Math 515
Professor Lieberman
December 6, 2004
HOMEWORK #14 SOLUTIONS
Chapter 7
48. (a) Because X is separable, there is a countable dense subset, which we denote by D. Now
for each point x ∈ X, there is an open set Ox with compact closure such that x ∈ Ox .
We will take {On } to be a subcollection of the family of open balls centered at points of
our countable dense subset with rational radii. To decide which ones to use, we fix a point
x ∈ X. Then there is a positive δ(x) such that B(x, δ(x)) has compact closure. Now let
r(x) be a rational number between 0 and δ(x). Then there is a point y(x) ∈ D such that
ρ(x, y(x)) < r(x)/2. Since B(y(x), r(x)/2) is a subset of B(x, δ(x)), it follows that the
closure of B(y(x), r(x)/2) is a subset of the closure of B(x, δ(x)), so it’s compact by Proposition 94. We then take {On } to be the collection of all balls of the form B(y(x), r(x)/2)
for some x ∈ X. Since the number of centers is countable and the number of different radii
is countable, it follows that this collection is countable, and, by construction X = ∪On .
(b) From Problem 7.11a, we know that
σ[f (x), h(x)]
σ[h(x), g(x)]
σ[f (x), g(x)]
≤
+
1 + σ[f (x), g(x)]
1 + σ[f (x), h(x)] 1 + σ[h(x), g(x)]
for any function h : X → Y , so σn∗ satisfies the triangle inequality, and hence so does σ ∗ .
It’s also clear that σ ∗ ≥ 0 and that σ ∗ (f, f ) = 0. If f 6= g, then there is a point x ∈ X such
that σ[f (x), g(x)] > 0. But, by part (a), this x is in some On , so σn∗ (f, g) > 0 and hence
σ ∗ (f, g) > 0.
50. It’s enough to show that this set, which I will label C1α , is sequentially compact, so let
hfn i be a sequence in C α . Then |fn (x)| ≤ 1 for each n ∈ N and x ∈ [0, 1]. In addition,
|fn (x) − fn (y)|/|x − y|α ≤ 1, so, given ε > 0, take δ = ε1/α . If |x − y| < δ, then |fn (x) −
fn (y)| ≤ |x − y|α < ε, so C α is also equicontinuous and Corollary 113 implies that hfn i has
a convergent subsequence, and the sequence converges uniformly because [0, 1] is compact.
Hence the subsequence converges in C[0, 1]. To see that the limit is in C1α , note that f ∈ C1α
if and only if
|f (y) − f (z)|
≤1
|f (x)| +
y − z|α
for all x ,y, and z in [0, 1] with y 6= z. If hfn i is a sequence in C1α which converges pointwise
(which is certainly true if the sequence converges uniformly) to some function f , then we
have
|fn (y) − fn (z)|
|fn (x)| +
≤1
y − z|α
for all n and sending n → ∞ gives
|f (y) − f (z)|
|f (x)| +
≤ 1,
y − z|α
so f ∈ C1α .
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