology is not metrizable.
24. (a) This is simple algebra:
∣f ∣
1 + ∣f ∣
<
= 1.
1 + ∣f ∣ 1 + ∣f ∣
(b) First, h is continuous because it’s the composition of the continuous function
g∶ R → R given by g(s) = s/(1 + ∣s∣) and the continuous function f . Because
h is continuous, the sets B and C are closed. Urysohn’s Lemma then gives a
continuous real-valued function H1 on X such that H1 (B) = {1} and H1 (C) =
{0} and H1 (x) ∈ [0, 1] for all x ∈ X. We then choose h1 (x) = (1 − 2H1 (x))/3. It
follows that h1 is a real-valued function on X. In addition h1 (x) = (1 − 2(1))/3 =
−1/3 for all x ∈ B, h1 (x) = (1 − 2(0))/3 = 1/3 for all x ∈ C. Furthermore, if x ∈ X,
then
1
1
1
h1 (x) = (1 − 2H1 (x)) ≥ (1 − 2(0)) = −
3
3
3
and
1
1
1
h1 (x) = (1 − 2H1 (x)) ≤ (1 − 2(1)) = ,
3
3
3
so ∣h1 (x)∣ ≤ 1/3. Finally, if x ∈ A, we consider three possibilities. If also x ∈ B,
then h(x) − h1 (x) ≤ − 31 − (− 13 ) = 0 ≤ 23 and h(x) − h1 (x) ≥ −1 − (− 13 ) = − 23 . If also
x ∈ C, then h(x) − h1 (x) ≤ 1 − ( 31 ) = 23 and h(x) − h1 (x) ≥ 13 − ( 31 ) = 0 ≥ − 23 . If
x ∈ A ∖ (B ∪ C), then ∣h(x) − h1 (x)∣ ≤ ∣h(x)∣ + ∣h1 (x)∣ < 31 + 31 = 23 . In all cases, we
have ∣h(x) − h1 (x)∣ ≤ 23 .
(c) We proceed by induction on n. Part (b) gives the case n = 1. Once we have
h1 , . . . , hn , we construct hn+1 as follows: Let
n
n
2n
2n
Bn = {x ∶ h(x) − ∑ hi (x) ≤ − n+1 }, Cn = {x ∶ h(x) − ∑ hi (x) ≥ n+1 },
3
3
i=1
i=1
∣h∣ =
and define the function Kn by
Kn (x) =
n
3n
(h(x)
−
hi (x)) .
∑
2n
i=1
Applying part (b) to the function Kn (with Bn in place of B and Cn in place of
C), we see that there is a continuous real-valued function kn such that kn (x) =
− 31 for all x ∈ Bn , kn (x) = 13 for all x ∈ Cn , ∣kn (x)∣ ≤ 13 for all x ∈ X, and
∣Kn (x)−kn (x)∣ < 23 for all x ∈ A. We now take hn+1 = (2n /3n )kn . Then ∣hn+1 (x)∣ ≤
(2n /3n ) 13 = 2n /3n+1 for all x ∈ X, and
n+1
∣h(x) − ∑ hi (x)∣ =
i=1
2n
2n+1
∣K
(x)
−
k
(x)∣
<
.
n
n
3n
3n+1
3
(d) This part is just the Weierstrass M -test. (For completeness, I will include a
proof here, but it isn’t really needed.) First, we show that the sequence of
partial sums (sn ), given by
n
sn (x) = ∑ hn (x),
k=1
is uniformly Cauchy. For ε > 0 be given, there is a positive integer N such that
∞
∑n=N 2n−1 /3n < ε (because this is the tail of a geometric series with ratio 2/3).
If n > m ≥ N , it follows that
n
n
n
i=m+1
i=m+1
i=m+1
∣sn (x) − sm (x)∣ = ∣ ∑ hi (x)∣ ≤ ∑ ∣hi (x)∣ ≤ ∑ 2i−1 3i < ε.
