Page 51 - DMTH503_TOPOLOGY

Unit 4: The Product Topology on X × Y

But  is obtained from the finite intersections of members of . Notes
It follows that  is a sub-base for the product topology T on X × Y.
Theorem 4: The product of two second axiom spaces is a second axiom space.

Proof: Let (X, T ) and (Y, T ) be two second countable spaces.
1 2
Let (X × Y × T) be the product topological space.
To prove that (X × Y × T) is second countable.

Our assumption implies that  countable bases.
B = {B : i  N} and {C : i  N}
1 i i
for X and Y respectively. Recall that
B = {G × G ; G  T , G  T }
1 2 1 1 2 2
is a base for the topology T on X × Y.
Write
C = {B × C i, j  N } = B × B
i j; 1 1 2
B and B are countable = B × B are countable
1 2 1 2
 C is countable
By definition of base B
any (x, y)  N  T   G × H  B s.t. (x, y)  G × H  N …(1)
 x  G  T , y  H  T
1 2
  B  B , C  B s.t. x  B  G, y  C  H
i 1 j 2 i j
This  (x, y)  B , × C  G × H  N.
i j
Thus any (x, y)  N  T   B × C  C s.t. (x, y)  B × B  N. By definition this proves that C is
i j i j
a base for the topology T on X × Y. Also C has been shown to be countable. Hence (X × Y, T) is
second countable.
Theorem 5: The product space of two Hausdorff space is Hausdorff space.

Proof: Let (X, T) be a product topological space of two Hausdorff space (X , T ) and (X , T ).
1 1 2 2
To prove that (X, T) is Hausdorff space.
Consider a pair of distinct elements (x , x ) and (y , y ) in X.
1 2 1 2
Case I. When x = y
1 1
then x  y ,  (x , x )  (y , y )
2 2 1 2 1 2
By the Hausdorff space property, given a pair of elements

x , y  X s.t. x  y , there are disjoint open sets
2 2 2 2 2
G , H  X s.t. x  G , y  H
2 2 2 2 2 2 2
Then X × G and X × H are disjoint open sets in X. for
1 2 2
x  X , x  G  (x , x )  X × G .
1 1 2 2 1 2 1 2
y  X , y  H  (y , y )  X × H .
1 1 2 2 1 2 1 2

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