Download p. 1 Math 490 Notes 14 We continue our discussion of metrics on

p. 1
Math 490 Notes 14
We continue our discussion of metrics on topological spaces. A topological space (X, τ ) is
defined to be metrizable iff the topology τ is induced by some metric on X. Of our seven
“standard” topologies on R, only two are metrizable; namely, τu and τd . τu is induced by
the “usual metric” d(x,
 y) = |x − y|, and τd is induced by the “discrete metric”
 1 if x 6= y
d(x, y) =
.
 0 if x = y
Indeed, the discrete metric is a well-defined metric which induces the discrete topology on
any set X, so all discrete topological spaces are metrizable. Of the remaining five topologies
on R, τcoc , τcof , τo and τi all fail to be T2 , and τcoc and τcof fail to be first countable as well.
The Sorgenfrey topology τs is both T2 and first countable, but it fails to be metrizable for
other reasons (to be given soon).
Prop N14.1 Metrizability is a topological property.
Proof : Let h : (X, τ ) → (Y, µ) be a homeomorphism, and assume (X, τ ) is metrizable by
some metric d. Then define d ′ : Y × Y → [0, ∞) by d ′ (y1 , y2 ) = d(h−1 (y1 ), h−1 (y2 )). Then
(as you may verify) d ′ is a metric on Y , and d ′ does induce the topology µ because the
homeomorphism h must map a basis for τ to a basis for µ, and it maps the τ -basis element
¯
¯
B(x0 , ǫ) = {x ∈ X ¯ d(x, x0 ) < ǫ} to the set B(f (x0 ), ǫ) = {y ∈ Y ¯ d ′ (y, f (x0 )) < ǫ}. So,
¯
denoting f (x0 ) by y0 , all the sets of the form B(y0 , ǫ) = {y ∈ Y ¯ d ′ (y, y0 ) < ǫ} must form a
basis for µ, which means µ is induced by the metric d ′ , and is hence metrizable.
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Prop N14.2 In a metric space (X, d), the metric d : X ×X → [ 0, ∞) is a continuous function.
Proof : Before beginning the proof, note that the topology on X × X is the product topology
of (X, τ ) × (X, τ ), where τ is the topology on X induced by d, and the topology on [0, ∞)
is that inherited (as a subspace) from (R, τu ).
p. 2
Proceeding with the proof, let (x1 , x2 ) ∈ X × X, with d(x1 , x2 ) = γ > 0. Any nbhd
of γ in [0, ∞) contains a basic nbhd of the form V = (γ − ǫ, γ + ǫ) with ǫ < γ. Then
¡
¡
ǫ¢
ǫ¢
× B x2 ,
is a nbhd of (x1 , x2 ) in X × X.
U = B x1 ,
2
2
Claim: d(U ) ⊆ V = (γ − ǫ, γ + ǫ).
¡
¡
ǫ¢
ǫ¢
To see this, note that for any (x, y) ∈ U , we have x ∈ B x1 ,
and y ∈ B y1 ,
, and so
2
2
(by the properties of a metric):
ǫ
ǫ
+γ+
= γ + ǫ;
(1) d(x, y) ≤ d(x, x1 ) + d(x1 , x2 ) + d(x2 , y) ≤
2
2
ǫ
ǫ
+ d(x, y) +
⇒ γ − ǫ < d(x, y).
(2) γ = d(x1 , x2 ) ≤ d(x1 , x) + d(x, y) + d(y, x2 ) ≤
2
2
This claim shows that d is continuous at any (x1 , x2 ) ∈ X × X with x1 6= x2 .
If x1 = x2 , then γ = 0, and any nbhd of γ then contains a basic nbhd of the form V = [0, ǫ).
Letting U be as above, then for any (x, y) ∈ U , (1) can be rewritten as
ǫ
ǫ
+γ+
= γ + ǫ,
0 ≤ d(x, y) ≤ d(x, x1 ) + d(x1 , x2 ) + d(x2 , y) ≤
2
2
which shows that we still have d(U ) ⊆ V . So d is continuous at any (x1 , x2 ) ∈ X × X.
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If (X, τ ) is any topological space and A, B are subsets of X, we define A to be dense in B
iff A ⊆ B ⊆ Clτ A. Note that A is dense in X iff Clτ A = X, in which case we simply say
that A is dense. A topological space which contains a countable, dense subset is said to be
separable.
