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Transcript
Introduction
Product Topology
Product Topology Vs Subspace
Product Topology in term of Subbasis
Lecture-11-12: The Product Topology
Dr. Sanjay Mishra
Department of Mathematics
Lovely Professional University
Punjab, India
October 18, 2014
Sanjay Mishra
The Product Topology
Introduction
Product Topology
Product Topology Vs Subspace
Product Topology in term of Subbasis
Outline
1
Introduction
2
Product Topology
3
Product Topology Vs Subspace
4
Product Topology in term of Subbasis
Projection Map
Sanjay Mishra
The Product Topology
Introduction
Product Topology
Product Topology Vs Subspace
Product Topology in term of Subbasis
Introduction I
Given two topological spaces X and Y , we would like to generate a
natural topology on the product, X × Y . Our first inclination might be
to take as the topology on X × Y the collection C of sets of the form
U × V where U is open in X and V is open in Y . But C is not a topology
since the union of two sets U1 × V1 and U2 × V2 need not be in the form
U × V for some U ⊂ X and V ⊂ Y .
Sanjay Mishra
The Product Topology
Introduction
Product Topology
Product Topology Vs Subspace
Product Topology in term of Subbasis
Introduction II
There are two way to define the product topology on X × Y .
1
First, product topology on X × Y with the help of open set of
given topologies on X and Y respectively
2
Other hand product topology on X × Y with the help of the bases
for the topologies on X and Y respectively.
Sanjay Mishra
The Product Topology
Introduction
Product Topology
Product Topology Vs Subspace
Product Topology in term of Subbasis
Product Topology in terms of Open Sets I
Definition
Let (X, τ1 ) and (Y, τ2 ) be topological spaces. The product topology (let
τ ) on X × Y is topology having as basis the collection B defined as
follows :
B = {U × V : U ∈ τ1 , V ∈ τ2 }
Of course, we must verify that B actually is a basis for a topology on
the product, X × Y .
Sanjay Mishra
The Product Topology
Introduction
Product Topology
Product Topology Vs Subspace
Product Topology in term of Subbasis
Product Topology in terms of Open Sets II
Theorem
The collection B is a basis for a topology on X × Y .
Proof:
1
2
Every point (x, y) ∈ X × Y , and X × Y ∈ B. Therefore the first
condition for a basis is satisfied.
Now suppose that (x, y) is in the intersection of two elements of B.
That is, (x, y) ∈ U1 × V1 ∩ U2 × V2 , where U1 and U2 are open sets
in X, and V1 and V2 are open sets in Y . Let U3 = U1 ∩ U2 and
V3 = V1 ∩ V2 . Then U3 is open in X, and V3 is open in Y , and
therefore U3 × V3 ∈ B. Also,
U3 × V3 = (U1 ∩ U2 ) × (V1 ∩ V2 ) = (U1 × V1 ) ∩ ((U2 × V2 )
and thus (x, y) ∈ U3 × V3 ⊂ (U1 × V1 ) ∩ (U2 × V2 ).
It follows that the second condition for a basis is satisfied.
Sanjay Mishra
The Product Topology
Introduction
Product Topology
Product Topology Vs Subspace
Product Topology in term of Subbasis
Examples I
Let X = {a, b, c} and Y = {1, 2} with topologies
{φ, X, {b}, {c}, {a, b}, {b, c}} and {φ, Y, {1}}
respectively. A basis for the product topology on X × Y is depicted in
Figure. Each nonempty open set in the product topology on X × Y is
a union of the basis elements shown.
Sanjay Mishra
The Product Topology
Introduction
Product Topology
Product Topology Vs Subspace
Product Topology in term of Subbasis
Product Topology in terms of Basis I
The basis B that as we used to define the product topology is relatively
large since we obtain it by pairing up every open set U in X with every
open set V in Y .
Now we can find a smaller basis for the product topology by using bases
for the topologies on X and Y , rather than using the whole topologies
themselves.
Sanjay Mishra
The Product Topology
Introduction
Product Topology
Product Topology Vs Subspace
Product Topology in term of Subbasis
Product Topology in terms of Basis II
Theorem
If C is a basis for X and D is a basis for Y , then
E = {C × D : C ∈ C and D ∈ D}
is a basis that generates the product topology on X × Y .
Proof:
Each set C × D ∈ E is an open set in the product topology, therefore, by
lemma 1 , it suffices to show that for every open set W in X × Y and every
point (x, y) ∈ W , there is a set C × D ∈ E such that (x, y) ∈ C × D ⊂ W .
But since W is open in X × Y , we know that there are open sets U in X
and V in Y such that (x, y) ∈ U × V ⊂ W . So x ∈ U and y ∈ V . Since
U is open in X, there is a basis element C ∈ C such that x ∈ C ⊂ U .
Sanjay Mishra
The Product Topology
Introduction
Product Topology
Product Topology Vs Subspace
Product Topology in term of Subbasis
Product Topology in terms of Basis III
Similarly, since V is open in Y , there is a basis element D ∈ D such that
y ∈ D ⊂ V . Thus (x, y) ∈ C × D ⊂ U × V ⊂ W . Hence, by lemma, it
follows that
E = {C × D : C ∈ C and D ∈ D}
is a basis for the product topology on X × Y .
