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

Phichet Chaoha 37
Lemma 2.32. A metric space is normal.
Proof. Let A and B be disjoint closed subsets of metric space X. For each
a ∈ A and each b ∈ B, let ϵ a > 0 and ϵ b > 0 be such that
B(a; ϵ a ) ∩ B = ∅ and B(b; ϵ b ) ∩ A = ∅.
Now, let U = ∪ a∈A B(a; ϵ a
2
), and V = ∪ b∈B B(b; ϵ b
2
). Clearly, U and V are open sets
containing A and B, respectively. To see that U ∩V = ∅, suppose there is z ∈ U ∩V .
Then, z ∈ B(a; ϵ a
2
) ∩ B(b; ϵ b
2
) for some a ∈ A and b ∈ B. WLOG, we may assume
that ϵ b ≤ ϵ a . So, we have d(a, b) ≤ d(a, z) + d(z, b) < ϵa 2 + ϵ b
2
≤ ϵa 2 + ϵa 2 = ϵ a which
is impossible because B(a; ϵ a ) ∩ B = ∅.
□
Corollary 2.33. A metric space is completely normal.
Proof. Since every subspace of a metric space X is also a metric space, and
hence normal, then X itself is completely normal.
□
Theorem 2.34. A second-countable regular space is normal.
Proof. Let A and B be disjoint closed subsets of a second-countable regular
space X, and B a countable basis of X. By regularity, we can find countable
subcollections {U 1 , U 2 , . . . } and {V 1 , V 2 , . . . } of B such that A ⊆ ∪ ∞ i=1 U i, B ⊆
∪ ∞ i=1 V i, U n ∩ B = ∅ and V n ∩ A = ∅ for all n ∈ N.
Since ∪ ∞ i=1 U i and ∪ ∞ i=1 V i may not be disjoint, then for each n ∈ N, we let
U n ′ = U n − ∪ n i=1 V i and V n ′ = V n − ∪ n i=1 U i. Clearly, U n’s ′ and V n’s ′ are open and
pairwise disjoint. Now, it is rather straightforward to check that U ′ = ∪ n≥1 U n ′ and
V ′ = ∪ n≥1 V n ′ are disjoint neighborhoods of A and B, respectively.
□
Theorem 2.35 (Urysohn Lemma). Let A and B be disjoint closed subsets of a
normal space X, and [a, b] a closed in interval in R. Then there exists a continuous
map f : X → [a, b] such that A ⊆ f −1 ({a}) and B ⊆ f −1 ({b}).
Proof. WLOG, we can assume that A ≠ ∅ ̸= B, a = 0 and b = 1.
We will construct a continuous map f : X → [0, 1] with the property that
f(A) = {0} and f(B) = {1}. First let P = Q ∩ [0, 1] and fix a representation
of P in such a way that the first two elements are 1 and 0 (for example, P =
{1, 0, 1 2 , 1 3 , 2 3 , . . . }). Let P n be the subset of the first n elements of P according to
the representation we have just fixed.
First let U 1 = X − B and, by normality, let U 0 be an open set such that
A ⊆ U 0 ⊆ U 0 ⊆ U 1 . Hence, we have defined U i for all i ∈ P 2 . Inductively, we
assume that for some n ≥ 2, U i is defined for all i ∈ P n so that U i ⊆ U j for i < j.
Now, suppose P n+1 = P n ∪ {r}. Since r /∈ {0, 1}, we can find in P n the immediate
predecessor and the immediate successor of r (according to the usual ordering of
R), says p and q, respectively. Again, by normality, there exists an open set U r
such that U p ⊆ U r ⊆ U r ⊆ U q ; i.e., U i is now defined for all i ∈ P n+1 . Hence, we
obtain the collection {U i } i∈P with the desired property. Moreover, we can extend
the definition of U i to all i ∈ Q simply by letting
{
∅ for i ∈ Q ∩ (−∞, 0)
U i =
X for i ∈ Q ∩ (1, ∞).
Finally, we define f : X → [0, 1] by
f(x) = inf{i | x ∈ U i }