Lecture: The Product Topology - · PDF fileIntroduction Product Topology Product Topology Vs...

IntroductionProduct Topology

Product Topology Vs SubspaceProduct Topology in term of Subbasis

Lecture: The Product Topology

Dr. Sanjay Mishra

Department of MathematicsLovely Professional University

Punjab, India

September 16, 2015

Sanjay Mishra The Product Topology



Outline

1 Introduction

2 Product Topology

3 Product Topology Vs Subspace

4 Product Topology in term of SubbasisProjection Map




Introduction I

Given two topological spaces X and Y , we would like to generate anatural topology on the product, X × Y . Our first inclination might beto take as the topology on X × Y the collection C of sets of the formU×V where U is open in X and V is open in Y . But C is not a topologysince the union of two sets U1× V1 and U2× V2 need not be in the formU × V for some U ⊂ X and V ⊂ Y .




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 ofgiven topologies on X and Y respectively

2 Other hand product topology on X × Y with the help of the basesfor the topologies on X and Y respectively.




Product Topology in terms of Open Sets I

Definition

Let (X, τ1) and (Y, τ2) be topological spaces. The product topology (letτ) on X × Y is topology having as basis the collection B defined asfollows :

B = {U × V : U ∈ τ1, V ∈ τ2}

Of course, we must verify that B actually is a basis for a topology onthe product, X × Y .




Product Topology in terms of Open Sets II

Theorem

The collection B is a basis for a topology on X × Y .

Proof:

1 Every point (x, y) ∈ X × Y , and X × Y ∈ B. Therefore the firstcondition for a basis is satisfied.

2 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 setsin X, and V1 and V2 are open sets in Y . Let U3 = U1 ∩ U2 andV3 = V1 ∩ V2. Then U3 is open in X, and V3 is open in Y , andtherefore 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.




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 inFigure. Each nonempty open set in the product topology on X × Y isa union of the basis elements shown.




Product Topology in terms of Basis I

The basis B that as we used to define the product topology is relativelylarge since we obtain it by pairing up every open set U in X with everyopen set V in Y .

Now we can find a smaller basis for the product topology by using basesfor the topologies on X and Y , rather than using the whole topologiesthemselves.




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 andD ∈ 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, bylemma 1, it suffices to show that for every open setW inX × Y and everypoint (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 Xand V in Y such that (x, y) ∈ U × V ⊂ W . So x ∈ U and y ∈ V . SinceU is open in X, there is a basis element C ∈ C such that x ∈ C ⊂ U .




Product Topology in terms of Basis III

Similarly, since V is open in Y , there is a basis element D ∈ D such thaty ∈ D ⊂ V . Thus (x, y) ∈ C × D ⊂ U × V ⊂ W . Hence, by lemma, itfollows that

E = {C ×D : C ∈ C andD ∈ D}

is a basis for the product topology on X × Y .

1Let X be a topological space and C is a collection of open sets of X such thatfor each open set U of X and each x ∈ U , there is an element C of C such thatx ∈ C ⊂ U , then C is a basis for the topology of X.




Product Topology Vs Subspace I

Theorem

Let X and Y be topological spaces, and assume that A ⊂ X andB ⊂ Y . Then the topology on A×B as a subspace of the productX × Y is the same as the product topology on A×B, where A has thesubspace topology inherited from X, and B has the subspace topologyinherited from Y .




Product Topology Vs Subspace II

Example

Let I = [0, 1] have the standard topology as a subspace of R. Theproduct space I × I is called the unit square. The product topology onI × I is the same as the standard topology on I × I as a subspace of R2.




Product Topology Vs Subspace III

Let S1 be the circle, and let I = [0, 1] have the standard topology.Then S1 × I appears as in Figure. We can think of it as a circle withintervals perpendicular at each point of the circle. Seen this way, it is acircle’s worth of intervals. Or it can be thought of as an interval withperpendicular circles at each point. Thus it is an interval’s worth ofcircles. The resulting topological space is called the annulus.The product space S1 × (0, 1) is the annulus with the innermost andoutermost circles removed. We refer to it as the open annulus.




Product Topology Vs Subspace IVConsider the product spaceS1 × S1, where S1 is the circle. Foreach point in the first S1, there is acircle corresponding to the secondS1. Since each S1 has a topologygenerated by open intervals in thecircle, it follows by Theorem thatS1 × S1 has a basis consisting ofrectangular open patches, as shownin the figure.




Product Topology Vs Subspace V

Let D be the disk as a subspace ofthe plane. The product spaceS1 ×D is called the solid torus. Ifwe think of the torus as the surfaceof a doughnut, then the solid torusis the whole doughnut itself.




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 ?




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 firstand second factors, respectively.




Projection Map

Projection Map II

The functions

πi : X1 × · · · ×Xn → Xi such thatπi(x1, · · · , xn) = xi

shall be referred to as the projection maps or simply as the projections.




Projection Map

Projection Map III

Remark

Here, π1 and π2 must be onto, otherwise spaces X or Y happens to beempty, in which case X × Y is empty and our whole discussion isempty.




Projection Map

Projection Map IV

Remark

If U is open in X, thenπ−11 (U) = U × Y , which is open inX × Y .Similarly, if V is open in Y , thenπ−12 (U) = X × V which is open inX × Y .And the intersection of these twosets is U × V .




Projection Map

Projection Map V

Theorem

The collection

S = {π−11 (U) : U is open inX} ∪ {π−12 (V ) : U is open inY }

is subbasis for the product topology on X × Y .

Proof:Let τ is product topology on X × Y and τ ′ is topology generated by S.Now our aim is show that τ = τ ′.Since every element of S belongs to τ , so do arbitrary unions of finiteintersections of elements of S. Thus τ ′ ⊂ τ .




Projection Map

Projection Map VI

Other hand, every basis element U × V for τ is a finite intersection ofelements of S, since

U × V = π−11 (U) ∩ π−12 (V )

Therefore, U × V ∈ τ ′. Hence τ ⊂ τ ′.


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