Download Theorem: let (X,T) and (Y,V) be two topological spaces... E={G×H:GT,HV} is a base for some topology X×Y.

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Theorem: let (X,T) and (Y,V) be two topological spaces then the collection
E={G×H:GT,HV} is a base for some topology X×Y.
Definition: let (X,T) and (Y,V) be two topological spaces then the topology
W whose base is E is called the product topology for X×Y and (X×Y , W)
is called the product of X and Y.
Theorem: let (X,T) and (Y,V) be two topological spaces and β be a base for
T and α is a base for Y then H={b×c : bβ , cα } is a base for X×Y.
Example: Take the topology T= { , {a, b}, {a} } on X = {a, b}.
Then the product topology on X ×X is { , X X, {(a, a)}, {(a, a), (a, b)},
{(a, a), (b, a)}, {(a, a), (a, b), (b, a)} }
Definition : the mapping πx:X×Y→X defined by πx(x,y)=x , for each
(x,y)X×Y , and πy:X×Y→Y defined by πy(x,y)=y , for every (x,y)X×Y
Theorem: let (X,T) and (Y,V) be two topological spaces and let (X×Y,W)
be the product space of X and Y then πx , πy are continuous and open
mapping.
Proof: If U ∈ T open set in X , then π-1(U) = U × Y is open in
X × Y , so πx is continuous. Similarly, πy is continuous.
now we prove that
is open mapping. Let U be open in the product.
Then it is a union of base elements Vi × Wi, where Vi,Wi are nonempty
open sets of X, Y respectively. So πx(U) = πx(∪(Vi ×Wi)) =∪ [πx(Vi ×Wi)]
= ∪Vi which is open in X
Theorem: Show that the product of two Hausdorff spaces is Hausdorff.
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Proof: Let X, Y be Hausdorff, and (a, b), (c, d) be disjoint points of X × Y.
If a = c then b ≠ d so there exist disjoint open U, V in Y containing them,
respectively. Then
X × U,X × V are disjoint open sets of the product containing the original
two points.
If a ≠ c then there are disjoint open U, V in X containing them, and U ×Y,
V ×Y are disjoint open sets of the product containing the original two
points
theorem : the product space of two T0space(T1 space) is T0 space(T1 space).
Theorem : the product space of two second countable spaces is second
countable space.