Math 516
Professor Lieberman
February 20, 2009
HOMEWORK #3 SOLUTIONS
Chapter 9
10. We will prove the implications (i) ⇒ (iii) ⇒ (ii) ⇒ (i).
(i) ⇒ (iii) Let f be a bounded, continuous real-valued function. Then f is upper
semicontinuous, so Proposition 19 implies that f attains its maximum.
(iii) ⇒ (ii) Let f be a continuous real-valued function and set g = |f |/(1 + |f |).
Then g is a bounded continuous real-valued function so it attains its maximum M .
Since 0 ≤ g < 1, it follows that 0 ≤ M < 1 as well. Hence |f | ≤ M (1 + |f |), so
|f | ≤ M/(1 − M ), and the number M/(1 − M ) is finite. Therefore f is bounded.
(ii) ⇒ (i) We prove the contrapositive of the implication. In other words, we’ll
show that, if X is a normal space that is not countably compact, then there is an
unbounded continuous real-valued function defined on X. If X is not countably
compact, then there is an infinite sequence (xn ) with no cluster point. To keep the
notation clear, we write S for the set of points in the sequence. Our first step is to
show that the complement of S is open, so let y ∈
/ {x1 , x2 , . . . }. Then there are an
open set O and a positive integer N such that y ∈ O and xn ∈
/ O if n ≥ N . But
X is normal, so X is T1 and therefore the set F = {x1 , . . . , xN −1 } is closed. We
now take U = O ∩ (X \ F ). Then U is open, y ∈ U and xn ∈
/ U for any n, so
U ⊂ (X \ S). Hence the complement of S is open, so S is closed. We now define f on
S by f (xn ) = n. (If some points are repeated, we have to be a little more careful and
define f (xn ) to be the smallest integer j such that xn = xj .) Our previous argument
shows that the subspace topology on S is just the discrete topology. (In other words,
every subset of S is relatively open in S.) Therefore f is continuous, and Tietze’s
Extension Theorem implies that f can be extended to an continuous function on X.
Since S is infinite, it follows that f is unbounded.
12. For simplicity, we write 1 for the first element of X, and we abbreviate our sets as
follows. We write [1, a) for {x : x < a}, (a, b) for {x : a < x < b}, and (a, Ω} for
{x : x > a}.
(a) To show that B is a base, we must show that each x ∈ X is contained in some
B in B and that, if x ∈ B1 ∩ B2 with B1 and B2 in B, then there is a B ∈ B
with x ∈ B ⊂ B1 ∩ B2 . If x 6= Ω, then x ∈ [1, Ω) and, if x = Ω, then x ∈ (1, Ω].
For the intersection condition, we consider various possibilities. First, we look
at [1, a) ∩ (b, c). If a ≤ b, then this intersection is empty. If b < a ≤ c, then the
intersection is (b, a). If a > c, then the intersection is (b, c). For [1, a) ∩ (b, Ω],
we note that the intersection is empty if a ≤ b. Otherwise, it’s (b, a). Next, we
look at (a, b) ∩ (c, d). If b ≤ c or if a ≥ d, then this intersection is empty. If b > c
and a < d, then the intersection is (max{a, c}, min{b, d}). Finally, we note that
[1, a) ∩ [1, b) = [1, min{a, b}) and (a, Ω] ∩ (b, Ω] = (max{a, b}, Ω].
1
2
(b) To see that X is compact, let I be an open cover of X, and define
Y = {x ∈ X : [1, x] can be covered by finitely many sets from I}.
We shall show that Y = X. First, 1 ∈ Y because there is a element O1 of I such
that 1 ∈ O1 . Next, for x ∈ X, we write the successor of x as x + 1. If x ∈ Y ,
then we can cover [1, x] by finitely many sets O1 , . . . , On+1 from I. Also there
is an element Ox ∈ I such that x + 1 ∈ Ox . It follows that x + 1 ∈ Y . If z is the
first upper bound of Y , then let Oz be an element of I such that z ∈ Oz . Then
z is an element of a set Uz of the form [1, a), (a, b), or (a, Ω] with Uz ⊂ Oz . If
Uz = [1, a), then z ∈ Y because we can cover [1, z] by Uz . If Uz = (a, b) or if
Uz = (a, Ω], then a < z, so a ∈ Y because otherwise a would be an upper bound
for Y . If [1, a] is covered by O1 , . . . , On , then {O1 , . . . , On , Oz } is a finite set of
elements of I which covers [1, z], so z ∈ Y . We have also shown that z can’t
have a successor, so z = Ω. To see that Y = X, we now take an element OΩ of
I which contains Ω. Then there is some a ∈ X such that (a, Ω] ⊂ OΩ (because
Ω isn’t in any set of the form [1, a) or (a, b)). We know that a ∈ Y , so [1, a] is
covered by finitely many elements O1 , . . . , On of I, and hence O1 , . . . , On , OΩ is
a cover of X by finitely elements of I.
