1. Urysohn Lemma
Class 15: Urysohn Lemma
Sebastian Vattamattam
23. November 2010
Sebastian Vattamattam Topology

2. Urysohn Lemma
Definition
X is normal if for disjoint closed sets A, B,
subsets of X, there exist disjoint open sets
U, V such that
A ⊂ U, B ⊂ V
Lemma
If X is normal, A is closed in X and open set
U contains A, then there exists open set V
such that
A ⊂ V ⊂ ¯V ⊂ U
Sebastian Vattamattam Topology

3. Urysohn Lemma
Theorem
Thm33.1 in [1]
Urysohn Lemma
X is normal, A, B disjoint, closed in X. [a, b]
a closed interval in R
Then there exists f : X → [a, b] such that
1
f is continuous.
2
f −1
(a) = A, f −1
(b) = B
Proof
We take a = 0, b = 1
Sebastian Vattamattam Topology

4. Urysohn Lemma
1
Step 1: Let P := Q [0, 1]
To show that, for p, q ∈ P, p < q
there exist open sets Up, Uq such
that
¯Up ⊂ Uq
Proof by induction
Since P is countable, let P = (xn), an
infinite sequence, where x1 = 1, x2 = 0.
For n ∈ N, let
Pn := {xk|k ≤ n}
Sebastian Vattamattam Topology

5. Urysohn Lemma
1 P1 = {1} and the result is vacuously true.
For P2 = {1, 0} , 0 < 1, let
U1 = X − B
U1 is an open neighborhood of A
By lemma 1.2, there is an open set U0 such that
A ⊂ U0 ⊂ ¯U0 ⊂ U1
Thus the result is true on P2
2 Suppose the result is true for p, q ∈ Pn, p < q
That is, there exist open sets Up, Uq such that
¯Up ⊂ Uq
Let xn+1 = r and Pn+1 := Pn {r}
Pn+1 is a finite subset of [0, 1]
With the usual order relation on [0, 1], Pn+1 is simply ordered.
Since x1 = 1, x2 = 0 ∈ Pn+1, 0 = min Pn+1, 1 = max Pn+1
In Pn+1 every number, other than 0 and 1, has an immediate
predecessor and an immediate successor.
Sebastian Vattamattam Topology

6. Urysohn Lemma
Since r /∈ {0, 1}, let p be the immediate predecessor and q be
the immediate successor of r in Pn+1
Since p < q in Pn, by hypothesis, there exist open sets Up, Uq
such that
¯Up ⊂ Uq (1.1)
Again by lemma 1.2, there exist open set Ur in X such that
¯Up ⊂ Ur ⊂ ¯Ur ⊂ Uq (1.2)
To show that the result holds for each pair of elements in Pn+1
By the hypothesis, it holds for every pair of elements in Pn
Consider the pair (r, s), where s ∈ Pn
If s ≤ p, then by 1.1 and 1.2, ¯Us ⊂ Up ⊂ ¯Up ⊂ Ur
If s ≥ q then r < q ≤ s and hence by 1.1 and 1.2,
¯Ur ⊂ Uq ⊂ Us
Thus for every pair of elements of Pn+1, the result holds.
The conclusion follows by induction.
2
Step 2
Sebastian Vattamattam Topology

7. Urysohn Lemma
To extend the result in Step 1 to all
p ∈ Q
Let
Up =
φ if p < 0;
X if p > 1
Now again, for p, q ∈ Q, p < q
¯Up ⊂ Uq (1.3)
3
Step 3
For x ∈ X, let
Q(x) := {p ∈ Q|x ∈ Up}
Sebastian Vattamattam Topology

8. Urysohn Lemma
If p < 0, Up = φ and hence
0 ≤ p ∀p ∈ Q(x)
Define
f : X → [0, 1], f (x) = inf Q(x)
= inf {p|x ∈ Up}
4
Step 4
To show that
1 f −1
(a) = A, f −1
(b) = B
2 f is continuous.
1 Let a ∈ A
Since A ⊂ U0 ⊂ ¯U0 ⊂ Up, ∀p ≥ 0
a ∈ Up, ∀p ≥ 0
Sebastian Vattamattam Topology

9. Urysohn Lemma
⇒ f (a) = inf Q(a) = 0, ∀a ∈ A
Let b ∈ B ⇒ b /∈ X − B = U1 ⊃ Up, ∀p ≤ 1
⇒ b /∈ Up, ∀p ≤ 1
f (b) = inf Q(b) = 1
2 To prove
x ∈ ¯Ur ⇒ f (x) ≤ r (1.4)
x ∈ Ur ⇒ f (x) ≥ r (1.5)
1 Let x ∈ ¯Ur
For r < s, x ∈ ¯Ur ⊂ Us
⇒ x ∈ Us , ∀s > r
Therefore
Q(x) ⊃ {p ∈ Q|p > r}
⇒ f (x) = inf Q(x) ≤ r
Sebastian Vattamattam Topology

10. Urysohn Lemma
2 Let x /∈ Ur
For s < r, Us ⊂ ¯Us ⊂ Ur
So, x /∈ Us , ∀s < r
f (x) = inf Q(x) ≥ r
3 To prove f is continuous.
Let x0 ∈ X, and c < f (x0) < d
Let p, q ∈ Q such that
c < p < f (x0) < q < q
Then ¯Up ⊂ Uq
Let
U := Uq − ¯Up, open
f (x0) < q ⇒ inf Q(x0) < q
⇒ x0 ∈ Uq
f (x0) > p ⇒ inf Q(x0) > p
⇒ x0 /∈ Up ⊂ ¯Up
Thus
x0 ∈ U
Sebastian Vattamattam Topology

11. Urysohn Lemma
To show that
f (U) ⊂ (c, d)
x ∈ U ⇒ x ∈ Uq ⊂ ¯Uq
By (1.5)
f (x) ≤ q
x /∈ ¯Up ⇒ x /∈ Up
By (1.6)
f (x) ≥ p
Thus
f (x) ∈ [p, q] ⊂ (c, d)
f (U) ⊂ (c, d)
Therefore f is continuous.
vattamattam@gmail.com
Sebastian Vattamattam Topology

12. Urysohn Lemma
James R. Munkres,Topology, Second
Edition, Prentice-Hall of India, New Delhi,
2002.
Sebastian Vattamattam Topology