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Urysohn Lemma
Class 15: Urysohn Lemma
Sebastian Vattamattam
23. November 2010
Sebastian Vattamattam Topology
Urysohn Lemma
DefinitionX 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
LemmaIf 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 ⊂ USebastian Vattamattam Topology
Urysohn Lemma
TheoremThm33.1 in [1]
Urysohn LemmaX is normal, A,B disjoint, closed in X . [a, b]
a closed interval in RThen 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
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 suchthat
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
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 ABy 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 < qThat 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 = minPn+1, 1 = maxPn+1
In Pn+1 every number, other than 0 and 1, has an immediatepredecessor and an immediate successor.
Sebastian Vattamattam Topology
Urysohn Lemma
Since r /∈ {0, 1}, let p be the immediate predecessor and q bethe 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 2Sebastian Vattamattam Topology
Urysohn Lemma
To extend the result in Step 1 to all
p ∈ QLet
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
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 that1 f −1(a) = A, f −1(b) = B2 f is continuous.
1 Let a ∈ ASince A ⊂ U0 ⊂ U0 ⊂ Up,∀p ≥ 0
a ∈ Up,∀p ≥ 0Sebastian Vattamattam Topology
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
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) < dLet 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 ∈ USebastian Vattamattam Topology
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.
Sebastian Vattamattam Topology
Urysohn Lemma
James R. Munkres,Topology, Second
Edition, Prentice-Hall of India, New Delhi,
2002.
Sebastian Vattamattam Topology