Page 87 - DMTH503_TOPOLOGY

Unit 8: The Product Topology

n
n
Since  is a finite product, the box and product topologies agree. Whenever we consider  , we Notes
will assume that it is given this topology, unless we specifically state otherwise.
Example 5: Consider  , the countably infinite product of  with itself. Recall that
w
w
 =  X ,
n
n

w
where X =  for each n. Let us define a function f :   by the equation
n
f (t) = (t, t, t, …);
the n coordinate function of f is the function f (t) = t. Each of the coordinate functions f :   
th
n
n
is continuous; therefore, the function f is continuous if  is given the product topology. But f is
w
w
not continuous if  is given the box topology. Consider, for example, the basic element.
1
1
B = (–1, 1) × (  1 2 , ) × (  1 3 , ) × …
3
2
–1
–1
for the box topology. We assert that f (B) is not open in . If f (B) were open in , it would
contain some interval (–, ) about the point 0. This would mean that f((–, ))  B so that,
applying  to both sides of the inclusion.
n
1
f ((–, )) = (–, )  (  1 , )
n n n
for all n, a contradiction.
Theorem 1: Let {X } be an indexed family of spaces; Let A  X for each . If X is given either
   
the product or the box topology, then
Õ A = Õ A .
 
Proof: Let x = (x ) be a point of Õ A ; we show that x  Õ A .
  
Let =  be a basis element for either the box or product topology that contains x. Since x 
 
A , we can choose a point y   A for each . Then y = (y ) belongs to both and A . Since
     
is arbitrary, it follows that x belongs to the closure of A .

Conversely, suppose x = (x ) lies in the closure of A , in either topology. We show that for any
 
given index , we have x  A . Let V be an arbitrary open set of X containing x . Since Õ - 1 (V )
      
is open in X in either topology, it contains a point y = (y ) of A . Then y belongs to V A .
     
It follows that x  A .
 
Theorem 2: Let f : A   X be given by the equation
  J 
f(a) = (f (a)) ,
   J
where f : AX for each. Let X have the product topology. Then the function f is continuous
  
if and only if each function f is continuous.

Proof: Let  be the projection of the product onto its th factor. The function  is continuous, for
 
if is open in X , the set  - 1 ( ) is a sub basis element for the product topology on X . Now
    
suppose that f : A  X is continuous. The function f equals the composite  of; being the
  
composite of two continuous functions, it is continuous.
Conversely suppose that each co-ordinate function f is continuous. To prove that f is continuous,

it suffices to prove that the inverse image under f of each sub-basis element is open in A; we
remarked on this fact when we defined continuous functions. A typical sub-basis element for the

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