Math 516
Professor Lieberman
January 23, 2009
HOMEWORK #1 SOLUTIONS
Chapter 7
48. (a) First, because X is locally compact, for each x ∈ X, there is an open set Ox with
Ox compact and x ∈ Ox . Hence there is a δ(x) > 0 such that B(x, δ(x)) ⊂ Ox .
It follows that B(x, δ(x)) ⊂ Ox and hence, by Proposition 7.22, B(x, δ(x)) is
compact. We now define the real-valued function r on X by
r(x) = sup{δ ∈ (0, 1) : B(x, δ) is compact}.
The preceding discussion shows that this set is non-empty, and it is bounded
by 1, so r is real-valued. In addition, if θ ∈ (0, r(x)), then B(x, θ) is compact
because there is a δ ∈ (θ, r(x)) such that B(x, δ) is compact and B(x, δ) ⊂
B(x, θ).
Because X is separable, there is a dense sequence (xn ) of points in X. We now
set Un = B(x
S n , r(xn )). By the definition of r, Un has compact closure. To see
that X = Un , let x ∈ X. Since (xn ) is dense, there is a positive integer n such
that ρ(x, xn ) < 12 r(x). Hence B(xn , 12 r(x)) ⊂ B(x, θ) for some θ ∈ (0, r(x)), so
B(xn , 21 r(x)) has compact
closure, which means that r(xn ) ≥ ρ(x, xn ). Therefore
S
x ∈ Un , and X = Un .
(b) The big step (to show that σ ∗ satisfies the triangle inequality) was done in my
515 class. Here’s a quick verification: Define h(t) = t/(1 + t) for t ≥ 0. We
want to show that h(s + t) − h(s) − h(t) ≥ 0 for all s, t ≥ 0. If s = 0, then
the inequality is true because h(s) = 0 and h(s + t) = h(t). If s > 0, we define
H(t) = h(s + t) − h(s) = h(t). Then H(0) = 0, and
H(t) = 1 −
so
H 0 (t) =
1
1
1
−
+
,
1+t 1+s 1+s+t
1
1
−
≥ 0.
2
(1 + t)
(1 + s + t)2
Hence H is increasing so H(t) ≥ 0, which is the same as h(s+t)−h(s)−h(t) ≥ 0.
Since σ ∗ (f, g) ∈ [0, 1) for all f and g, it follows that σ ∗ is real-valued and
nonnegative for all f and g. In addition, σ ∗ (f, g) = 0 if f = g. On the other
hand, if σ ∗ (f, g) S
= 0, then, because each σn∗ ≥ 0, it follows that σn∗ (f, g) = 0 for
all n. But X = On , so σ(f (x), g(x)) = 0 for all x ∈ X and hence f = g (as
functions). Therefore σ ∗ is a metric on the set of all functions.
1
2
50. Write A for the set of functions f with kf kα ≤ 1. If f ∈ A, then
|f (x) − f (y)|
max |f (x)| + sup
≤ 1,
|x − y|α
so max |f (x)| ≤ 1, which means that A is bounded, and
|f (x) − f (y)|
≤1
|x − y|α
for all x and y in [0, 1]. Therefore A is bounded and equicontinuous. The ArzelaAscoli Theorem implies that any sequence in A has a pointwise convergent sequence
and that the convergence is uniform on any compact set. Since [0, 1] is compact, it
follows that any sequence in A has a uniformly convergent subsequence. Hence the
limit function is continuous.
We now show that the limit function is in A. Let’s write (fn ) for the uniformly
convergent subsequence and f for the limit function. Then, for any x 6= y in [0, 1],
there is a positive integer n such that |fn (s) − f (s)| < |x − y|α for all s ∈ [0, 1]. It
follows that
|f (x) − f (y)|
|fn (x) − fn (y)|
≤
+ 1 ≤ 2.
|x − y|α
|x − y|α
In addition, there is a positive integer M such that |f (x) − fM (x)| < 1 for any
x ∈ [0, 1]. Therefore, for any x ∈ [0, 1], we have
|f (x)| ≤ |fM (x)| + 1 ≤ 2.
Hence kf kα ≤ 4. Now let ε > 0 be given. Then there is a positive integer N1 such
that |fn (x) − f (x)| ≤ 41 ε if n ≥ N1 for all x ∈ [0, 1]. In addition, there are points s,
r, and t in [0, 1] such that
|f (r) − f (t)|
|f (x) − f (y)| 1
≥ sup
− ε
α
|r − t|
|x − y|α
4
and
|f (s)| = max |f (x)|.
Finally, there is a positive integer N2 such that |fn (s) − f (s)| ≤ 12 ε|r − t|α for all
s ∈ [0, 1] and all n ≥ N2 . Hence, if n = max{N1 , N2 }, we have
|f (x) − f (y)|
kf kα = max |f (x)| + sup
|x − y|α
|f (r) − f (t)| 1
≤ |f (s)| +
+ ε
|r − t|α
4
|fn (r) − fn (t)|
+ε
≤ |fn (s)| +
|r − t|α
≤ 1 + ε.
Since ε > 0 is arbitrary, it follows that kf kα ≤ 1. Hence A is compact.