Hence, for each x ∈ X, there is a number k(x) such that sn (x) → k(x) because
R is complete. Now, let ε > 0 be given and let N be the positive integer such
that ∣sn (x) − sm (x)∣ < ε/2 for all n, m ≥ N and all x ∈ X. Fix an x ∈ X and note
that there is a positive integer N (x) such that ∣sn (x) − k(x)∣ < ε/2 if n ≥ N (x).
If n = max{N, N (x)} and if m ≥ N , it follows that
∣sm (x) − k(x)∣ ≤ ∣sm (x) − sn (x)∣ + ∣sn (x) − k(x)∣ < ε.
Hence the sequence (sn ) converges uniformly to k. Finally, we show that k is
continuous at any x ∈ X. Let ε > 0 and x ∈ X be given. Then there is a positive
integer N such that ∣sn (t) − k(t)∣ < ε/3 for any t ∈ X and there is a neighborhood
O of x such that ∣sN (x) − sN (y)∣ < ε/3 for all y ∈ O. If y ∈ O, it follows that
∣k(x) − k(y)∣ ≤ ∣k(x) − sN (x)∣ + ∣sN (x) − sN (y)∣ + ∣sN (y) − k(y)∣ < ε.
Hence k is continuous at x.
(e) Since k is continuous, it follows that {k(x) = 1} is closed. Hence Urysohn’s
Lemma gives a continuous function ϕ on X which is 1 on A and 0 on {k(x) = 1}.
(f) Since ϕ = 0 when k = 1, it follows that ϕ∣k∣ < 1 on X. Therefore, g = ϕk/(1−ϕ∣k∣)
is continuous on X. In addition, on A, we have
k
f /(1 + ∣f ∣)
f
g=
=
=
= f.
1 − ∣k∣ 1 − (∣f ∣/(1 + ∣f ∣)) 1 + ∣f ∣ − ∣f ∣
25. Here is some notation that makes the proof easier to read. For ε > 0, y ∈ X, and F ∗
a finite subset of F , we write
B(ε, y, F ∗ ) = {x ∶ ∣f (x) − f (y)∣ < ε for all f ∈ F ∗ }.
To show that B, the collection of all such B(ε, y, F ∗ ) is a base for the weak topology,
we must show that every one of these sets is in the weak topology and that, if U
is in the weak topology and x ∈ U , then there is a choice of ε, y and F ∗ so that
x ∈ B(ε, y, F ∗ ) ⊂ U .
The first step is easy. Fix ε > 0, y ∈ X and F ∗ a finite subset of F . For each
fi ∈ F ∗ , we set Bi = B(ε, y, {fi }), so B(ε, y, F ∗ ) = ⋂ Bi . For Oi = (f (y) − ε, f (y) + ε),
we then have that Bi = fi−1 (Oi ). Hence Bi is in subbase denoted by C (on page 179)
for the weak topology, and therefore B(ε, y, F ∗ ) is also in the weak topology.
4
For the second step, we show that B is a base for some topology T . According to
Proposition 5, we must show that each x ∈ X is in some B(ε, y, F ∗ ) and that, if x ∈ X
is in the intersection of B(ε1 , y1 , F1∗ ) and B(ε2 , y2 , F3∗ ), then there is a B(ε3 , y3 , F3∗ )
such that x ∈ B(ε3 , y3 , F3∗ ) ⊂ B(ε1 , y1 , F1∗ ) ∩ B(ε2 , y2 , F2∗ ). The first statement is
proved by noting that, for any x ∈ X and any f ∈ F , we have x ∈ B(1, x, {f }). To
prove the second statement, we choose y3 = x and F3∗ = F1∗ ∪ F2∗ . Then, for any
ε3 > 0, we have x ∈ B(ε3 , y3 , F3∗ ). We now set
η1 = min∗ {∣f (x) − f (y)∣},
f ∈F1
η2 = min∗ {∣f (x) − f (y)∣}.
f ∈F2
Since x, y1 , and y2 are fixed and F1∗ and F2∗ are finite, η1 ∈ (0, ε1 ) and η2 ∈ (0, ε2 ).
