Page 55 - DMTH503_TOPOLOGY

Unit 4: The Product Topology on X × Y

 (x , x )   (U × U )   (G) Notes
2 1 2 2 1 2 2
 x  U   (G) and x  U   (G) ...(1)
1 1 1 2 2 2
For  (U × U ) = { (x , x ) : (x , x )  U × U }
1 1 2 1 1 2 1 2 1 2
= {x : x  U , x  U }
1 1 1 2 2
= U
1
Similarly,  (U × U ) = U
2 1 2 2
( o f)(y) =  (f(y))
1 1
=  (x , x )
1 1 2
= x
1
Similarly, ( o f) (y) = x
2 2
Thus, ( o f)(y) = x , ( o f)(y) = x
1 1 2 2
In this event (1) takes the form

( o f)(y)  U   1 (G)ü

1
1
ý ...(2)

( o f)(y)  U   (G)
1 2 2 þ
This y  ( o f) (U ) and
–1
1 1
–1
y  ( o f) (U )
2 2
–1
–1
 y  [( o f) (U )] [( o f) (U )] ...(3)
1 1 2 2
–1
 o f,  o f are given to be continuous and hence ( o f) (U ) and ( o f) (U ) are open
–1
1 2 1 1 2 2
in Y.
–1
 [( o f) (U ) [( o f) (U )] is open in Y.
–1
1 1 2 2
–1
–1
On taking ( o f) (U ) = V , [ o f) (U ) = V .
1 1 1 2 2 2
We have V V as an open set in Y.
1 2
According to (3), y  V V = V (say)
1 2
any v  V  v  V and v  V
1 2
–1
 v  ( o f) (U ), v = ( o f) (U )
–1
1 1 2 2
 ( o f) (v)  U , ( o f) (v)  U
1 1 2 2
 ( o f) (v)  U   (G) and
1 1 1
( o f) (v)  U   (G) [from (2)]
2 2 2
 v  ( o f) [ (G)] and v  ( o f) [ (G)]
–1
–1
1 1 2 2

 v  (f o  ) [ (G)] and
1
–1
1 1

1
–1
v  (f o  ) [ (G)]
2 2
–1
 v  f (G) and V  f (G)
–1
–1
 any v  V  v  f (G)
 V  f (G)
–1
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