To see that X is not separable, let S = {xn } be a countable set of points in X
and set S 0 = S \ {Ω}. For each xn ∈ S 0 , let Sn = {x ∈ X : x < xn }. From
Proposition 1.8, we know that Sn is countable for each n and hence
S 00 = {x : x < xn for at least one xn ∈ S 0 }
is countable, so there is a point z ∈ X \ {Ω} such that z ≥ xn for all xn ∈ S 0 .
But then z + 1, the successor of z is in the interval (z, z + 2), which doesn’t
contain any point of S. Hence S is not dense.
To see that X is not first countable, let (On ) be a countable collection of open
sets containing Ω. Then, for each n ∈ N, there is an an ∈ X \ {Ω} such that
(an , Ω] ⊂ On . The preceding argument shows that there is a element z ∈ X \{Ω}
which is greater than all the an ’s, so (z, Ω] is an open set containing Ω which is
not a subset of any On .
18. Since X is locally compact, for each x ∈ K, there is an open set Ox with compact
closure. Then (Ox )x∈K is an open cover of K, so it has a finite subcover Ox1 , . . . , Oxn .
It follows that O = Ox1 ∪ . . . Oxn is an open set with K ⊂ O. In addition,
n
[
Oxi ,
Ō =
i=1
so Ō is a finite union of compact sets. Hence Ō is compact.
19. (a) From Problem 18, there is an open set O such that Ō is compact and K ⊂ O.
Since X is Hausdorff, it follows that Ō is a compact Hausdorff space, so Ō is
compact. Then Ō \O is a closed subset of Ō and K is a closed subset of Ō (again
we use the fact that X is Hausdorff to apply Proposition 9.2), and K ∩ (Ō \ O)
is empty, so Urysohn’s Lemma gives a continuous real-valued function g on Ō
3
such that g(x) = 0 for all x ∈ Ō \ O and g(x) = 1 for all x ∈ K. We now define
f on all of X by the formula
(
g(x) if x ∈ Ō,
f (x) =
0
if x ∈
/ Ō.
It follows that f|Ō is continuous because this restriction is just g and f|X\O is
continuous because this restriction is just the zero function. Because Ō and
X \ O are closed subsets of X whose union is X, it follows from Proposition 8.3
that f is continuous.
(b) The only part of Proposition
T 9.15 left to prove is the statement about K being
a Gδ , so suppose that K = ∞
n=1 Un , where Un are open sets. Let O be an open
set such
T that Ō is compact and K ⊂ O, and set On = Un ∩ O. It follows that
K = ∞
n=1 On , and we know from part (a) that there is a continuous function
fn on X such that fn (x) = 1 if x ∈ K and fn (x) = 0 if x ∈
/ On . In addition, we
note from Urysohn’s Lemma that 0 ≤ f (x) ≤ 1 for all x ∈ X. We now set
f=
∞
X
2−n fn .
n=1
Then the series converges uniformly by the Weierstrass M -test (see Homework
#2), so f is continuous. If x ∈ K, then fn (x) = 1 for all n, so f (x) = 1. On the
other hand, if x ∈
/ K, then there is some positive integer m such that x ∈
/ Om .
Hence fm (x) = 0, so
X
2−n fn (x) ≤ 1 − 2−m < 1.
f (x) =
n6=m
44. Write G for the set of all polynomials in a finite number of functions of F . Then G
separates points and includes the constants (because 1 is a polynomial). In addition
G is an algebra because any sum or product of polynomials is a polynomial. We can
therefore apply the Stone-Weierstrass theorem to conclude that G is dense in C(X),
which is the same thing as saying that every continuous real-valued function on X
can be uniformly approximated by an element of G .
Chapter 10
8. We write x1 , . . . , xn for some basis of M , which we will keep fixed. Suppose that (yk )
is a sequence with
n
X
aik xi
yk =
i=1
for some real numbers aik . We then set
zk =
1
xk ,
max{a1k , . . . , ank }
4
so
zk =
n
X
bik xi
i=1
for bik = aik / max{a1k , . . . , ank }, and we can then extract a subsequence (zk(m) ) so
that bjk(m) = 1 for some j and all k(m). A further subsequence (still labeled (yk(m) ))
can be extracted so that, for each i, (bik(m) ) is a Cauchy
P sequence. Hence each
sequence (bik(m) ) has a limit bi , and bj = 1. If we set z =
bi xi , then
kz − zk(m) k ≤ n max{|bi − bik(m) |} max{kxi k}.
Hence zn(k) → z. It follows that, if the set of coefficients {aik } is unbounded, that
(yk ) can’t have a limit in X.
To show that the subspace is closed, we have to show that any sequence in the
subspace which converges to a point in the normed linear space actually converges
to a point in the subspace, so let (yk ) be a sequence in the subspace and define aik
by
n
X
yk =
aik xi .
i=1
Our previous step shows that each sequence (aik )∞
n=1 is bounded, so we can extract a
subsequence (yk(m) ) such that (aik(m) ) is a Cauchy (and hence convergent) sequence
for each i. With ai = lim aik(m) , we conclude that
yk(m) → y =
n
X
ai xi .
i=1
In other words, a subsequence of (yk ) converges to the point y in the subspace. On
the other hand, the sequence (yk ) converges to a point, so the limit of the whole
sequence must be y. (This result is Problem 15 from Chapter 7.)