Of the seven standard topologies on R, the set Q of rationals is dense in all but two: τd and
τcoc . Thus, the remaining five (τs , τu , τcof , τo and τi ) are separable. There is no countable
dense set in R relative to τd or τcoc (why not?), so these two topologies are not separable.
Since (R, τd ) is metrizable, it follows that metrizable spaces need not be separable.
Prop N14.3 Let (X, τ ) be a topological space.
(a) If (X, τ ) is second countable, then (X, τ ) is separable.
(b) If (X, τ ) is metrizable, then (X, τ ) is second countable iff (X, τ ) is separable.
p. 3
Before proving Prop N14.3, we state the following Lemma, whose proof is left as an exercise.
Lemma N14.1: Let (X, τ ) be a topological space, let B be any basis for τ , and let D ⊆ X.
Then D is dense (in X) iff whenever φ 6= B ∈ B, D ∩ B 6= φ.
Proof of Prop N14.3 :
(a) Assume (X, τ ) is 2nd countable. Then τ has a countable basis of non-empty sets
¯
{Un ¯ n ∈ N}. By the Axiom of Choice, we can choose an xn from each nonempty Un . By
¯
the previously stated lemma, the set D = {xn ¯ n ∈ N} is a countable, dense subset of X,
and therefore (X, τ ) is separable.
(b) Assume (X, τ ) is metrizable with metric d. By part (a), if (X, τ ) is 2nd countable,
then (X, τ ) must be separable. It remains to show that if (X, τ ) is separable, then it is 2nd
countable.
¯
So assume (X, τ ) is separable (in addition to being metrizable). Let D = {xn ¯ n ∈ N} be a
¯
countable dense subset of X. For any k ∈ N, define the collection Bk = {B(xn , k1 ) ¯ n ∈ N},
[
and let B =
Bk . Note that B is just the countable collection of all balls of radius 1/k (for
k∈N
all k ∈ N) centered at every xn ∈ D.
Claim: B is a basis for τ .
To prove this, let x ∈ U ∈ τ . Then ∃ǫ > 0 such that B(x, ǫ) ⊆ U . Choose k ∈ N such
that
1
k
<
ǫ
2
. Since D is dense and B(x, k1 ) is an element of a known basis for τ , the lemma
tells us that there must be an xn ∈ D ∩ B(x, k1 ). Now, B(xn , k1 ) is a nbhd of x, since
xn ∈ B(x, k1 ) ⇒ d(xn , x) < 1/k ⇒ x ∈ B(xn , k1 ). Also, B(xn , k1 ) is contained by U , since
y ∈ B(xn , k1 ) ⇒ d(x, y) ≤ d(xn , y) + d(xn , y) <
1
k
+
1
k
<
ǫ
2
+
ǫ
2
= ǫ ⇒ y ∈ B(x, ǫ) ⊆ U .
Therefore B is a countable basis for τ , which makes (X, τ ) 2nd countable.
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p. 4
Remark: We showed previously that (R, τs ) is separable but not 2nd countable. Thus (R, τs )
is not metrizable. Recall that a hereditary property is one that is preserved by subspaces.
Prop N14.4 Metrizability is hereditary, but separability is generally not. However, any subspace of a separable, metrizable space is separable.
Proof : If (X, τ ) is metrizable by some metric d and (A, τA ) is a subspace, then the metric
d|A×A = dA defined by dA (x, y) = d(x, y) for all x, y ∈ A is a metric on A which induces the
topology τA on A (verify).
To prove the last assertion in Prop N14.4, let (X, τ ) be separable and metrizable. then
by Prop N14.3, (X, τ ) is second countable. Since second-countability and metrizability are
hereditary, any subspace of (X, τ ) is second countable, and hence, by Prop N14.3(a), separable.
To show that separability is not hereditary in general, consider the product space (R, τs ) ×
(R, τs ), for which basic open sets in the product topology have the form [a, b) × [c, d). Q × Q
is dense in (R, τs ) × (R, τs ), so this product space is separable. Now consider the subset
¯
A = {(x, −x) ∈ R2 ¯ x ∈ R}. Then (A, τA ) is an uncountable discrete space, which is not
separable.
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