We now take ε3 = min{ε1 − η1 , ε2 − η2 }. If z ∈ B(ε3 , y3 , F3∗ ), then, for any f ∈ F1∗ , we
have
∣f (z) − f (y1 )∣ ≤ ∣f (z) − f (x)∣ + ∣f (x) − f (y1 )∣ < ε3 + ∣f (x) − f (y1 )∣ ≤ ε3 + η1 ≤ ε1 ,
and for any f ∈ F2∗ , we have
∣f (z) − f (y2 )∣ ≤ ∣f (z) − f (x)∣ + ∣f (x) − f (y2 )∣ < ε3 + ∣f (x) − f (y2 )∣ ≤ ε3 + η2 ≤ ε2 .
Hence B(ε3 , y3 , F3∗ ) ⊂ B(ε1 , y1 , F1∗ ) ∩ B(ε2 , y2 , F2∗ ).
Now, suppose that the weak topology is Hausdorff. Then for any pair of distinct
points x and y in X, there are disjoint open sets (in the weak topology) O1 and O2
such that x ∈ O1 and y ∈ O2 . Hence there are ε1 and ε2 , y1 and y2 , and F1∗ and F2∗
such that x ∈ B(ε1 , y1 , F1∗ ), y ∈ B(ε2 , y2 , F2∗ ), and B(ε1 , y1 , F1∗ ) ∩ B(ε1 , y1 , F1∗ ) =
∅. Suppose that, for every function f ∈ F , we had f (x) = f (y). It follows that
∣f (x) − f (y1 )∣ < ε1 for all f ∈ F1∗ and ∣g(y) − g(y2 )∣ < ε2 for all g ∈ F2∗ . But then,
g(x) = g(y) for all g ∈ F2 so x ∈ B(ε2 , y2 , F2∗ ) which contradicts our assumption
that B(ε1 , y1 , F1∗ ) ∩ B(ε1 , y1 , F1∗ ) = ∅. Therefore there is some function f ∈ F with
f (x) ≠ f (y).
Conversely, if, for every x ≠ y in X, there is a function f ∈ F such that f (x) ≠
f (y), let ε = 21 ∣f (x) − f (y)∣. Then B(ε, x, {f }) and B(ε, y, {f }) are open sets with
x ∈ B(ε, x, {f }) and y ∈ B(ε, y, {f }). If z ∈ B(ε, x, {f }), then ∣f (x) − f (z)∣ < ε, so
∣f (z) − f (y)∣ ≥ ∣f (x) − f (y)∣ − ∣f (y) − f (z)∣ = 2ε − ∣f (y) − f (z)∣ > ε.
Hence z ∉ B(ε, y, {f }), which means that B(ε, x, {f }) and B(ε, y, {f }) are disjoint.
Therefore the weak topology is Hausdorff in this case.
32. Let A and B be non-empty open subsets of G such that G = A ∪ B. Then, for each α,
we have Cα ⊂ A∪B. If A∩Cα = ∅ for some α, then Cα ⊂ B. Since G∩A is non-empty,
it follows that Cβ ∩ A is non-empty for some β. Since Cα ∩ Cβ is non-empty, there is
a point x ∈ Cα ∩ Cβ , so x ∈ B. It follows that Cβ ∩ A and Cβ ∩ B are non-empty open
subsets of Cβ and hence, because Cβ is connected, A ∩ B is non-empty. Hence G is
connected, too.
42. Let x = (xα ) and y = (yα ) be disjoint points in the direct product. Then there is
an index α0 such that xα0 ≠ yα0 . Because Xα0 is Hausdorff, there are disjoint open
subsets Uα0 and Vα0 of Xα0 with xα0 ∈ Uα0 and yα0 ∈ Vα0 .
5
We now set
Uα = {
Uα0
Xα
if α = α0 ,
otherwise,
Vα
Vα = { 0
Xα
if α = α0 ,
otherwise.
It follows that U = ⨉ Uα and V = ⨉ Vα are disjoint open subsets of ⨉ Xα with x ∈ U
α
α
and y ∈ V . Hence the product is Hausdorff.