3
Chapter 8
6. Write f and g for the two functions. We will use a modified version of the definition
of continuity. Specifically, we will show first that a real-valued function is continuous
if and only if, for any ε > 0 and any x ∈ X, there is an open set U such that x ∈ U
and |f (y) − f (x)| < ε for all y ∈ U .
First, if f is continuous, then (f (x) − ε, f (x) + ε) is an open subset of R, so
U = f −1 (f (x) − ε, f (x) + ε) is an open set and x ∈ U . Moreover, if y ∈ U , then
f (y) ∈ (f (x) − ε, f (x) + ε), so |f (y) − f (x)| < ε.
Conversely, suppose that for any ε > 0 and any x ∈ X, there is an open set U such
that x ∈ U and |f (y) − f (x)| < ε for all y ∈ U . Now let V be an open subset of R
and set O = f −1 (V ). To show that f is continuous, we must show that O is open, so
let x ∈ O. Then f (x) ∈ V , so there is an ε > 0 such that (f (x) − ε, f (x) + ε) ⊂ V .
Then there is an open set Ux ⊂ O with |f (y) − f (x)| < ε for all y ∈ Ux and x ∈ Ux .
Hence x is an interior point of O. Because every point of V is an interior point, it
follows from Proposition 2 that O is open. Therefore f is continuous.
Now, fix x ∈ X and let ε > 0 be given.
First, there are open sets U1 and U2 such that x ∈ U1 and x ∈ U2 with |f (x) −
f (y)| < ε for all y ∈ U1 and |g(x) − g(y)| < ε for all y ∈ U2 . Then U = U1 ∩ U2 is an
open set with x ∈ U , and
|[f + g](x) − [f + g](y)| ≤ |f (x) − f (y)| + |g(x) − g(y)| < ε
for all y ∈ U . Hence f + g is continuous.
Next, there is an open set V such that x ∈ V with |f (x) − f (y)| < 1 and |g(x) −
g(y)| < 1 for all y ∈ V . Also, there is an open set U1 with x ∈ U1 such that
|f (x) − f (y)| < ε/2(|g(x)| + 1) and |g(x) − g(y)| < ε/(2|f (x)| + 1) in U1 . Hence, if
y ∈ U = V ∩ U1 , we have
|f g(x) − f g(y)| ≤ |f (x)g(x) − f (x)g(y) + f (x)g(y) − f (y)g(y)|
≤ |f (x)||g(x) − g(y)| + |g(y)|f (x) − f (y)| < ε.
8. To make the solution easier to read, I will write sets in terms of closure, complement,
and interior. Note that, for any set A, we have Ao = {({A), so we don’t introduce
any new sets with this notation.
(a) The fourteen sets are
E, E o , E o , (E o )o ,
Ē, (Ē)o , (Ē)o ,
o
({E), ({E)o , ({E)o ), (({E)o ) ,
{E, ({E)o , ({E)o .
To see that there are no more sets in the list, we observe first that if you take the
complement of any set in this list, you get a set in the list by using the formula
({A)o = {(Ā) for any set A.
4
Next, if you take the closure of any set in this list, there are three possibilities
to examine. First, if the set is the closure of some set in the list (for example,
E o , then its closure is the same set. Second, if the set is not at the end of a
line in the display and not the closure of a set on the list, then its closure is the
next set on the list. Finally, let’s look at A = (E o )o and its closure Ā. Then
A ⊂ E o , so Ā ⊂ E o . On the other hand, we have E o is an open subset of (E o ),
so E o ⊂ A. Hence E o ⊂ Ā, and therefore Ā = (E o ). A similar argument shows
o
that the closure of (({E)o ) is ({E)o )
(b) There are many different sets E which have these fourteen sets all different.
Here’s one simple choice with a quick verification. Set
1
3
A1 = {(x, y) : < x2 + y 2 ≤ 1, (x, y) 6= (0, )},
4
4
1
2
2
A2 = {(x, y) : x + y < , x ∈ Q},
4
and E = A1 ∪ A2 ∪ {(2, 2)}.
To simplify notation, we write our fourteen sets as E1 , . . . , E14 . First, because
E is bounded and {E is unbounded, we see that Ei 6= Ej if i ≤ 7 and j > 7.
In addition, E is neither open (because (0, 0) is a point of E which is not an
interior point) nor closed (because (0, 21 ) is a point of closure of E which is not
in E. Hence E1 6= Ej if j > 1. It’s also easy to check that the sets that are
interiors of other sets are not closed and that the sets that are closures of other
sets are not open, so we only have to compare the following pairs of sets:
(i) E o and (E o )o ,
(ii) E o and (E)o ,
(iii) (E o )o and (E)o ,
(iv) E o and Ē,
(v) E o and (E)o ,
(vi) Ē and (E)o .
The first two comparisons are easy. (0, 34 ) is not in E and hence not in E o but
it is in (E o )o and (E)o . The sets in the third pair are different because (0, 0) is
in (E)o but not in (E o )o . For the fourth and fifth pairs, we note that (0, 0) is
in Ē and (E)o but not in E o . Finally (2, 2) is in Ē but not in (E)o . (Note that
comparison of sets with subscripts larger than 7 can be done by pointing out
that the complements are set with subscripts less than or equal to 7.)