Page 176 - DMTH503_TOPOLOGY

Topology

Notes and FF* g(x) = f(x) = 1  xF
 g(F) = {1}.
Hence for every T -closed subset F of Y and for each point yY – F, there exists a continuous
Y
mapping g of Y into [0, 1] such that
g(y) = 0 and g(F) = {1}.
Hence (Y, T ) is also completely regular.
Y
Theorem 4: A completely regular space is regular.
Proof: Let (X, T) be a completely regular space, then given any closed set FX and pX s.t.
pF; continuous map f : X[0, 1] with the property that

f(p) = 0, f(F) = {1}.
To prove that (X, T) is a regular space.

Consider the set [0, 1] with usual topology. It is easy to verity that [0, 1] is a T -space, then we can
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find out disjoint open sets G, H in [0, 1] s.t. 0G, 1H.
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By hypothesis, f is continuous, hence f (G), f (H) are open in X.
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f (G)f (H) = f (HG) = f (), = 
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f (G) = {xX : f(x)G}.
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Furthermore, f(p) = 0Gf(p)Gpf (G)
f(F) = 1Hf(F) ={1}H
 f(F) H
 F f (H).
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Given any closed set F X and p X s.t. p F; disjoint open sets f (G), f (H) in X s.t. p f (G),
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F f (H), in X s.t. p  f (G), F  f (H), showing thereby X is regular.
Theorem 5: A Tychonoff space is a T -space. Or Completely regular space regular space.
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Proof: Let (X, T) be a Tychonoff space, then
(i) X is a T -space
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(ii) X is a completely regular space.
To prove that (X, T) is a T -space, it suffices to show that
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(iii) X is a T -space.
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(iv) X is a regular space

Evidently (i) (iii)
Prove as in Theorem (1)
Hence the result.

Example 8: Prove that a topological space (X, T) is completely regular iff for every xX
and every open set G containing x there exists a continuous mapping f of X into [0, 1] such that
f(x) = 0 and f(Y) = 1  y  X – G.

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