Page 180 - DMTH503_TOPOLOGY

Topology

Notes Proof:
(1) Let R denote the set of all real numbers lying in the closed interval [0, 1] with usual topology.
Let (X, T) be a topological space and let given a pair of disjoint closed sets A, B  X ;  a
continuous map f : X  R s.t.
f(A) = {0}, f(B) = {1}.
To prove that (X, T) is a normal space.
Let a, b  R be arbitrary s.t. a  b

write G = [0, a), H = (b, 1].
Then G and H are disjoint open sets in R.
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Continuity of f implies that f (G) and f (H) are open in X.
Then our assumption says that
f(A) = {0}, f(B) = {1}
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f(A) = {0}  f ({0} = f (f(A))  A
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 f ({0})  A  A  f ({0})
Similarly B  f {1}.
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Evidently
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{0}  [0, a)  f ({0})  f ([0, a))
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 A  f ({0})  f ([0, a)]
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 A  f ([0, a))  A  f (G)
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{1}  (b, 1]  B  f ({1})  f ((b, 1])
 B  f ((b, 1])  B  f (H)
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f (G) f (H) = f (G H) = f () = 
Given a pair of disjoint closed sets A, B  X, we are able to discover a pair of disjoint open sets,
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f (G), f (H)  X s.t. A  f (G), B  f (H).
This proves that (X, T) is a normal space.
(2) Conversely, suppose that R is a set of real numbers lying in the interval [0, 1] with usual
topology. Also suppose that A, B are disjoint closed subsets of a normal space (X, T).
To prove that  a continuous map.
f : (X, T)  R s.t. f(A) = {0}, f(B) = {1}.
Step (i): Firstly, we shall prove that  a map
f : (X, T)  R s.t. f(A) = {0}, f(B) = {1}.
 m n 

Write T  t : t  ,where m,n N s.t. m  2
2 n
Throughout the discussion we treat t  T.
n
Making use of the fact that m takes 2 values for a given value of n, we have
 m  1 2 n
sup(T) = sup(t) = sup  n   sup(m)   1 sup(T)  1
 2  2 n 2 n

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