Page 181 - DMTH503_TOPOLOGY

Unit 19: The Urysohn Lemma

1 Notes
t  ,1, for n  1
2
1 2 3 4
t  , , , , for n  2
2 2 2 2 2 2 2 2
A B =   A  X – B

X – B is an open set containing a closed set A. Using the normality, we can find an open set
G  X s.t.

A  G  G  X – B ...(1)

Writing G = H , X – B = H , we get
1/2 1
A  H  H 1/2  H
1/2 1
This is the first stage of our construction

Consider the pairs of sets (A, H ), ( H 1/2 , H )
1/2 1
Using normality, we obtain open sets, H , H  X s.t.
1/4 3/4
A  H  H 1/4  H
1/4 1/2
H 1/2  H  H 3/4  H
3/4 1
Combining the last two relations, we have

A  H  H 1/4  H  H 1/2  H  H 3/4  H
1/4 1/2 3/4 1
This is the second stage of our construction.

n
If we continue this process of each dyadic rational m of the function t = m/2 , where
n
n = 1, 2, ... and m = 1, 2 ... 2 – 1,
Then open sets H will have the following properties:
t
(i) A  H  H  H  t  T
t
t 1
(ii) Given t , t  T s.t.
1 2
t < t  A  H  1 t H  H  2 t H  H
1 2 1 t 2 t 2
Construct a function f : X  R s.t. f(x) = 0  x  H
t
and f(x) = {t : x  H } otherwise
t
In both cases x  X.

f(x) = sup{t : x  H } = sup{t} = sup(T) = 1
t
Thus f(x) = 0  x  H
t
and f(x) = 1  x  H otherwise
t
f(x) = 0  x  H , A  H  t  T  f(x) = 0  x  A  f(A) = {0}
t t
f(x) = 1  x  H , H  H  t  T  f(x) = 1  x  H
t t 1 1

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