Page 53 - DMTH503_TOPOLOGY

Unit 4: The Product Topology on X × Y

Then  and  both are called projection maps on the first and second coordinate spaces Notes
1 2
respectively.
Step (i): To prove: projection maps are continuous maps.

Firstly, we shall show that  is continuous.
1
Let G  X be an arbitrary open set.
1
  1 (G) = {(x , x )  X :  (x , x )  G}
1 1 2 1 1 2
= {(x , x )  X : x  G}
1 2 1
= {(x , x )  X × X : x  G}
1 2 1 2 1
= G × X
2
= An open set in X.
For G is open in X , X is open in X
1 2 2
 G × X is open in X.
2
   1 (G) is open in X.
1
Thus, we have prove that
G
any open set G  X    1   is open in X.
1 1
  is continuous.
1
Similarly, we can prove that  is continuous map. Consequently, projection maps are continuous
2
maps.
Step (ii): To prove that projection maps are open maps. We shall first show that  is an open map.
2
Let G  X be an arbitrary open set.
Let x   [G] be arbitrary.
2 2
x   [G]  (u , u )  G s.t.  (u , u ) = x
2 2 1 2 2 1 2 2
 u = x [  (u , u ) = u ]
2 2 2 1 2 2
Now (u , x )  G
1 2
Let  be the base for the topology T on X.
By definition of base,
(u , x )  G  T   U × U   s.t. (u , x )  U × U  G
1 2 1 2 1 2 1 2
  (u , x )   (U × U )   (G)
2 1 2 2 1 2 2
 x   (U × U )   (G)
2 2 1 2 2
 x  U   (G).
2 2 2
For  (U × U ) = { (x , x ) : (x , x )  U × U }
2 1 2 2 1 2 1 2 1 2
= {x : x  U , x  U } = U .
2 1 1 2 2 2
 Given any x   [G]   open set U  X s.t. x  U   [G].
2 2 2 2 2 2 2
This proves that x is an interior point of  [G]. But x is an arbitrary point of  [G].
2 2 2 2

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