Page 57 - DMTH503_TOPOLOGY

Unit 4: The Product Topology on X × Y

Notes
Example 4: Let B be a member of the defining base for the product space X = x X , show
i
i
that the projection of B into any coordinate space is open.
or
Each projection is a continuous map.
Solution: Let B be a member of the defining base for the product space X = x X so that B is
i i
expressible as

B = x{X : i  j , j , ..., j } × G  ... G

i 1 2 m i j m j
where G is an open subset of X .
k i k i
The projection map  is defined as

 : X  X
 
ì X if   j , j ,...j
 (B) = í  1 2 m

î G  if  {j , j ,...j }
1
m
2
In either case,  (B) is an open set.

Theorem 9: Let y be a fixed element of Y and let A = X × {y }. Then the restriction f or  to A is
o o x x
a homeomorphism of the subspace A of X × Y onto X. Also the restriction f of  to B = {x } × Y
y y o
into Y is a homeomorphism, where x  X.
o
Proof: Let (X × Y, T) be the product topological space of (X, T ) and (X, T ). Let x  X and y  Y be
1 2
arbitrary. Then the projection maps are defined as
 : X × Y  X s.t.  (x, y) = x
x x
and  : X × Y  X s.t.  (x, y) = y.
y y
Let x  X and y  Y be fixed elements.
o o
Let f be the restriction of  to A so that f is a map s.t. f : A  X
x x x x
s.t. f (x, y ) = x.
x o
To prove that f is a homeomorphism, we have to prove that
x
(i) f is one-one onto
x
(ii) f is continuous
x
–1
(iii) f is continuous
x
f (x , y ) = f (x , y )  x = x , by definition of f
x 1 o x 2 o 1 2 x
 (x , y ) = (x , y ).
1 o 2 o
Hence f is one-one.
x
Given any x  X,  (x, y )  A s.t. f (x, y ) = x.
o x o
This proves that f is onto. Hence the result (i).
x
 is a projection map   is continuous.
x x
Also f is its restriction  f is continuous. Hence (ii).
x x

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