Page 58 - DMTH503_TOPOLOGY

Topology

Notes (iv) To prove f : X  A is continuous. We have to prove: given any V open subset of A.
–1
x
[f ] (V) = f (V) is open in X.
–1 –1
x x
Now V is expressible as V = A B, where B  T.
Let B be a base for T. Then
B = {G × H : G  T , H  T }
1 2
By definition of base,
B  T   B  B s.t.
1
B = {G × H  T × T : G × H  B }
l 2 1
Then A B = {A (G × H) : G × H  B }
1
= {(X × {y }) (G × H) : G × H  B }
o 1



ì {G {y }) : G H B } if y  H
o
1
o
= í


î or {G   : G H B } if y Ï H
o
1
ì {G y } : G H B } if y  H



o
o
1
= í
î or  : if y Ï H
o
Moreover  is an open set and an arbitrary union of open sets is open.
In either case, f (A B) is open in X, i.e., f (V) is open in X.
x x
Self Assessment
2. Prove that the collection
  1 (U)|U open in X     1 (V)|V open in  Y
S =  1 2
is a sub basis for the product topology on X × Y.
3. A map f : X  Y is said to be an open map if for every open set U of X, the set f(U) is open
in Y, show that  : X × Y  X and  : X × Y  Y are open maps.
1 2
4.3 Summary
 If X and Y are topological spaces, the product topology on X × Y is the topology whose
basis is {A × B : A  T , B  T }.
X Y
 Given any product of sets X × Y, there are projections maps  and  from X × Y to X and
x y
to Y given by (x, y)  x and (x, y)  y.
 If (X, Y) is the product space if topological spaces (X , T ) and (X , T ), then the projection
1 1 2 2
maps  and  are continuous and open.
1 2
4.4 Keywords

Basis: A collection  of open sets in a topological space X is called a basis for the topology if
every open set in X is a union of sets in .

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