Page 183 - DMTH503_TOPOLOGY

Unit 19: The Urysohn Lemma

19.1.2 Solved Examples Notes

Example 1: If F and F are T-closed disjoint subsets of a normal space (x, T), then there
1 2
exist a continuous map g of X into [0, 1] such that


 0 if x F 1
g(x) = 

 1 if x F 2
f(F ) = {0} and g(F ) = {1}.
1 2
Solution: Here write the proof of step II of the Urysohn’s Theorem.

Example 2: If F and F are T-closed disjoint subsets of a normal space (X, T) and [a, b] is
1 2
any closed interval on the real line, then there exists a continuous map f of X into [a, b] such that

 a if x F 1
f(x) = 

 b if x F 2
i.e., f(F ) = {a}, f(F ) = {b}
1 2
This problem in known as general form of Urysohn’s lemma.
Solution: Let F and F be disjoint closed subset of (X, T).
1 2
To prove that  a continuous map
f : X  [a, b] s.t. f(F ) = {a}, f(F ) = {b}
1 2
By Urysohn’s lemma,  a continuous map

g : X  [0, 1] s.t. g(F ) = {0}, g(F ) = {1}.
1 2
Define a map h : [0, 1]  [a, b] s.t.

(b a)x

h(x) =  a
1 0

i.e., h(x) = x(b – a) + a
[This is obtained by writing the equation of the straight line joining (0, a) and (1, b) and then
putting y = h(x)].
Evidently h(0) = a, h(1) = b – a + a = b

Also h is continuous
Write f = hg

g : X  [0, 1], h : [0, 1]  [a, b]
 hg : X  [a, b]  f : X  [a, b]
Product of continuous functions is continuous

Therefore f(F ) = (hg)(F ) = h[g(F )] = h({0}) = {a}
1 1 1
f(F ) = (hg)(F ) = h[g(F )] = h({1}) = {b}
2 2 2

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