11. First, we show that kxk1 is a pseudonorm. Let x and y be points in X and let α ∈ R.
First, we fix ε > 0. Then there are m1 and m2 in M such that kx − m1 k ≤ kxk1 + ε/2
and ky − m2 k ≤ kyk1 + ε/2. Then m1 + m2 ∈ M , so
kx + yk1 ≤ k(x + y) − (m1 + m2 )k ≤ kx − m1 k + ky − m2 k ≤ kxk1 + kyk1 + ε.
Since this chain of inequalities is true for any ε > 0, it follows that kx + yk1 ≤
kxk1 + kyk1 . If α = 0, then αx = 0. Because 0 ∈ M , it follows that k0k1 = 0, so
kαxk1 = |αkxk1 for α = 0. If α 6= 0, then αm and m/α are in M if m ∈ M . For any
ε > 0, there are m1 and m2 in M so that
ε
, kαx − m2 k < kαxk1 + ε.
kx − m1 k < kxk1 +
|α|
Hence
kαxk1 ≤ kαx − αm1 k = |α|kx − m1 k < |α|kxk1 + ε
and
|α|kxk1 ≤ |α|kx −
1
m2 k = kαx − m2 k < kαxk1 + ε.
α
5
Since these inequalities are true for all ε > 0, we conclude that kαxk1 = |α|kxk1 .
Next, we show that ϕ takes open sets into open sets. Let U be an open set in X
and let x ∈ U . Then there is a positive number δ such that y ∈ U if kx − yk < δ.
Now suppose that z ∈ X 0 and kz − ϕ(x)k < δ. Then z = ϕ(y1 ) for some y1 ∈ X and
ky1 − xk1 < δ. Hence, there is a vector m ∈ M so that k(y1 − x) − mk < δ. It follows
that y = y1 − m is in U . Hence ϕ(U ) is open.
16. There are several steps to prove that Y is a normed linear space. First, we show that
linear combinations of functions in Y are in Y . For this step, we let f and g be in
Y and we let α and β be real numbers. Then αf (x0 ) + βg(x0 ) = α0 + β0 = 0. In
addition, for any x and y in X, we have
|(αf (x) + βg(x)) − (αf (y) + βg(y))| ≤ |α||f (x) − f (y)| + |β||g(x) − g(y)|
≤ (|α|kf k + |β|kgk)ρ(x, y).
Hence αf + βg ∈ Y and
(1)
kαf + βgk ≤ |α|kf k + |β|kgk.
Now we check that k · k is a norm on Y . It’s easy to see that kf k ≥ 0 for all f and
that k0k = 0. In addition, if kf k = 0, then f (x) = f (y) for all x and y in X, and
f (x0 ) = 0, so f must be the zero function. Next, if α ∈ R, then
|αf (x) − αf (y)|
|α||f (x) − f (y)|
kαf k = sup
= sup
ρ(x, y)
ρ(x, y)
|f (x) − f (y)|
= |α| sup
ρ(x, y)
= |α|kf k.
We use (1) with α = β = 1 to infer the triangle inequality.
To show that Fx is bounded, we need to show that |Fx (f )| ≤ M kf k for some
number M . But
|Fx (f )| = |f (x)| ≤ |f (x0 )| + |f (x) − f (x0 )| ≤ kf kρ(x, x0 )
because f (x0 ) = 0. To show that Fx is a linear functional, we observe first that f (x)
is real, so Fx is real-valued. Next, if f and g are in Y and α and β are in R, then
Fx (αf + βg) = αf (x) + βg(x) = αFx (f ) + βFx (g).
In addition
kFx − Fy k = sup |Fx (f ) − Fy (f )| = sup |f (x) − f (y)| ≤ ρ(x, y).
On the other hand, if we fix x and define f by f (z) = ρ(z, x) − ρ(x, x0 ), then f ∈ Y
and
|Fx (f ) − Fy (f )| = |[ρ(x, x) − ρ(x, x0 )] − [ρ(y, x) − ρ(x, x0 )]| = |ρ(y, x)| = ρ(x, y),
so kFx − Fy k = ρ(x, y). The function F : X → Y ∗ defined by F (x) = Fx is then
an isometry (using the metric on Y ∗ which comes from the norm), so X is isometric
to a subset of Y ∗ . Since R is a Banach space, Proposition 10.3 implies that Y ∗ is
a Banach space, and any closed subset S of a complete metric space Z is complete
6
because, if (zn ) is a Cauchy sequence in S, then it has a limit z which is a cluster
point of the sequence. Hence z ∈ S̄ by Problem 14 of Chapter 9, and S = S̄ because
S is closed.