Page 50 - DMTH503_TOPOLOGY

Topology

Notes Thus, we have shown that
(x , x ) G T
1 2
  B × B   s.t. (x , x )  B × B  G
1 2 1 2 1 2
By definition,
This proves that  is base for T on X.
Remark: From the theorems (1) and (2), it is clear that

 = {B × B : B   , B   },
1 2 1 1 2 2
 = {G × G : G  T , G  T }
1 2 1 1 2 2
both are bases for the same topology T on X.

Theorem 3: Let (X, T) and (Y, ) be any two topological spaces and let L and M be sub-bases for T
1
and  respectively. Then the collection  of all subsets of the form L × Y and X × M, is a sub-base
for the product topology T on X × Y, where L  , M  .
Proof: Now in order to prove that  is a sub-base for T on X × Y, we are to prove that: the
collection  of finite intersections of members of  form a base for T on X × Y.

Since the intersection of empty sub collection of  is X × Y and so X × Y  G.
Next let {L × Y, L × Y, ..., L × Y} {X × M , X × M , ..., X × M } be a non empty finite sub-collection
1 2 p 1 2 q
of . This intersection of these elements belong to , by construction of . This element of  is
(L × Y) (L × Y) ... (L × Y) (X × M ) (X M ) ... (X × M )
1 2 p 1 2 q
= [(L L ... L ) × Y] [X × (M M ... M )]
1 2 p 1 2 q
[For A × (B C) = (A × B) (A C)]

= [(L L ... L ) X] × [(M M ... M ) Y]
1 2 p 1 2 q
[For (A × B) (C × D) = (A C) × (B D)]

= (L L ... L ) × (M M ... M )
1 2 p 1 2 q
[For L  X and M  Y  n]
n n
 p   q 
=  L  M r  ...(1)
r  
 n 1   r 1 


We suppose that  is base for T on X generated by the elements of  and  is a base for  on Y
1
generated by the elements of .
As we know that the finite intersections of sub-base form the base for that topology.
In view of the above statements,
p q
L  M 
r
r

r 1 r 1

From (i), it follows that  is expressible as
 = {B × C : B  , C  }
Then  is a base for the product topology T on X × Y. (Refer Theorem 2).

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