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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