Page 188 - DMTH503_TOPOLOGY

Topology

Notes Since continuous map f can be determined corresponding to every element (U, V) of . Take  as
the collection of all such continuous maps.
 is countable   is countable.

To prove that  distinguishes points and closed sets. For this, let H is closed subset of X and
x  X – H.
Now X – H is a nhd of x so that

 B   s.t. x  B  X – H
j j
Regularity of X   G  B s.t. x  G  G  B.
j
By definition of base, we can choose B   s.t. x  B  G
j j
Thus, x  B  B  B  X – H
i
i j
or x  B  B  X – H
i
j
This implies (B , B)  
i j
If f be corresponding member of , then
f (B ) = {0}, f(X – B) = {1}
i
j
B  X – H
j
 H  X – B
j
 f(H)  f(X – B) = {1}
j
 f(H)  {1}

 f(H)  {1}  {1}.

{For {1} is closed in I = [0, 1] for the usual topology on I and so {1}  {1}].

This implies f(H)  {1}  f(H) = {1}

Also f(X – B) = {1}.
j
Hence, f(X – B) = {1} = f(H)
j
Also f(B ) = {0}
i
f(x) = 0  {1} = f(X – B) = f(H)
j
 f(x)  f(H) ...(1)

f(H) is closed subset of X.
Equation (1) shows that  distinguishes points and closed sets. Also, we have seen that  is
countable family of continuous maps f : X  [0, 1].
N
If follows that X can be embedded as a subspace of the Hilbert Cube I which is metrizable.
Also, every subspace of metrizable space is metrizable.

This proves that (X, T) is metrizable.

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