Page 54 - DMTH503_TOPOLOGY

Topology

Notes  Every point of  [G] is an interior point.
2
This proves that  [G] is open in X .
2 2
 Any open set G  X

  (G) is open in X .
2 2
This proves that the map  : X  X is an open map. Similarly, we can show that  is an open
2 2 1
map. Consequently, projection maps are open maps.
This completes the proof of the theorem.
Theorem 7: Let (X, T) be the product topological space of (X , T ) and (X , T ).
1 1 2 2
Let  : X  X ,  : X  X
1 1 2 2
be the projection maps on the first and second co-ordinate spaces respectively.
Let f : Y  X be another map, where Y is another topological space. Show that f is continuous iff
 o f and  o f are continuous maps.
1 2
Proof: Let X = X × X
1 2
Let (X, T) be the product topological space of (X , T ) and (X , T ).
1 1 2 2
Let (Y, U) be another topological space.
Let  be the base for the topology T on X.

Let  : X  X ,
1 1
 : X  X be projection maps.
2 2
Let f : Y  X be another map.

Then  o f : Y  X
1 1
 o f : Y  X are also maps.
2 2
Let f be continuous.

To prove that  o f and  o f are continuous maps.
1 2
By theorem 6, projection maps are continuous, i.e.  and  are continuous maps.
1 2
Also f is given to be continuous.

This means that  o f,  o f are continuous maps. Conversely, suppose that  o f,  o f are
1 2 1 2
continuous maps.
To show that f is continuous.

Let G  X be an arbitrary open set.
–1
If we prove that f (G) is open in Y, the result will follow.
Let y  f (G) be an arbitrary, then f(y)  G.
–1
 f(y) is an element of X = X × X and hence it can be taken as f(y) = (x , x )  G
1 2 1 2
By definition of base,
(x , x )  G  T   U × U   s.t. (x , x )  U × U  G
1 2 1 2 1 2 1 2
  (x , x )   (U × U )   (G) and
1 1 2 1 1 2 1

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