Page 175 - DMTH503_TOPOLOGY

Unit 18: Normal Spaces, Regular Spaces and Completely Regular Spaces

6. Give an example to show that a normal space need not be a regular. Notes
7. Prove that regularity is a topological property.

18.3 Completely Regular Space

A topological space (X, T) is called a completely regular space if given a closed set FX and a
point xX s.t. xF,a continuous map f : X[0, 1] with the property,
f(x) = 0, f(F) = {1}

Example 7: Every metric space is a completely regular space.
Solution: Let (X, d) be a metric space.

Let aX and F be a closed set in X not containing a.
Define F : X R by
d (x,a)
f(x) = xX,
d(x,a) + d(x,F)

where d(x,F) = inf{d(x, y) : yF},
d(x, F) = 0x F = F,

Consequently d(x, a) + d(x, F)0 as aF.
Thus we see that f (X, R), 0f(x)1 for every xX, f(a) = 0 and f(F) = {1}.
Theorem 3: Every subspace of a completely regular space is completely regular i.e. complete
regularity is hereditary property.
Proof: Let (Y, T ) be a subspace of a completely regular space (X, T).
Y
Let F be a T -closed subset of Y and yY – F. Since F is a T -closed, there exists a T-closed subset
Y Y
F* of X such that
F = YF*

Also yFyYF*
 yF* ( yY)
and yYyX.
It follows that F* is a T-closed subset of X and yX – F*.

Since X is completely regular, there exists a continuous real valved function f : X[0, 1], such
that

f(y) = 0 and f(F*) = {1}.
Let g denote the restriction of f to Y. Then g is a continuous mapping of Y into [0, 1].
Now by the definition of g.

g(x) = f(x)  xY.
Hence f(y) = 0g(y) = 0
and f(x) = 1  xF*

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