Page 193 - DMTH503_TOPOLOGY

Unit 21: The Tietze Extension Theorem

(ii) Conversely, suppose that (X, T) is a normal space. Notes
Let f : F [a, b] be a continuous map. F being a closed subset of X.
To prove that a continuous extension of f over X. For convenience, we take a = – 1, b = 1
Now we define a map f = F [–1, 1] s.t.
0
f (x) = f(x)  x F.

0
Suppose A and B are two subsets of F. s.t.
0 0
0  1   1 
A = x : f (x)   , B  x : f (x) 
0
0
0
3 3
Then A and B are closed in X.
0 0
For F is closed in X. Applying general from of Urysohn’s lemma,  a continuous function
 1 1  1 1
g : X   ,  s.t. g (A ) =  , g (B ) 
0
0

0
0
0
 3 3 3 3
Write f = f – g
1 0 0
2

Then f (x) = (f  g )(x)  f (x) g (x) 
1
0
0
0
0
3
2 
  1   
,
Let A =  x : f (x)      , 
1 1   3   
 3 
1  1 2 
B = x : f (x) , ,
1
3 3
Then A , B are non-empty disjoint closed sets in X and hence a continuous function s.t.
1 1
 1 2 1 2 
g : X   , , ,
1   3 3 3 3  
1 2 1 2
g (A ) =  , ,g (B )  ,
1
1
1 1 3 3 3 3
Again we define a function f and F s.t.
2
f = f  g  f  g  g  f  (g  g )
2 1 1 0 0 1 0 0 1
2
 2 

Then f (x) = f (x) (g  g )(x)   
2
0
1
0
3 

Continuing this process, we get a sequence of function.
f ,f ,f ,..., f ,...
0
n
2
1
n
 2 
defined on F s.t. f (x)   

3 
n
and a sequence g , g ,g ,....
1
2
0
n
2
1  
defined on X s.t. g (x)  3  
.  
n
3
LOVELY PROFESSIONAL UNIVERSITY 187