Lecture Notes Topology (2301631) Phichet Chaoha Department of ...

30 Topology (2301631)
d x
X ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣✲♣ ♣ R + 0
◗
ι ◗◗ ✑ ✑✸
d {x}×X
{x} × X
where ι(y) = (x, y) for each y ∈ X. Clearly, d x is continuous and hence B d (x; r) =
d −1
x ([0, r)) is open in X; i.e., B d (x; r) ∈ τ. □
Theorem 1.142. The converse of the sequence lemma holds in a metric space.
Proof. Let (X, d) be a metric space, A ⊆ X and a ∈ A. Then, we have
B d (a; 1 n )∩A ≠ ∅ for each n ∈ N. By the axiom of choice, we obtain a sequence (a n)
in A such that a n ∈ B d (a; 1 n ) ∩ A for each n ∈ N. To show that (a n) converges to a,
we let U be an neighborhood of a. Since the metric topology on X is generated by
open balls, there exits ϵ > 0 such that B d (a; ϵ) ⊆ U. By choosing N ∈ N such that
1
N < ϵ, we have a n ∈ B d (a; 1 n ) ∩ A ⊆ B d(a; 1 N ) ∩ A ⊆ B d(a; ϵ) ∩ A ⊆ U ∩ A ⊆ U for
all n ≥ N. This proves that (a n ) → a as desired.
□
Theorem 1.143. Let f : X → Y be a map of spaces, where X satisfies the
converse of the sequence lemma. Then, f is continuous iff for any x ∈ X and any
sequence (x n ) converging x, the sequence (f(x n )) converges to f(x).
Proof. (⇒) Follows from Exercise 1.95(1).
(⇐) Let A ⊆ X and y ∈ f(A). Then y = f(x) for some x ∈ A. Since X
satisfies the converse of the sequence lemma, there exists a sequence (a n ) in A
converging to x. By the assumption, the sequence (f(a n )) (in f(A)) converges
to f(x). Therefore, by the sequence lemma, y = f(x) ∈ f(A). It follows that
f(A) ⊆ f(A) which implies the continuity of f.
□
Corollary 1.144. The previous theorem is true when X is metrizable.
Example 1.145. The space S Ω is not metrizable because it does not satisfy
the converse of the sequence lemma.
Definition 1.146. Two metrics d 1 and d 2 on a set X are said to be (topologically)
equivalent if d 1 and d 2 induce the same metric topology on X.
Exercise 1.147. Let d and ρ be metrics on a set X. If there exist positive real
numbers a, b and c such that for any x, y ∈ X,
prove that d and ρ are equivalent.
ad(x, y) ≤ bρ(x, y) ≤ cd(x, y),
Exercise 1.148. On R n , we can define the square metric ρ and the taxicab
metric d t by
• ρ(x, y) = max{|x 1 − y 1 |, . . . , |x n − y n |}
• d t (x, y) = ∑ n
i=1 |x i − y i |
for any x, y ∈ R n . Prove that ρ and d t are really metrics on R n , and both are
equivalent to the euclidean metric d. [Hint : use the previous exercise]
Unfortunately, for the metrizability of R ω , we do not have obvious generalizations
of the d or d t because of the convergence issue. However, we can modify ρ to
get the bounded version by letting
ρ(x, y) = sup{d(x i , y i ) | i ∈ N}