Page 49 - DMTH503_TOPOLOGY

Unit 4: The Product Topology on X × Y

So, we have proved that Notes
 W × W  B s.t (x , x )  W × W
1 2 1 2 1 2
Now, it remains to prove that
W × W  (U × U ) (V × V )
1 2 1 2 1 2
Let (y , y )  W × W be arbitrary.
1 2 1 2
(y , y )  W × W  y  W , y  W
1 2 1 2 1 1 2 2
 y  U V , y  U V
1 1 1 2 2 2
 y  U , y  V and y  U , y  V
1 1 1 1 2 2 2 2
 (y , y )  U × U and (y , y )  V × V
1 2 1 2 1 2 1 2
 (y , y )  (U × U ) (V × V )
1 2 1 2 1 2
Finally, any (y , y )  W × W
1 2 1 2
 (y , y )  (U × U ) (V × V )
1 2 1 2 1 2
This proves that
W × W  (U × U ) (V × V )
1 2 1 2 1 2
It immediately follows from (i) and (ii) that B is a base for some topology, say, T on X.
Theorem 2: Let (X , T ) and (X , T ) be two topological spaces and let  ,  be bases for T and T
1 1 2 2 1 2 1 2
respectively.
Let X = X × X
1 2
Then  = { ×  : B   , B   } is a base for the product topology T on X.
1 2 1 1 2 2
Proof: Let  = {G × G : G  T , G  T }.
1 2 1 1 2 2
Then  is a base for the topology T on X (refer theorem 1)

We are to prove that  is a base for T on X.
By definition of base,
for (x , x )  G  T
1 2
  G × G   s.t. (x , x )  G × G  G ...(1)
1 2 1 2 1 2
Again (x , x )  G × G  C
1 2 1 2
 x  G  T , x  G  T .
1 1 1 2 2 2
Applying definition of base,
x  G  T   B   , s.t. x  B  G ...(2)
1 1 1 1 1 1 1 1
x  G  T   B   , s.t. x  B  G ...(3)
2 2 2 2 2 2 2 2
B  , B    B × B  .
1 2 2 1 2
Now (2) and (3)   B × B  B s.t.
1 2
(x , x )  B × B  G × G  G
1 2 1 2 1 2
or (x , x )  B × B  G
1 2 1 2

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