α
6
Chapter 9
4. Since the proof of this fact is so similar to the proof of problem 3, I will prove both
results, using X to denote the compact Hausdorff space. If x ∈ X and F is a nonempty closed subset of X such that x ∉ F , then, for each y ∈ F , there are disjoint
open sets Uy and Vy such that x ∈ Uy and y ∈ Vy . (This follows from the Hausdorff
property.) Hence (Vy )y∈F is an open cover of F . From Proposition 12, F is compact,
and hence this open cover has a finite subcover Vy1 , . . . , Vyn . We then set
U = ∩ni=1 Uyi ,
V = ∪ni=1 Vyi .
Because Vy1 , . . . , Vyn is an open cover of F , it follows that V is open and F ⊂ V .
Because x ∈ Uy for all y ∈ F , it follows that x ∈ U ; in addition, U is open because it’s
a finite intersection of open sets. Finally, if z ∈ U , then, for each i, z ∈ Uyi so z ∉ Vyi
and therefore z ∉ V . Therefore U and V are disjoint, so X is regular.
Now, if A and B are disjoint, non-empty closed subsets of X, then, for each x ∈ A,
the previous step shows that there are disjoint open sets Ux and Vx such that x ∈ Ux
and B ⊂ Vx . Since A is compact, we can take a finite subcover Ux1 , . . . , Uxm . If we
set
U = ∪m
V = ∩m
i=1 Uxi ,
i=1 Vxi ,
then U and V are disjoint open sets with A ⊂ U and B ⊂ V . Therefore X is normal.
8. (a) Fix x ∈ X and let O be a neighborhood of f (x). Then {x} is compact, so there
is a positive integer M such that n ≥ M implies that fn ∈ N{x},O . In other words,
fn (x) ⊂ O. Therefore, lim fn (x) = f (x).
(b) Suppose fn → f in the compact open topology, and let C be a compact subset
of X. We first show that f∣C is continuous. Fix a point x ∈ C and let ε > 0 be
given. Write Ox for the open ball B(f (x), ε/2). Then there is a positive integer
N1 such that ρ(f (x), fn (x)) < ε/2 by part (a). Let K= fN−11 (B(f (x), ε/4)). Note
that K is closed because it’s the inverse image of a closed set. Moreover, K ∩ C
is compact, by Proposition 12, because it’s a closed subset of a compact set.
Since fn → f in the compact-open topology, there is a positive integer N2 such
that fn (K) ⊂ Ox . Then U = fN−11 (O) ∩ C is an open subset of C and U ⊂ K, so
fn (U ) ⊂ B(f (x), ε/2) for n ≥ N2 . Hence, if y ∈ U and n ≥ N2 , it follows that
ρ(fn (y), f (x)) < ε/2, In addition, there is a positive integer N3 (y) such that
ρ(fn (y), f (y)) < ε/2 if n ≥ N3 (y). Now, we fix y ∈ U and n = max{N2 , N3 (y)} to
conclude that
ρ(f (x), f (y)) ≤ ρ(f (x), fn (y)) + ρ(fn (y), f (y)) < ε.
Hence f∣C is continuous at x. To show that fn → f uniformly on a compact set
C, we first fix x ∈ C and ε > 0. Then we set O = B(f (x), ε/2), U = f −1 (O) ∩ C,
and K = f −1 (B(f (x), ε/4)). As before U is open and K is compact with U ⊂ K.
Hence there is a number N4 such that ρ(f (x), fn (y)) < ε/2 for n ≥ N3 and y ∈ K.
Hence for y ∈ K and n ≥ N4 , we have
ρ(fn (y), f (y)) ≤ ρ(fn (y), f (x)) + ρ(f (x), f (y)) < ε.
7
Therefore, for any x ∈ C, there are an open set U (x) and a positive integer M (x)
such that x ∈ U (x) and ρ(fn (y), f (y)) < ε for all y ∈ U (x) and n ≥ M (x). Hence
(U (x))x∈C is an open cover of C so it has a finite subcover: U (x1 ), . . . U (xm ).