1
Let X be a topological space and C is a collection of open sets of X such that
for each open set U of X and each x ∈ U , there is an element C of C such that
x ∈ C ⊂ U , then C is a basis for the topology of X.
Sanjay Mishra
The Product Topology
Introduction
Product Topology
Product Topology Vs Subspace
Product Topology in term of Subbasis
Product Topology Vs Subspace I
Theorem
Let X and Y be topological spaces, and assume that A ⊂ X and
B ⊂ Y . Then the topology on A × B as a subspace of the product
X × Y is the same as the product topology on A × B, where A has the
subspace topology inherited from X, and B has the subspace topology
inherited from Y .
Sanjay Mishra
The Product Topology
Introduction
Product Topology
Product Topology Vs Subspace
Product Topology in term of Subbasis
Product Topology Vs Subspace II
Example
Let I = [0, 1] have the standard topology as a subspace of R. The
product space I × I is called the unit square. The product topology on
I × I is the same as the standard topology on I × I as a subspace of R2 .
Sanjay Mishra
The Product Topology
Introduction
Product Topology
Product Topology Vs Subspace
Product Topology in term of Subbasis
Product Topology Vs Subspace III
Let S 1 be the circle, and let I = [0, 1] have the standard topology.
Then S 1 × I appears as in Figure. We can think of it as a circle with
intervals perpendicular at each point of the circle. Seen this way, it is a
circle’s worth of intervals. Or it can be thought of as an interval with
perpendicular circles at each point. Thus it is an interval’s worth of
circles. The resulting topological space is called the annulus.
The product space S 1 × (0, 1) is the annulus with the innermost and
outermost circles removed. We refer to it as the open annulus.
Sanjay Mishra
The Product Topology
Introduction
Product Topology
Product Topology Vs Subspace
Product Topology in term of Subbasis
Product Topology Vs Subspace IV
Consider the product space
S 1 × S 1 , where S 1 is the circle. For
each point in the first S 1 , there is a
circle corresponding to the second
S 1 . Since each S 1 has a topology
generated by open intervals in the
circle, it follows by Theorem that
S 1 × S 1 has a basis consisting of
rectangular open patches, as shown
in the figure.
Sanjay Mishra
The Product Topology
Introduction
Product Topology
Product Topology Vs Subspace
Product Topology in term of Subbasis
Product Topology Vs Subspace V
Let D be the disk as a subspace of
the plane. The product space
S 1 × D is called the solid torus. If
we think of the torus as the surface
of a doughnut, then the solid torus
is the whole doughnut itself.
Sanjay Mishra
The Product Topology
Introduction
Product Topology
Product Topology Vs Subspace
Product Topology in term of Subbasis
Product Topology Vs Subspace VI
Now thing about these questions.
How can define the product topology in term of subbasis?
How can define product topology for more than two spaces ?
How can define product topology for infinite number of spaces ?
Sanjay Mishra
The Product Topology
Introduction
Product Topology
Product Topology Vs Subspace
Product Topology in term of Subbasis
Projection Map
Projection Map I
Definition (Projection Map)
Let π1 X × Y → X be defined by the equation
π1 (x, y) = X
and let π2 X × Y → Y be defined by the equation
π1 (x, y) = Y
The maps π1 and π2 are called the projection of X × Y onto its first
and second factors, respectively.
Sanjay Mishra
The Product Topology
Introduction
Product Topology
Product Topology Vs Subspace
Product Topology in term of Subbasis
Projection Map
Projection Map II
Remark
Here, π1 and π2 must be onto, otherwise spaces X or Y happens to be
empty, in which case X × Y is empty and our whole discussion is
empty.
Sanjay Mishra
The Product Topology
Introduction
Product Topology
Product Topology Vs Subspace
Product Topology in term of Subbasis
Projection Map
Projection Map III
Remark
If U is open in X, then
π1−1 (U ) = U × Y , which is open in
X ×Y.
Similarly, if V is open in Y , then
π2−1 (U ) = X × V which is open in
X ×Y.
And the intersection of these two
sets is U × V .
Sanjay Mishra
The Product Topology
Introduction
Product Topology
Product Topology Vs Subspace
Product Topology in term of Subbasis
Projection Map
Projection Map IV
Theorem
The collection
S = {π1−1 (U ) : U is open in X} ∪ {π2−1 (V ) : U is open in Y }
is subbasis for the product topology on X × Y .
Proof:
Let τ is product topology on X × Y and τ 0 is topology generated by S.
Now our aim is show that τ = τ 0 .
Since every element of S belongs to τ , so do arbitrary unions of finite
intersections of elements of S. Thus τ 0 ⊂ τ .
Sanjay Mishra
The Product Topology
Introduction
Product Topology
Product Topology Vs Subspace
Product Topology in term of Subbasis
Projection Map
Projection Map V
Other hand, every basis element U × V for τ is a finite intersection of
elements of S, since
U × V = π1−1 (U ) ∩ π2−1 (V )
Therefore, U × V ∈ τ 0 . Hence τ ⊂ τ 0 .
Sanjay Mishra
The Product Topology